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G.f.: Sum_{n>=0} x^n * Product_{k=0..n-1} (1 + (2*k+1)*x + x^2).
4

%I #7 Sep 13 2017 17:01:03

%S 1,1,2,6,15,47,150,522,1903,7319,29396,122988,534141,2400061,11136516,

%T 53220492,261576725,1319629445,6825232486,36137198722,195664517227,

%U 1082169511883,6108213101658,35153836421302,206126910439763,1230477025952427,7473067121404104,46146114390128888,289554642297817561,1845220293901278041,11936266843924805064

%N G.f.: Sum_{n>=0} x^n * Product_{k=0..n-1} (1 + (2*k+1)*x + x^2).

%C Antidiagonal sums of irregular triangle A291845, which has row sums equal to the odd double factorials A001147.

%H Paul D. Hanna, <a href="/A291848/b291848.txt">Table of n, a(n) for n = 0..300</a>

%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 15*x^4 + 47*x^5 + 150*x^6 + 522*x^7 + 1903*x^8 + 7319*x^9 + 29396*x^10 + 122988*x^11 + 534141*x^12 +...

%e which equals the series:

%e A(x) = 1 + x*(1+x+x^2) + x^2*(1+x+x^2)*(1+3*x+x^2) + x^3*(1+x+x^2)*(1+3*x+x^2)*(1+5*x+x^2) + x^4*(1+x+x^2)*(1+3*x+x^2)*(1+5*x+x^2)*(1+7*x+x^2) +...

%t nmax = 30; CoefficientList[Series[Sum[2^n*x^(2*n)*Pochhammer[(1 + x + x^2)/(2*x), n], {n, 0, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Sep 13 2017 *)

%o (PARI) {a(n)=sum(k=0, n, polcoeff(prod(j=0, n-k-1, 1+(2*j+1)*x+x^2), k))}

%o for(n=0,30,print1(a(n),", "))

%o (PARI) {a(n)=polcoeff(sum(m=0, n, x^m*prod(j=0, m-1, 1+(2*j+1)*x+x^2))+x*O(x^n), n)}

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

%Y Cf. A291845, A201951.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 03 2017