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%I #48 Mar 27 2021 08:09:17
%S 1,1,7,25,151,751,4411,24697,146455,862351,5195257,31392967,191815339,
%T 1177508515,7276161907,45154764025,281492498455,1761076827895,
%U 11055132835705,69600761349175,439370198255401,2780265190892641,17631718101804517,112038660509078695
%N G.f.: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1).
%C Compare g.f. to: Sum_{n>=0} (3*n)!/n!^3 * x^(2*n)/(1-2*x)^(3*n+1), which is a g.f. of the Franel numbers (A000172).
%C From _Zhi-Wei Sun_, Nov 12 2016: (Start)
%C Conjecture: (i) For any prime p > 3 and positive integer n, the number (a(p*n)-a(n))/(p*n)^3 is always a p-adic integer.
%C (ii) For any prime p == 1 (mod 3), we have Sum_{k=0..p-1}a(k) == C(2(p-1)/3,(p-1)/3) (mod p^2). For any prime p == 2 (mod 3), we have Sum_{k=0..p-1}a(k) == 2p/C(2(p+1)/3,(p+1)/3) (mod p^2).
%C We have proved part (i) of this conjecture for n = 1. (End)
%C Diagonal of rational functions 1/(1 - x*y - y*z - x*z - x*y*z), 1/(1 - x*y + y*z + x*z - x*y*z). - _Gheorghe Coserea_, Jul 03 2018
%C Number of paths from (0,0,0) to (n,n,n) using steps (1,1,0), (1,0,1), (0,1,1), and (1,1,1). - _William J. Wang_, Dec 07 2020
%H Vaclav Kotesovec, <a href="/A208425/b208425.txt">Table of n, a(n) for n = 0..1000</a>
%H A. Bostan, S. Boukraa, J.-M. Maillard and J.-A. Weil, <a href="http://arxiv.org/abs/1507.03227">Diagonals of rational functions and selected differential Galois groups</a>, arXiv preprint arXiv:1507.03227 [math-ph], 2015.
%H Hao Pan and Zhi-Wei Sun, <a href="https://arxiv.org/abs/2012.05121">Supercongruences for central trinomial coefficients</a>, arXiv:2012.05121 [math.NT], 2020.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1610.03384">Supercongruences involving Lucas sequences</a>, arXiv:1610.03384 [math.NT], 2016.
%F Conjecture: n^2*(3*n-5)*a(n) +(-9*n^3+24*n^2-17*n+4) *a(n-1) -(3*n-4) *(24*n^2-56*n+27)*a(n-2) -(3*n-2)*(n-2)^2*a(n-3)=0. - _R. J. Mathar_, Mar 10 2016
%F a(n) ~ sqrt(1/2 + sqrt(13)*cos(arctan(53*sqrt(3)/19)/3)/6) * (1 + 6*cos(Pi/9))^n / (Pi*n). - _Vaclav Kotesovec_, Jul 05 2016
%F It is easy to show that a(n) = Sum_{k=0..n}C(n,k)*C(n-k,k)*C(n+k,k) = Sum_{k=0..n}C(n+k,k)*C(n,2k)*C(2k,k). By this formula and the Zeilberger algorithm, we confirm the recurrence conjectured by _R. J. Mathar_. - _Zhi-Wei Sun_, Nov 12 2016
%F G.f. y=A(x) satisfies: 0 = x*(x + 2)*(x^3 + 24*x^2 + 3*x - 1)*y'' + (3*x^4 + 56*x^3 + 147*x^2 + 12*x - 2)*y' + (x^3 + 9*x^2 + 42*x + 2)*y. - _Gheorghe Coserea_, Jul 03 2018
%e G.f.: A(x) = 1 + x + 7*x^2 + 25*x^3 + 151*x^4 + 751*x^5 + 4411*x^6 +...
%e where
%e A(x) = 1/(1-x) + 6*x^2/(1-x)^4 + 90*x^4/(1-x)^7 + 1680*x^6/(1-x)^10 + 34650*x^8/(1-x)^13 + 756756*x^10/(1-x)^16 +...
%p series(hypergeom([1/3, 2/3],[1],27*x^2/(1-x)^3)/(1-x), x=0, 25); # _Mark van Hoeij_, May 20 2013
%t nmax = 20; CoefficientList[Series[Sum[(3*n)!/n!^3 * x^(2*n)/(1-x)^(3*n+1), {n, 0, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Jul 05 2016 *)
%o (PARI) {a(n)=polcoeff(sum(m=0,n, (3*m)!/m!^3*x^(2*m)/(1-x+x*O(x^n))^(3*m+1)),n)}
%o for(n=0,25,print1(a(n),", "))
%Y Cf. A000172, A208426, A244973.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 26 2012
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