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A195443
Self-convolution square equals A195442.
2
1, 2, 17, 268, 6218, 191092, 7331943, 338203880, 18267488524, 1132962942756, 79464790795184, 6226443470442668, 539554771302358998, 51273487668881045852, 5305090125544931178395, 593952571797108683880684, 71571203087218842477715328, 9238664328667537859230164188
OFFSET
0,2
COMMENTS
Sequence A195442 satisfies the inversion:
1 = Sum_{n>=0} A195442(n)*x^n / Product_{k=1..n+1} (1+k*x)^4.
CONJECTURES. (Start)
_ This sequence consists entirely of integers.
_ a(2*(2^n-1)) is odd; odd numbers occur only at positions {2^(n+1)-2, n>=0}.
_ In the self-convolution square of this sequence (A195442), odd numbers occur only at positions {2^(n+2)-4, n>=0}. (End)
LINKS
EXAMPLE
O.g.f.: A(x) = 1 + 2*x + 17*x^2 + 268*x^3 + 6218*x^4 + 191092*x^5 +...
This sequence has odd terms at [0,2,6,14,30,62,126,...,2^(n+1)-2,...].
The square of the g.f. yields the g.f. of A195442:
A(x)^2 = 1 + 4*x + 38*x^2 + 604*x^3 + 13797*x^4 + 416168*x^5 +...
where the coefficients in A(x)^2 satisfy:
1 = 1/(1+x)^4 + 4*x/((1+x)^4*(1+2*x)^4) + 38*x^2/((1+x)^4*(1+2*x)^4*(1+3*x)^4) + 604*x^3/((1+x)^4*(1+2*x)^4*(1+3*x)^4*(1+4*x)^4) +...
Further, A195442 has odd terms at [0,4,12,28,60,124,...,2^(n+2)-4,...].
MATHEMATICA
(* b = A195442 *) b[n_] := b[n] = If[n == 0, 1, SeriesCoefficient[1 - Sum[b[k] x^k/Product[1 + j x, {j, 1, k+1}]^4, {k, 0, n-1}], {x, 0, n}]];
a[n_] := SeriesCoefficient[(1+Sum[b[m] x^m, {m, 1, n}])^(1/2), {x, 0, n}];
a /@ Range[0, 17] (* Jean-François Alcover, Nov 03 2019 *)
PROG
(PARI) {A195442(n)=if(n==0, 1, polcoeff(1-sum(k=0, n-1, A195442(k)*x^k/prod(j=1, k+1, 1+j*x+x*O(x^n))^4), n))}
{a(n)=polcoeff((1+sum(m=1, n, A195442(m)*x^m)+x*O(x^n))^(1/2), n)}
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Sep 18 2011
STATUS
approved