A376527 - OEIS

%I #14 Oct 11 2024 16:03:03

%S 1,1,2,7,34,216,1610,13461,122254,1183568,12054498,127960158,

%T 1405852350,15901061916,184381675404,2184565641269,26375002217314,

%U 323767457670588,4033503712929478,50917059047932592,650430305337318538,8398511711996887848,109507259507469905574,1440631950110092280386

%N a(n) = Sum_{k=0..n*(n-1)/2} A227543(n,k)^2 for n >= 0.

%C Compare to binomial(2*n,n)/(n+1) = Sum_{k=0..n*(n-1)/2} A227543(n,k) for n >= 0; that is, the row sums of A227543 equals the Catalan numbers (A000108).

%C G.f. F(x,q) of triangle A227543 satisfies F(x,q) = 1 + x*F(x,q)*F(q*x,q).

%C Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

%H Paul D. Hanna, <a href="/A376527/b376527.txt">Table of n, a(n) for n = 0..301</a>

%F a(n) ~ c * 16^n / n^(9/2), where c = 0.430217025951475334005904244213062400539... - _Vaclav Kotesovec_, Oct 11 2024

%e G.f.: A(x) = 1 + x + 2*x^2 + 7*x^3 + 34*x^4 + 216*x^5 + 1610*x^6 + 13461*x^7 + 122254*x^8 + 1183568*x^9 + 12054498*x^10 + ...

%e where coefficient a(n) of x^n in A(x) equals the sum of the square of the terms in row n of triangle A227543, as follows.

%e a(0) = 1^2 = 1;

%e a(1) = 1^2 = 1;

%e a(2) = 1^2 + 1^2 = 2;

%e a(3) = 1^2 + 2^2 + 1^2 + 1^2 = 7;

%e a(4) = 1^2 + 3^2 + 3^2 + 3^2 + 2^2 + 1^2 + 1^2 = 34;

%e a(5) = 1^2 + 4^2 + 6^2 + 7^2 + 7^2 + 5^2 + 5^2 + 3^2 + 2^2 + 1^2 + 1^2 = 216;

%e a(6) = 1^2 + 5^2 + 10^2 + 14^2 + 17^2 + 16^2 + 16^2 + 14^2 + 11^2 + 9^2 + 7^2 + 5^2 + 3^2 + 2^2 + 1^2 + 1^2 = 1610;

%e ...

%o (PARI) \\ From g.f. of A227543, F(x, q) = 1 + x*F(q*x, q)*F(x, q)

%o {A227543(n, k) = my(F=1); for(i=1, n, F = 1 + x*F*subst(F, x, q*x) +x*O(x^n)); polcoef(polcoef(F, n, x), k, q)}

%o {a(n) = sum(k=0, n*(n-1)/2, A227543(n, k)^2)}

%o for(n=0,25, print1(a(n),", "))

%o (PARI) \\ faster (using program by Joerg Arndt in A227543)

%o N=30;

%o VP=vector(N+1); VP[1] = VP[2] = 1;  \\ one-based; memoization

%o P(n) = VP[n+1];

%o for (n=2, N, VP[n+1] = sum( i=0, n-1, P(i) * P(n-1 -i) * x^((i+1)*(n-1-i)) );print1(n,",") );

%o for(n=0,N, AV=Vec(P(n)); print1(AV*AV~,", "))

%Y Cf. A227543.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 11 2024