Mixing
Concerning this sum $sum_{k=1}^{infty}{2k choose k}^2 4^{-2k}cdot frac{1}{(ak)^3-ak}$
$begingroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
asked Jan 3 at 9:21
user583851user583851
4507
$endgroup$
add a comment |
$begingroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
asked Jan 3 at 9:21
user583851user583851
4507
$endgroup$
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
$begingroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
asked Jan 3 at 9:21
user583851user583851
4507
$endgroup$
I was looking at this Ramanujan phi-Function
Let:
$$R(a)=sum_{k=1}^{infty}frac{1}{(ak)^3-ak}tag1$$
and this paper of the form,
$$sum_{k=1}^{infty}frac{{2k choose k}}{4^k}tag2$$ and
$$sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2tag 3$$
By combining them together we have
$$S(a)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(ak)^3-ak}tag4$$ and
$$T(a)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(ak)^3-ak}tag5$$
The conjectured closed form for $(4)$ and $(5)$, where $a=2$ are
$$S(2)=sum_{k=1}^{infty}frac{{2k choose k}}{4^k}cdot frac{1}{(2k)^3-2k}=frac{pi}{4}-ln(2)tag6$$ and
$$T(2)=sum_{k=1}^{infty}left[frac{{2k choose k}}{4^k}right]^2cdot frac{1}{(2k)^3-2k}=frac{6G-piln(4)-1}{pi}tag7$$
Where G is the Catalan's constant.
How can we prove these conjectures?
sequences-and-series
sequences-and-series
asked Jan 3 at 9:21
user583851user583851
4507
asked Jan 3 at 9:21
user583851user583851
4507
edited Jan 3 at 10:10
user583851
asked Jan 3 at 9:21
user583851user583851
4507
asked Jan 3 at 9:21
user583851user583851
4507
asked Jan 3 at 9:21
user583851user583851
4507
4507
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060392%2fconcerning-this-sum-sum-k-1-infty2k-choose-k2-4-2k-cdot-frac1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
Let us tackle $(4)$ first. The ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]$ is $frac{1}{sqrt{1-x}}$ and
$$ frac{1}{(an)^3-an}=frac{1}{(an-1)an(an+1)} = int_{0}^{1} x^{an}frac{(1-x)^2}{2x^2},dx $$
hence
$$ S(a) = int_{0}^{1}left(frac{1}{sqrt{1-x^a}}-1right)cdotfrac{(1-x)^2}{2x^2},dx $$
and similarly, since the ordinary generating function for $left[frac{1}{4^n}binom{2n}{n}right]^2$ is $frac{2}{pi}K(x)$,
$$ T(a) = frac{2}{pi}int_{0}^{1}left(K(x^a)-frac{pi}{2}right)frac{(1-x)^2}{2x^2},dx. $$
For specific values of $a$ (like $a=2$ or $a=4$) integration by parts, suitable substitutions and known FL-expansions greatly simplify the underlying hypergeometric structure of these integrals. I won't expect a simple closed form in the general case.
(6) and (7) are proved in the mentioned paper, not just conjectured.
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
edited Jan 3 at 17:22
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
answered Jan 3 at 17:16
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3060392%2fconcerning-this-sum-sum-k-1-infty2k-choose-k2-4-2k-cdot-frac1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Is it ${2n choose n}^2 4^{-2n}$ or ${2k choose k}^2 4^{-2k}$ in the headline?
$endgroup$
– Andreas
Jan 3 at 9:43
$begingroup$
Why don't you try the methods of the paper? ;)
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:04
$begingroup$
Please notice that $(2)$ is actually a divergent series, since $frac{1}{4^n}binom{2n}{n}simfrac{1}{sqrt{pi n}}$, so "this paper of the form divergent series" does not make much sense. Better to say "this paper about the weight $frac{1}{4^n}binom{2n}{n}$ and its square".
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:20
$begingroup$
$(3)$ is also a divergent series (same reason), while $$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^3$$ is related to the lemniscate constant (hence to $Gammaleft(frac{1}{4}right)$) by many good reasons; among them, Clausen's formula for the square of particular $phantom{}_2 F_1$s.
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:32
$begingroup$
$$sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^4$$ is much more obscure: see math.stackexchange.com/questions/2506266/…
$endgroup$
– Jack D'Aurizio
Jan 3 at 17:34