A118023 Column 0 of triangle A118022, where the matrix square of A118022 shifts each column up 1 row, dropping the main diagonal of powers of 2.
1, 1, 3, 19, 243, 6227, 319251, 32737427, 6714170259, 2754046149011, 2259333156408723, 3706972573115098515, 12164337831474297132435, 79833941280970262512121235, 1047892334589811621056371520915
Offset: 0
Keywords
Examples
1 = (1-x) + 1*x*(1-x)*(1-2*x) + 3*x^2*(1-x)*(1-2*x)*(1-4*x) + 19*x^3*(1-x)*(1-2*x)*(1-4*x)*(1-8*x) + 243*x^4*(1-x)*(1-2*x)*(1-4*x)*(1-8*x)*(1-16*x) + ...
Crossrefs
Cf. A118022.
Programs
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PARI
{a(n)=if(n==0, 1, polcoeff(1-sum(k=0, n-1, a(k)*x^k*prod(j=0, k, 1-2^j*x+x*O(x^n))), n))} {a(n)=local(CF=1+x*O(x^n)); for(k=1, n, CF=1/(1-x/2^(n-k+1)*CF)); 2^(n*(n+1)/2)*polcoeff(CF, n)} \\ Paul D. Hanna, Sep 28 2012
Formula
G.f.: 1 = Sum_{n>=0} a(n)*x^n*prod_{k=0, n} (1-2^k*x) with a(0)=1.
a(n) = 2^(n*(n-1)/2)*b(n) where b(0)=1 and b(n)=sum(i=0,n-1,b(i)*b(n-1-i)/2^i). - Benoit Cloitre, Oct 25 2006
G.f.: Sum_{n>=0} a(n)*x^n/2^(n*(n+1)/2) = 1/(1 - (x/2)/(1 - (x/2^2)/(1 - (x/2^3)/(1 - (x/2^4)/(1 - (x/2^5)/(1 - ...)))))), a continued fraction. - Paul D. Hanna, Sep 28 2012
Comments