A141057 - OEIS

%I #32 Jun 18 2022 14:19:17

%S 1,3,27,381,6219,111753,2151549,43497891,912018123,19671397617,

%T 434005899777,9754118112951,222621127928109,5147503311510927,

%U 120355825553777043,2841378806367492381,67648182142185172683,1622612550613755130497,39178199253650491044441

%N Number of Abelian cubes of length 3n over an alphabet of size 3. An Abelian cube is a string of the form x x' x'' with |x| = |x'| = |x''| and x is a permutation of x' and x''.

%C Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(3*k)) hold for primes p >= 5 and positive integers n and k. Extending the sequence to negative n via a(-n) = Sum_{k = 0..n} C(-n,k)^3 * Sum_{j = 0..k} C(k,j)^3 produces the sequence [-1, 255, -53893, 14396623, -4388536251, 1461954981315, -518606406878589, ...] that appears to satisfy the same supercongruences. - _Peter Bala_, Apr 27 2022

%H Alois P. Heinz, <a href="/A141057/b141057.txt">Table of n, a(n) for n = 0..250</a>

%F a(n) = sum of (n!/(n1)! (n2)! (n3!))^3 over all nonnegative n1, n2, n3 such that n1+n2+n3 = n.

%F G.f.: Sum_{n>=0} a(n)*x^n/n!^3 = [ Sum_{n>=0} x^n/n!^3 ]^3. - _Paul D. Hanna_, Jan 19 2011

%F a(n) = Sum_{k=0..n} C(n,k)^3 * Sum_{j=0..k} C(k,j)^3 = Sum_{k=0..n} C(n,k)^3*A000172(k). - _Paul D. Hanna_, Jan 20 2011

%F a(n) ~ 3^(3*n+2) / (4 * Pi^2 * n^2). - _Vaclav Kotesovec_, Sep 04 2014

%F a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^3. - _Peter Luschny_, May 31 2017

%e a(1) = 3 as the Abelian cubes are aaa, bbb, ccc.

%e G.f.: A(x) = 1 + 3*x + 27*x^2/2!^3 + 381*x^3/3!^3 + 6219*x^4/4!^3 +...

%e A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +...]^3. - _Paul D. Hanna_

%p a:= proc(n) option remember; `if`(n<3, [1, 3, 27][n+1],

%p      ((567*n^6-3213*n^5+7083*n^4-7920*n^3+4968*n^2-1680*n+240)*a(n-1)

%p       -3*(3*n-4)*(63*n^5-399*n^4+1039*n^3-1380*n^2+920*n-240)*a(n-2)

%p       +729*(21*n^2-35*n+15)*(n-2)^4*a(n-3))/(n^4*(21*n^2-77*n+71)))

%p     end:

%p seq(a(n), n=0..20); # _Alois P. Heinz_, May 25 2013

%p A141057_list := proc(len) series(hypergeom([], [1, 1], x)^3, x, len);

%p seq((n!)^3*coeff(%, x, n), n=0..len-1) end:

%p A141057_list(19); # _Peter Luschny_, May 31 2017

%t a[n_] := Sum[Binomial[n, k]^3 HypergeometricPFQ[{-k, -k, -k}, {1, 1}, -1], {k, 0, n}]; Table[a[n], {n, 0, 18}] (* _Jean-François Alcover_, Jun 27 2019 *)

%o (PARI) {a(n)=if(n<0,0,n!^3*polcoeff(sum(m=0,n,x^m/m!^3+x*O(x^n))^3,n))}

%o (PARI) {a(n)=sum(k=0,n,binomial(n,k)^3*sum(j=0,k,binomial(k,j)^3))}

%o (PARI) N=33; x='x+O('x^N)

%o Vec(serlaplace(serlaplace(serlaplace(sum(n=0,N,x^n/(n!^3)))^3))) /* show terms */

%Y Cf. A000172 (Franel numbers), A002893.

%K nonn

%O 0,2

%A _Jeffrey Shallit_, Aug 01 2008

%E Extended by _Paul D. Hanna_, Jan 19 2011

%E Offset corrected by _Alois P. Heinz_, May 25 2013