A141057 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''.

1, 3, 27, 381, 6219, 111753, 2151549, 43497891, 912018123, 19671397617, 434005899777, 9754118112951, 222621127928109, 5147503311510927, 120355825553777043, 2841378806367492381, 67648182142185172683, 1622612550613755130497, 39178199253650491044441

Offset: 0

Author

Jeffrey Shallit, Aug 01 2008

Keywords

• nonn

Comments

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

Examples

			a(1) = 3 as the Abelian cubes are aaa, bbb, ccc.
G.f.: A(x) = 1 + 3*x + 27*x^2/2!^3 + 381*x^3/3!^3 + 6219*x^4/4!^3 +...
A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +...]^3. - _Paul D. Hanna_
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..250

Crossrefs

Cf. A000172 (Franel numbers), A002893.

Programs

Maple

a:= proc(n) option remember; `if`(n<3, [1, 3, 27][n+1],
     ((567*n^6-3213*n^5+7083*n^4-7920*n^3+4968*n^2-1680*n+240)*a(n-1)
      -3*(3*n-4)*(63*n^5-399*n^4+1039*n^3-1380*n^2+920*n-240)*a(n-2)
      +729*(21*n^2-35*n+15)*(n-2)^4*a(n-3))/(n^4*(21*n^2-77*n+71)))
    end:
seq(a(n), n=0..20); # Alois P. Heinz, May 25 2013
A141057_list := proc(len) series(hypergeom([], [1, 1], x)^3, x, len);
seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
A141057_list(19); # Peter Luschny, May 31 2017

Mathematica

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 *)

PARI

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

PARI

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

PARI

N=33; x='x+O('x^N)
Vec(serlaplace(serlaplace(serlaplace(sum(n=0,N,x^n/(n!^3)))^3))) /* show terms */

Formula

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

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

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

a(n) ~ 3^(3*n+2) / (4 * Pi^2 * n^2). - Vaclav Kotesovec, Sep 04 2014

a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^3. - Peter Luschny, May 31 2017

Extensions

Extended by Paul D. Hanna, Jan 19 2011

Offset corrected by Alois P. Heinz, May 25 2013