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 A186239 G.f. A(x) satisfies A(x) = 1+x*A(x)+x^2*A(x)^2+2*x^3*A(x)^3. 0

%I #29 Aug 08 2018 15:11:28

%S 1,1,2,6,17,51,163,533,1779,6055,20908,73052,257863,918139,3293538,

%T 11891778,43183835,157616827,577902846,2127539802,7861397403,

%U 29145629141,108385383175,404184619545,1511132059333,5663069529201,21269203639596,80044555061812

%N G.f. A(x) satisfies A(x) = 1+x*A(x)+x^2*A(x)^2+2*x^3*A(x)^3.

%H Vladimir Kruchinin, D. V. Kruchinin, <a href="http://arxiv.org/abs/1103.2582">Composita and their properties</a>, arXiv:1103.2582 [math.CO], 2011-2013.

%F a(n) = 1/n * sum(j=0..n, C(n,j) * sum(i=j..n+j-1, C(j,i-j) * C(n-j,3*j-n-i-1) * 2^(3*j-n-i-1))), n>0.

%F Conjecture: 4*(2*n+3)*(n+1)*a(n) +(113*n^2-91*n-72)*a(n-1) + 3*(-135*n^2+263*n-108)*a(n-2) -3*(107*n-119)*(n-2)*a(n-3) -1411*(n-2)*(n-3)*a(n-4)=0. - _R. J. Mathar_, Nov 14 2011

%F a(n) is the top left term of M^n, M = an infinite matrix with (1,1,1,...) as diagonals starting at positions (1,2), (1,1), and (2,1); with a diagonal of (2,2,2,...) starting at (3,1). - _Gary W. Adamson_, Nov 25 2011

%e a(3) = 6 since the top row of M^3 = (6, 5, 3, 1, 0, 0, ...).

%t terms = 28;

%t A[_] = 0;

%t Do[A[x_] = 1 + x A[x] + x^2 A[x]^2 + 2 x^3 A[x]^3 + O[x]^terms, {terms}];

%t CoefficientList[A[x], x] (* _Jean-François Alcover_, Aug 08 2018 *)

%K nonn

%O 0,3

%A _Vladimir Kruchinin_, Feb 15 2011

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Last modified August 15 15:45 EDT 2024. Contains 375173 sequences. (Running on oeis4.)