Mixing
General method to find sum of binomial
Specifically I want to ask the method to solve
$$sum_{k=0}^n binom{4n+b}{4k}; b=[0,1,2,3]$$
And how to solve series of type
$$sum_{k=0}^n binom{an+b}{ak}; a=[1,2,3,...], b=[0,1,2,...,a-1]$$
combinatorics complex-numbers binomial-coefficients
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
add a comment |
Specifically I want to ask the method to solve
$$sum_{k=0}^n binom{4n+b}{4k}; b=[0,1,2,3]$$
And how to solve series of type
$$sum_{k=0}^n binom{an+b}{ak}; a=[1,2,3,...], b=[0,1,2,...,a-1]$$
combinatorics complex-numbers binomial-coefficients
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38
add a comment |
Specifically I want to ask the method to solve
$$sum_{k=0}^n binom{4n+b}{4k}; b=[0,1,2,3]$$
And how to solve series of type
$$sum_{k=0}^n binom{an+b}{ak}; a=[1,2,3,...], b=[0,1,2,...,a-1]$$
combinatorics complex-numbers binomial-coefficients
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
Specifically I want to ask the method to solve
$$sum_{k=0}^n binom{4n+b}{4k}; b=[0,1,2,3]$$
And how to solve series of type
$$sum_{k=0}^n binom{an+b}{ak}; a=[1,2,3,...], b=[0,1,2,...,a-1]$$
combinatorics complex-numbers binomial-coefficients
combinatorics complex-numbers binomial-coefficients
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
edited Nov 22 '18 at 7:38
Saurav Singh
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
asked Nov 22 '18 at 7:27
Saurav SinghSaurav Singh
183
183
series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38
add a comment |
series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38
series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38
add a comment |
1 Answer
1
active
oldest
votes
I am sure that you will enjoy the result !
If $$S_n=sum_{k=0}^n binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{frac{4 n+b}{2} } left((-1)^b, 2^{frac{4 n+b}{2} } cos (pi b)+cos
left(frac{ (4 n+b)pi}{4} right)+(-1)^b cos left(frac{3(4
n+b)pi}{4} right)right)-$$ $$4 binom{4 n+b}{4 n+4} ,
_5F_4left(1,frac{4-b}{4},frac{5-b}{4},frac{6-b}{4},frac
{7-b}{4};frac{4n+5}{4},frac{4n+6}{4},frac{4n+7}{4},frac{4n+8}{4};1right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$left(
begin{array}{cc}
b & text{OEIS sequence} \
0 & text{A070775} \
1 & text{A090407} \
2 & text{A001025} \
3 & text{A090408}
end{array}
right)$$
Be sure that I did not make this by myself.
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I am sure that you will enjoy the result !
If $$S_n=sum_{k=0}^n binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{frac{4 n+b}{2} } left((-1)^b, 2^{frac{4 n+b}{2} } cos (pi b)+cos
left(frac{ (4 n+b)pi}{4} right)+(-1)^b cos left(frac{3(4
n+b)pi}{4} right)right)-$$ $$4 binom{4 n+b}{4 n+4} ,
_5F_4left(1,frac{4-b}{4},frac{5-b}{4},frac{6-b}{4},frac
{7-b}{4};frac{4n+5}{4},frac{4n+6}{4},frac{4n+7}{4},frac{4n+8}{4};1right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$left(
begin{array}{cc}
b & text{OEIS sequence} \
0 & text{A070775} \
1 & text{A090407} \
2 & text{A001025} \
3 & text{A090408}
end{array}
right)$$
Be sure that I did not make this by myself.
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
I am sure that you will enjoy the result !
If $$S_n=sum_{k=0}^n binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{frac{4 n+b}{2} } left((-1)^b, 2^{frac{4 n+b}{2} } cos (pi b)+cos
left(frac{ (4 n+b)pi}{4} right)+(-1)^b cos left(frac{3(4
n+b)pi}{4} right)right)-$$ $$4 binom{4 n+b}{4 n+4} ,
_5F_4left(1,frac{4-b}{4},frac{5-b}{4},frac{6-b}{4},frac
{7-b}{4};frac{4n+5}{4},frac{4n+6}{4},frac{4n+7}{4},frac{4n+8}{4};1right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$left(
begin{array}{cc}
b & text{OEIS sequence} \
0 & text{A070775} \
1 & text{A090407} \
2 & text{A001025} \
3 & text{A090408}
end{array}
right)$$
Be sure that I did not make this by myself.
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
add a comment |
I am sure that you will enjoy the result !
If $$S_n=sum_{k=0}^n binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{frac{4 n+b}{2} } left((-1)^b, 2^{frac{4 n+b}{2} } cos (pi b)+cos
left(frac{ (4 n+b)pi}{4} right)+(-1)^b cos left(frac{3(4
n+b)pi}{4} right)right)-$$ $$4 binom{4 n+b}{4 n+4} ,
_5F_4left(1,frac{4-b}{4},frac{5-b}{4},frac{6-b}{4},frac
{7-b}{4};frac{4n+5}{4},frac{4n+6}{4},frac{4n+7}{4},frac{4n+8}{4};1right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$left(
begin{array}{cc}
b & text{OEIS sequence} \
0 & text{A070775} \
1 & text{A090407} \
2 & text{A001025} \
3 & text{A090408}
end{array}
right)$$
Be sure that I did not make this by myself.
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
I am sure that you will enjoy the result !
If $$S_n=sum_{k=0}^n binom{4n+b}{4k}$$ then a CAS gives
$$4S_n=2^{frac{4 n+b}{2} } left((-1)^b, 2^{frac{4 n+b}{2} } cos (pi b)+cos
left(frac{ (4 n+b)pi}{4} right)+(-1)^b cos left(frac{3(4
n+b)pi}{4} right)right)-$$ $$4 binom{4 n+b}{4 n+4} ,
_5F_4left(1,frac{4-b}{4},frac{5-b}{4},frac{6-b}{4},frac
{7-b}{4};frac{4n+5}{4},frac{4n+6}{4},frac{4n+7}{4},frac{4n+8}{4};1right)$$ where appears the generalized hypergeometric function.
For the first values of $b$, the sequences can be found in $OEIS$.
$$left(
begin{array}{cc}
b & text{OEIS sequence} \
0 & text{A070775} \
1 & text{A090407} \
2 & text{A001025} \
3 & text{A090408}
end{array}
right)$$
Be sure that I did not make this by myself.
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
edited Nov 22 '18 at 12:08
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
answered Nov 22 '18 at 10:40
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
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series multisection
– Lord Shark the Unknown
Nov 22 '18 at 7:38
For $4k$, see Arthur T. Benjamin, Jacob N. Scott, Third and fourth binomial coefficients. For $ak$, there is a formula resulting in an $a$-term sum, involving roots of unity; but that's not very explicit in my opinion.
– darij grinberg
Nov 22 '18 at 7:38