cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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''.

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%I A141057 #32 Jun 18 2022 14:19:17
%S A141057 1,3,27,381,6219,111753,2151549,43497891,912018123,19671397617,
%T A141057 434005899777,9754118112951,222621127928109,5147503311510927,
%U A141057 120355825553777043,2841378806367492381,67648182142185172683,1622612550613755130497,39178199253650491044441
%N 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''.
%C A141057 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 A141057 Alois P. Heinz, <a href="/A141057/b141057.txt">Table of n, a(n) for n = 0..250</a>
%F A141057 a(n) = sum of (n!/(n1)! (n2)! (n3!))^3 over all nonnegative n1, n2, n3 such that n1+n2+n3 = n.
%F A141057 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 A141057 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 A141057 a(n) ~ 3^(3*n+2) / (4 * Pi^2 * n^2). - _Vaclav Kotesovec_, Sep 04 2014
%F A141057 a(n) = (n!)^3 * [x^n] hypergeom([], [1, 1], x)^3. - _Peter Luschny_, May 31 2017
%e A141057 a(1) = 3 as the Abelian cubes are aaa, bbb, ccc.
%e A141057 G.f.: A(x) = 1 + 3*x + 27*x^2/2!^3 + 381*x^3/3!^3 + 6219*x^4/4!^3 +...
%e A141057 A(x) = [1 + x + x^2/2!^3 + x^3/3!^3 + x^4/4!^3 +...]^3. - _Paul D. Hanna_
%p A141057 a:= proc(n) option remember; `if`(n<3, [1, 3, 27][n+1],
%p A141057      ((567*n^6-3213*n^5+7083*n^4-7920*n^3+4968*n^2-1680*n+240)*a(n-1)
%p A141057       -3*(3*n-4)*(63*n^5-399*n^4+1039*n^3-1380*n^2+920*n-240)*a(n-2)
%p A141057       +729*(21*n^2-35*n+15)*(n-2)^4*a(n-3))/(n^4*(21*n^2-77*n+71)))
%p A141057     end:
%p A141057 seq(a(n), n=0..20); # _Alois P. Heinz_, May 25 2013
%p A141057 A141057_list := proc(len) series(hypergeom([], [1, 1], x)^3, x, len);
%p A141057 seq((n!)^3*coeff(%, x, n), n=0..len-1) end:
%p A141057 A141057_list(19); # _Peter Luschny_, May 31 2017
%t A141057 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 A141057 (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 A141057 (PARI) {a(n)=sum(k=0,n,binomial(n,k)^3*sum(j=0,k,binomial(k,j)^3))}
%o A141057 (PARI) N=33; x='x+O('x^N)
%o A141057 Vec(serlaplace(serlaplace(serlaplace(sum(n=0,N,x^n/(n!^3)))^3))) /* show terms */
%Y A141057 Cf. A000172 (Franel numbers), A002893.
%K A141057 nonn
%O A141057 0,2
%A A141057 _Jeffrey Shallit_, Aug 01 2008
%E A141057 Extended by _Paul D. Hanna_, Jan 19 2011
%E A141057 Offset corrected by _Alois P. Heinz_, May 25 2013