The On-Line Encyclopedia of Integer Sequences (OEIS)

Revision History for A300058

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Showing entries 1-10 | older changes

A300058

a(n) = binomial(3*n,n)/(2*Pi)*Integral_{x=0..2*Pi} (12*cos^2(x)*sin(x) + 20*sin^3(x))^(2*n) dx.
(history; published version)

#33 by Joerg Arndt at Wed Sep 12 02:18:10 EDT 2018

STATUS

reviewed

approved

#32 by Michel Marcus at Wed Sep 12 01:44:35 EDT 2018

STATUS

proposed

reviewed

#31 by Michael De Vlieger at Tue Sep 11 22:13:28 EDT 2018

STATUS

editing

proposed

#30 by Michael De Vlieger at Tue Sep 11 22:13:26 EDT 2018

LINKS

Brad Klee, <a href="http://demonstrations.wolfram.com/DerivingHypergeometricPicardFuchsEquations/">Deriving Hypergeometric Picard-Fuchs Equations</a>, Wolfram Demonstrations Project (2018).

STATUS

approved

editing

#29 by N. J. A. Sloane at Thu Apr 19 13:12:19 EDT 2018

STATUS

proposed

approved

#28 by Peter Luschny at Thu Apr 19 05:00:07 EDT 2018

STATUS

editing

proposed

#27 by Peter Luschny at Thu Apr 19 04:59:54 EDT 2018

MAPLE

a := n -> 36^n*(3*n)!/n!^3*hypergeom([-2*n, n+1/2], [n+1], -2/3):

seq(simplify(a(n)), n=0..10); # Peter Luschny, Apr 19 2018

STATUS

proposed

editing

#26 by Vaclav Kotesovec at Thu Apr 19 03:22:23 EDT 2018

STATUS

editing

proposed

#25 by Vaclav Kotesovec at Thu Apr 19 03:22:07 EDT 2018

MATHEMATICA

b[NN_]:=Expand[Total[Flatten[#]]&/@Table[Binomial[3*n, n]*Binomial[2*n, k2]*Binomial[2*n, k1]*Binomial[2*n, 3*n-k1-k2]*((4+Sqrt[15])^(2*n-k1))*((4-Sqrt[15])^(2*n-k2)), {n, 0, NN}, {k1, 0, 2*n}, {k2, 0, 2*n}]]; ({#, SameQ[#, a/@Range[0, 10]]}&@b[10])[[1]]

STATUS

proposed

editing

#24 by Vaclav Kotesovec at Wed Apr 18 18:36:05 EDT 2018

STATUS

editing

proposed

Discussion

Wed Apr 18

19:24

Bradley Klee: Why should we remove validation? Compare with A295870. Sloane did not hassle me for similar programming there, and published right away. In case anyone is worried about the bigger issue--if the Picard-Fuchs equation is actually correct--here's a certified derivation: https://ptpb.pw/Qs8s.png .

Thu Apr 19

03:21

Vaclav Kotesovec: A295870 has the same problem. The solution is easy (...)[[1]].