Mixing
For every odd $ninmathbb{N}$, is it true that $sigma(n) < 2n$?
$begingroup$
Is the following proposition true?
Let $n in mathbb{N}$ be an odd number, then $sigma(n) < 2n$ .
For $n=p_1^alpha p_2^beta$ it is true :
$$sigma(n)=left(frac{p_1^{alpha+1}-1}{P_1-1}right)left(frac{p_2^{beta+1}-1}{P_2-1}right) < 2p_1^{alpha} p_2^{beta}
Longleftrightarrow $$
$$2 < frac{p_1^{alpha+1} p_2^{beta+1}+p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}=frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$
$$Longrightarrow 2< frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$ This is true beacuse of $frac{{p_1}{p_2}}{p_1+p_2-1} geq 2$ and $frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)} > 0$
Can somebody help me for $n=p_1^{alpha_1} p_2^{alpha_2}...p_m^{alpha_m}$ ?
Thank you.
number-theory divisor-sum perfect-numbers
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
$endgroup$
add a comment |
$begingroup$
Is the following proposition true?
Let $n in mathbb{N}$ be an odd number, then $sigma(n) < 2n$ .
For $n=p_1^alpha p_2^beta$ it is true :
$$sigma(n)=left(frac{p_1^{alpha+1}-1}{P_1-1}right)left(frac{p_2^{beta+1}-1}{P_2-1}right) < 2p_1^{alpha} p_2^{beta}
Longleftrightarrow $$
$$2 < frac{p_1^{alpha+1} p_2^{beta+1}+p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}=frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$
$$Longrightarrow 2< frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$ This is true beacuse of $frac{{p_1}{p_2}}{p_1+p_2-1} geq 2$ and $frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)} > 0$
Can somebody help me for $n=p_1^{alpha_1} p_2^{alpha_2}...p_m^{alpha_m}$ ?
Thank you.
number-theory divisor-sum perfect-numbers
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
$endgroup$
add a comment |
$begingroup$
Is the following proposition true?
Let $n in mathbb{N}$ be an odd number, then $sigma(n) < 2n$ .
For $n=p_1^alpha p_2^beta$ it is true :
$$sigma(n)=left(frac{p_1^{alpha+1}-1}{P_1-1}right)left(frac{p_2^{beta+1}-1}{P_2-1}right) < 2p_1^{alpha} p_2^{beta}
Longleftrightarrow $$
$$2 < frac{p_1^{alpha+1} p_2^{beta+1}+p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}=frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$
$$Longrightarrow 2< frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$ This is true beacuse of $frac{{p_1}{p_2}}{p_1+p_2-1} geq 2$ and $frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)} > 0$
Can somebody help me for $n=p_1^{alpha_1} p_2^{alpha_2}...p_m^{alpha_m}$ ?
Thank you.
number-theory divisor-sum perfect-numbers
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
$endgroup$
Is the following proposition true?
Let $n in mathbb{N}$ be an odd number, then $sigma(n) < 2n$ .
For $n=p_1^alpha p_2^beta$ it is true :
$$sigma(n)=left(frac{p_1^{alpha+1}-1}{P_1-1}right)left(frac{p_2^{beta+1}-1}{P_2-1}right) < 2p_1^{alpha} p_2^{beta}
Longleftrightarrow $$
$$2 < frac{p_1^{alpha+1} p_2^{beta+1}+p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}=frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$
$$Longrightarrow 2< frac{{p_1}{p_2}}{p_1+p_2-1}+frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)}$$ This is true beacuse of $frac{{p_1}{p_2}}{p_1+p_2-1} geq 2$ and $frac{p_1^{alpha+1}+p_2^{beta+1}-1}{p_1^{alpha} p_2^{beta}(p_1+p_2-1)} > 0$
Can somebody help me for $n=p_1^{alpha_1} p_2^{alpha_2}...p_m^{alpha_m}$ ?
Thank you.
number-theory divisor-sum perfect-numbers
number-theory divisor-sum perfect-numbers
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
edited May 30 '15 at 21:03
Zev Chonoles
111k16233430
111k16233430
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
asked May 30 '15 at 20:48
MojtabaMojtaba
421215
421215
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
No, that's false. The numbers for which $sigma(n) = 2n$ are called perfect, while $sigma(n) > 2n$ are called abundant.
It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is $945$.
Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every integer multiple (so in particular every odd multiple) of an abundant number is again abundant.
edited May 30 '15 at 21:42
user84413
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
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add a comment |
$begingroup$
For a counterexample, look at $n=945=3^3cdot 5cdot 7$. It turns out that $sigma(n)=1920gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying $945$ by any odd number. There are other families.
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
$endgroup$
add a comment |
$begingroup$
An efficient way to produce infinitely many numbers with increasing values of $frac{sigma(n)}{n}$ while staying with odd numbers, let $n$ be the least common multiple of the consecutive odd numbers $1,3,5,7,9,11,ldots.$ Not the product, the LCM. This method should produce ratios that slowly grow. Let me work up a table, give me a few minutes.
odd lcm sigma ratio
1 1 1 1
3 3 4 1.333333333333333
5 15 24 1.6
7 105 192 1.828571428571429
9 315 624 1.980952380952381
11 3465 7488 2.161038961038961
13 45045 104832 2.327272727272727
15 45045 104832 2.327272727272727
17 765765 1886976 2.464171122994653
19 14549535 37739520 2.593864339994371
21 14549535 37739520 2.593864339994371
23 334639305 905748480 2.706641050428909
It is not difficult to prove that the ratio stays the same if the new odd number is divisible by two or more different primes, but the ratio increases if the new odd number is a prime or prime power. In the long run, the exponent for some prime factor $p$ in the LCM is roughly proportional to $frac{1}{log p}.$
Another sequence, less efficient but easier to describe and to program, just takes the ordinary product of the consecutive prime numbers, beginning with $3.$ It is not even necessary to find the product or its sum of divisors, every time we use a new prime we just multiply the existing ratio by $(p+1)/p.$ This increases without bound, because the sum of prime reciprocals diverges. Note how the first three ratios in this list are the same as the first three in the earlier list.
May 30, 2015
3 1.333333333333333
5 1.6
7 1.828571428571429
11 1.994805194805195
13 2.148251748251748
17 2.27461949814891
19 2.394336313840958
23 2.498437892703608
29 2.584590923486491
31 2.66796482424412
37 2.740071981656123
41 2.806903005598956
43 2.872179819682652
47 2.93329002861207
53 2.988635123491544
59 3.039289956093095
61 3.089114381602818
67 3.13522056640286
71 3.179378602549379
73 3.222931734091151
79 3.263728338320153
83 3.303050366492685
89 3.340163291958895
97 3.374597965071873
101 3.408009826112189
103 3.441097300152113
107 3.473257088004002
109 3.505121831930644
113 3.536140609204367
127 3.563984236048496
131 3.591190222583217
137 3.617403289901343
139 3.643427774001352
149 3.667880309397335
151 3.692170907472814
157 3.715687919622322
163 3.738483551031048
167 3.760869680079138
173 3.782608811177862
179 3.803740703977738
181 3.824755845988665
191 3.844780745705883
193 3.864701889466017
197 3.884319665554677
199 3.903838859853947
211 3.922340465824818
223 3.939929436523585
227 3.957285953865098
229 3.974566678554465
233 3.991624904642682
239 4.008326264076334
241 4.024958323263372
251 4.040994013794302
257 4.056717725910233
263 4.072142508138029
269 4.08728058437646
271 4.102362800554971
277 4.117172774564195
281 4.131824634971897
283 4.146424722021268
293 4.160576342232945
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
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add a comment |
$begingroup$
Let $O_n$ be the product of the first $n$ odd integers.
Notice that $sigma(O_n)geq frac{O_n}{1} + frac{O_n}{3} + dots+ frac{O_n}{n} = O_n(1+1/3 + dots 1/n)$.
Recall $sumlimits_{i=1}^infty frac{1}{2i-1}$ diverges.
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
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add a comment |
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4 Answers
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4 Answers
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$begingroup$
No, that's false. The numbers for which $sigma(n) = 2n$ are called perfect, while $sigma(n) > 2n$ are called abundant.
It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is $945$.
Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every integer multiple (so in particular every odd multiple) of an abundant number is again abundant.
edited May 30 '15 at 21:42
user84413
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
$endgroup$
add a comment |
$begingroup$
No, that's false. The numbers for which $sigma(n) = 2n$ are called perfect, while $sigma(n) > 2n$ are called abundant.
It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is $945$.
Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every integer multiple (so in particular every odd multiple) of an abundant number is again abundant.
edited May 30 '15 at 21:42
user84413
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
$endgroup$
add a comment |
$begingroup$
No, that's false. The numbers for which $sigma(n) = 2n$ are called perfect, while $sigma(n) > 2n$ are called abundant.
It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is $945$.
Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every integer multiple (so in particular every odd multiple) of an abundant number is again abundant.
edited May 30 '15 at 21:42
user84413
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
$endgroup$
No, that's false. The numbers for which $sigma(n) = 2n$ are called perfect, while $sigma(n) > 2n$ are called abundant.
It is currently unknown whether there are odd perfect numbers or not, while the smallest odd abundant number is $945$.
Even more: there are infinitely many odd abundant numbers (cfr. OEIS sequence A005231). This is because every integer multiple (so in particular every odd multiple) of an abundant number is again abundant.
edited May 30 '15 at 21:42
user84413
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
edited May 30 '15 at 21:42
user84413
23k11848
edited May 30 '15 at 21:42
user84413
23k11848
edited May 30 '15 at 21:42
user84413
23k11848
23k11848
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
answered May 30 '15 at 20:56
A.P.A.P.
8,19021840
8,19021840
add a comment |
add a comment |
$begingroup$
For a counterexample, look at $n=945=3^3cdot 5cdot 7$. It turns out that $sigma(n)=1920gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying $945$ by any odd number. There are other families.
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
$endgroup$
add a comment |
$begingroup$
For a counterexample, look at $n=945=3^3cdot 5cdot 7$. It turns out that $sigma(n)=1920gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying $945$ by any odd number. There are other families.
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
$endgroup$
add a comment |
$begingroup$
For a counterexample, look at $n=945=3^3cdot 5cdot 7$. It turns out that $sigma(n)=1920gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying $945$ by any odd number. There are other families.
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
$endgroup$
For a counterexample, look at $n=945=3^3cdot 5cdot 7$. It turns out that $sigma(n)=1920gt 1890$. There are infinitely many other counterexamples. One family is obtained by multiplying $945$ by any odd number. There are other families.
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
edited May 30 '15 at 21:02
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
answered May 30 '15 at 20:53
André NicolasAndré Nicolas
455k36432820
455k36432820
add a comment |
add a comment |
$begingroup$
An efficient way to produce infinitely many numbers with increasing values of $frac{sigma(n)}{n}$ while staying with odd numbers, let $n$ be the least common multiple of the consecutive odd numbers $1,3,5,7,9,11,ldots.$ Not the product, the LCM. This method should produce ratios that slowly grow. Let me work up a table, give me a few minutes.
odd lcm sigma ratio
1 1 1 1
3 3 4 1.333333333333333
5 15 24 1.6
7 105 192 1.828571428571429
9 315 624 1.980952380952381
11 3465 7488 2.161038961038961
13 45045 104832 2.327272727272727
15 45045 104832 2.327272727272727
17 765765 1886976 2.464171122994653
19 14549535 37739520 2.593864339994371
21 14549535 37739520 2.593864339994371
23 334639305 905748480 2.706641050428909
It is not difficult to prove that the ratio stays the same if the new odd number is divisible by two or more different primes, but the ratio increases if the new odd number is a prime or prime power. In the long run, the exponent for some prime factor $p$ in the LCM is roughly proportional to $frac{1}{log p}.$
Another sequence, less efficient but easier to describe and to program, just takes the ordinary product of the consecutive prime numbers, beginning with $3.$ It is not even necessary to find the product or its sum of divisors, every time we use a new prime we just multiply the existing ratio by $(p+1)/p.$ This increases without bound, because the sum of prime reciprocals diverges. Note how the first three ratios in this list are the same as the first three in the earlier list.
May 30, 2015
3 1.333333333333333
5 1.6
7 1.828571428571429
11 1.994805194805195
13 2.148251748251748
17 2.27461949814891
19 2.394336313840958
23 2.498437892703608
29 2.584590923486491
31 2.66796482424412
37 2.740071981656123
41 2.806903005598956
43 2.872179819682652
47 2.93329002861207
53 2.988635123491544
59 3.039289956093095
61 3.089114381602818
67 3.13522056640286
71 3.179378602549379
73 3.222931734091151
79 3.263728338320153
83 3.303050366492685
89 3.340163291958895
97 3.374597965071873
101 3.408009826112189
103 3.441097300152113
107 3.473257088004002
109 3.505121831930644
113 3.536140609204367
127 3.563984236048496
131 3.591190222583217
137 3.617403289901343
139 3.643427774001352
149 3.667880309397335
151 3.692170907472814
157 3.715687919622322
163 3.738483551031048
167 3.760869680079138
173 3.782608811177862
179 3.803740703977738
181 3.824755845988665
191 3.844780745705883
193 3.864701889466017
197 3.884319665554677
199 3.903838859853947
211 3.922340465824818
223 3.939929436523585
227 3.957285953865098
229 3.974566678554465
233 3.991624904642682
239 4.008326264076334
241 4.024958323263372
251 4.040994013794302
257 4.056717725910233
263 4.072142508138029
269 4.08728058437646
271 4.102362800554971
277 4.117172774564195
281 4.131824634971897
283 4.146424722021268
293 4.160576342232945
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
$endgroup$
add a comment |
$begingroup$
An efficient way to produce infinitely many numbers with increasing values of $frac{sigma(n)}{n}$ while staying with odd numbers, let $n$ be the least common multiple of the consecutive odd numbers $1,3,5,7,9,11,ldots.$ Not the product, the LCM. This method should produce ratios that slowly grow. Let me work up a table, give me a few minutes.
odd lcm sigma ratio
1 1 1 1
3 3 4 1.333333333333333
5 15 24 1.6
7 105 192 1.828571428571429
9 315 624 1.980952380952381
11 3465 7488 2.161038961038961
13 45045 104832 2.327272727272727
15 45045 104832 2.327272727272727
17 765765 1886976 2.464171122994653
19 14549535 37739520 2.593864339994371
21 14549535 37739520 2.593864339994371
23 334639305 905748480 2.706641050428909
It is not difficult to prove that the ratio stays the same if the new odd number is divisible by two or more different primes, but the ratio increases if the new odd number is a prime or prime power. In the long run, the exponent for some prime factor $p$ in the LCM is roughly proportional to $frac{1}{log p}.$
Another sequence, less efficient but easier to describe and to program, just takes the ordinary product of the consecutive prime numbers, beginning with $3.$ It is not even necessary to find the product or its sum of divisors, every time we use a new prime we just multiply the existing ratio by $(p+1)/p.$ This increases without bound, because the sum of prime reciprocals diverges. Note how the first three ratios in this list are the same as the first three in the earlier list.
May 30, 2015
3 1.333333333333333
5 1.6
7 1.828571428571429
11 1.994805194805195
13 2.148251748251748
17 2.27461949814891
19 2.394336313840958
23 2.498437892703608
29 2.584590923486491
31 2.66796482424412
37 2.740071981656123
41 2.806903005598956
43 2.872179819682652
47 2.93329002861207
53 2.988635123491544
59 3.039289956093095
61 3.089114381602818
67 3.13522056640286
71 3.179378602549379
73 3.222931734091151
79 3.263728338320153
83 3.303050366492685
89 3.340163291958895
97 3.374597965071873
101 3.408009826112189
103 3.441097300152113
107 3.473257088004002
109 3.505121831930644
113 3.536140609204367
127 3.563984236048496
131 3.591190222583217
137 3.617403289901343
139 3.643427774001352
149 3.667880309397335
151 3.692170907472814
157 3.715687919622322
163 3.738483551031048
167 3.760869680079138
173 3.782608811177862
179 3.803740703977738
181 3.824755845988665
191 3.844780745705883
193 3.864701889466017
197 3.884319665554677
199 3.903838859853947
211 3.922340465824818
223 3.939929436523585
227 3.957285953865098
229 3.974566678554465
233 3.991624904642682
239 4.008326264076334
241 4.024958323263372
251 4.040994013794302
257 4.056717725910233
263 4.072142508138029
269 4.08728058437646
271 4.102362800554971
277 4.117172774564195
281 4.131824634971897
283 4.146424722021268
293 4.160576342232945
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
$endgroup$
add a comment |
$begingroup$
An efficient way to produce infinitely many numbers with increasing values of $frac{sigma(n)}{n}$ while staying with odd numbers, let $n$ be the least common multiple of the consecutive odd numbers $1,3,5,7,9,11,ldots.$ Not the product, the LCM. This method should produce ratios that slowly grow. Let me work up a table, give me a few minutes.
odd lcm sigma ratio
1 1 1 1
3 3 4 1.333333333333333
5 15 24 1.6
7 105 192 1.828571428571429
9 315 624 1.980952380952381
11 3465 7488 2.161038961038961
13 45045 104832 2.327272727272727
15 45045 104832 2.327272727272727
17 765765 1886976 2.464171122994653
19 14549535 37739520 2.593864339994371
21 14549535 37739520 2.593864339994371
23 334639305 905748480 2.706641050428909
It is not difficult to prove that the ratio stays the same if the new odd number is divisible by two or more different primes, but the ratio increases if the new odd number is a prime or prime power. In the long run, the exponent for some prime factor $p$ in the LCM is roughly proportional to $frac{1}{log p}.$
Another sequence, less efficient but easier to describe and to program, just takes the ordinary product of the consecutive prime numbers, beginning with $3.$ It is not even necessary to find the product or its sum of divisors, every time we use a new prime we just multiply the existing ratio by $(p+1)/p.$ This increases without bound, because the sum of prime reciprocals diverges. Note how the first three ratios in this list are the same as the first three in the earlier list.
May 30, 2015
3 1.333333333333333
5 1.6
7 1.828571428571429
11 1.994805194805195
13 2.148251748251748
17 2.27461949814891
19 2.394336313840958
23 2.498437892703608
29 2.584590923486491
31 2.66796482424412
37 2.740071981656123
41 2.806903005598956
43 2.872179819682652
47 2.93329002861207
53 2.988635123491544
59 3.039289956093095
61 3.089114381602818
67 3.13522056640286
71 3.179378602549379
73 3.222931734091151
79 3.263728338320153
83 3.303050366492685
89 3.340163291958895
97 3.374597965071873
101 3.408009826112189
103 3.441097300152113
107 3.473257088004002
109 3.505121831930644
113 3.536140609204367
127 3.563984236048496
131 3.591190222583217
137 3.617403289901343
139 3.643427774001352
149 3.667880309397335
151 3.692170907472814
157 3.715687919622322
163 3.738483551031048
167 3.760869680079138
173 3.782608811177862
179 3.803740703977738
181 3.824755845988665
191 3.844780745705883
193 3.864701889466017
197 3.884319665554677
199 3.903838859853947
211 3.922340465824818
223 3.939929436523585
227 3.957285953865098
229 3.974566678554465
233 3.991624904642682
239 4.008326264076334
241 4.024958323263372
251 4.040994013794302
257 4.056717725910233
263 4.072142508138029
269 4.08728058437646
271 4.102362800554971
277 4.117172774564195
281 4.131824634971897
283 4.146424722021268
293 4.160576342232945
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
$endgroup$
An efficient way to produce infinitely many numbers with increasing values of $frac{sigma(n)}{n}$ while staying with odd numbers, let $n$ be the least common multiple of the consecutive odd numbers $1,3,5,7,9,11,ldots.$ Not the product, the LCM. This method should produce ratios that slowly grow. Let me work up a table, give me a few minutes.
odd lcm sigma ratio
1 1 1 1
3 3 4 1.333333333333333
5 15 24 1.6
7 105 192 1.828571428571429
9 315 624 1.980952380952381
11 3465 7488 2.161038961038961
13 45045 104832 2.327272727272727
15 45045 104832 2.327272727272727
17 765765 1886976 2.464171122994653
19 14549535 37739520 2.593864339994371
21 14549535 37739520 2.593864339994371
23 334639305 905748480 2.706641050428909
It is not difficult to prove that the ratio stays the same if the new odd number is divisible by two or more different primes, but the ratio increases if the new odd number is a prime or prime power. In the long run, the exponent for some prime factor $p$ in the LCM is roughly proportional to $frac{1}{log p}.$
Another sequence, less efficient but easier to describe and to program, just takes the ordinary product of the consecutive prime numbers, beginning with $3.$ It is not even necessary to find the product or its sum of divisors, every time we use a new prime we just multiply the existing ratio by $(p+1)/p.$ This increases without bound, because the sum of prime reciprocals diverges. Note how the first three ratios in this list are the same as the first three in the earlier list.
May 30, 2015
3 1.333333333333333
5 1.6
7 1.828571428571429
11 1.994805194805195
13 2.148251748251748
17 2.27461949814891
19 2.394336313840958
23 2.498437892703608
29 2.584590923486491
31 2.66796482424412
37 2.740071981656123
41 2.806903005598956
43 2.872179819682652
47 2.93329002861207
53 2.988635123491544
59 3.039289956093095
61 3.089114381602818
67 3.13522056640286
71 3.179378602549379
73 3.222931734091151
79 3.263728338320153
83 3.303050366492685
89 3.340163291958895
97 3.374597965071873
101 3.408009826112189
103 3.441097300152113
107 3.473257088004002
109 3.505121831930644
113 3.536140609204367
127 3.563984236048496
131 3.591190222583217
137 3.617403289901343
139 3.643427774001352
149 3.667880309397335
151 3.692170907472814
157 3.715687919622322
163 3.738483551031048
167 3.760869680079138
173 3.782608811177862
179 3.803740703977738
181 3.824755845988665
191 3.844780745705883
193 3.864701889466017
197 3.884319665554677
199 3.903838859853947
211 3.922340465824818
223 3.939929436523585
227 3.957285953865098
229 3.974566678554465
233 3.991624904642682
239 4.008326264076334
241 4.024958323263372
251 4.040994013794302
257 4.056717725910233
263 4.072142508138029
269 4.08728058437646
271 4.102362800554971
277 4.117172774564195
281 4.131824634971897
283 4.146424722021268
293 4.160576342232945
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
edited May 30 '15 at 22:43
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
answered May 30 '15 at 21:11
Will JagyWill Jagy
104k5103202
104k5103202
add a comment |
add a comment |
$begingroup$
Let $O_n$ be the product of the first $n$ odd integers.
Notice that $sigma(O_n)geq frac{O_n}{1} + frac{O_n}{3} + dots+ frac{O_n}{n} = O_n(1+1/3 + dots 1/n)$.
Recall $sumlimits_{i=1}^infty frac{1}{2i-1}$ diverges.
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
add a comment |
$begingroup$
Let $O_n$ be the product of the first $n$ odd integers.
Notice that $sigma(O_n)geq frac{O_n}{1} + frac{O_n}{3} + dots+ frac{O_n}{n} = O_n(1+1/3 + dots 1/n)$.
Recall $sumlimits_{i=1}^infty frac{1}{2i-1}$ diverges.
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
add a comment |
$begingroup$
Let $O_n$ be the product of the first $n$ odd integers.
Notice that $sigma(O_n)geq frac{O_n}{1} + frac{O_n}{3} + dots+ frac{O_n}{n} = O_n(1+1/3 + dots 1/n)$.
Recall $sumlimits_{i=1}^infty frac{1}{2i-1}$ diverges.
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
$endgroup$
Let $O_n$ be the product of the first $n$ odd integers.
Notice that $sigma(O_n)geq frac{O_n}{1} + frac{O_n}{3} + dots+ frac{O_n}{n} = O_n(1+1/3 + dots 1/n)$.
Recall $sumlimits_{i=1}^infty frac{1}{2i-1}$ diverges.
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
answered Feb 1 at 2:29
Jorge Fernández HidalgoJorge Fernández Hidalgo
77.1k1394195
77.1k1394195
add a comment |
add a comment |
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