A202989
E.g.f: Sum_{n>=0} 3^(n^2) * exp(3^n*x) * x^n/n!.
Original entry on oeis.org
1, 4, 100, 21952, 45212176, 864866612224, 151334226289000000, 240066313618039143841792, 3437872835498096514323500400896, 443629285048033016198674962874808664064, 515464807019389919369209932597753906250000000000
Offset: 0
E.g.f.: A(x) = 1 + 4*x + 100*x^2/2! + 21952*x^3/3! + 45212176*x^4/4! +..
By the series identity, the e.g.f.:
A(x) = exp(x) + 3*exp(3*x)*x + 3^4*exp(3^2*x)*x^2/2! + 3^9*exp(3^3*x)*x^3/3! +...
expands into:
A(x) = 1 + 4*x + 10^2*x^2/2! + 28^3*x^3/3! + 82^4*x^4/4! + 244^5*x^5/5! +...+ (3^n+1)^n*x^n/n! +...
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{a(n, q=3, m=1, b=1)=(m*q^n + b)^n}
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{a(n, q=3, m=1, b=1)=n!*polcoeff(sum(k=0, n, m^k*q^(k^2)*exp(b*q^k*x+x*O(x^n))*x^k/k!), n)}
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{a(n, q=3, m=1, b=1)=polcoeff(sum(k=0,n,m^k*q^(k^2)*x^k/(1-b*q^k*x +x*O(x^n))^(k+1)),n)}
A243918
a(n) = Sum_{k=0..n} binomial(n,k) * (1 + 2^k)^k.
Original entry on oeis.org
1, 4, 32, 814, 86600, 39560554, 75654970772, 594996059517934, 19035905851947436400, 2460857798358946973785234, 1280109151917797032199865564812, 2672783800502564772495577135824089014, 22366199286781599568269093307412768076442280
Offset: 0
O.g.f.: A(x) = 1 + 4*x + 32*x^2 + 814*x^3 + 86600*x^4 + 39560554*x^5 +...
where the g.f. may be expressed by the series identity:
A(x) = 1/(1-x) + 3*x/(1-x)^2 + 5^2*x^2/(1-x)^3 + 9^3*x^3/(1-x)^4 + 17^4*x^4/(1-x)^5 + 33^5*x^5/(1-x)^6 + 65^6*x^6/(1-x)^7 +...
A(x) = 1/(1-2*x) + 2*x/(1-3*x)^2 + 2^4*x^2/(1-5*x)^3 + 2^9*x^3/(1-9*x)^4 + 2^16*x^4/(1-17*x)^5 + 2^25*x^5/(1-33*x)^6 + 2^36*x^6/(1-65*x)^7 +...
Illustration of initial terms:
a(0) = 1;
a(1) = 1 + (1+2);
a(2) = 1 + 2*(1+2) + (1+2^2)^2;
a(3) = 1 + 3*(1+2) + 3*(1+2^2)^2 + (1+2^3)^3;
a(4) = 1 + 4*(1+2) + 6*(1+2^2)^2 + 4*(1+2^3)^3 + (1+2^4)^4;
a(5) = 1 + 5*(1+2) + 10*(1+2^2)^2 + 10*(1+2^3)^3 + 5*(1+2^4)^4 + (1+2^5)^5; ...
Also, by a binomial identity we have
a(0) = 1;
a(1) = 2 + 2;
a(2) = 2^2 + 2*(1+2)*2 + 2^4;
a(3) = 2^3 + 3*(1+2)^2*2 + 3*(1+2^2)*2^4 + 2^9;
a(4) = 2^4 + 4*(1+2)^3*2 + 6*(1+2^2)^2*2^4 + 4*(1+2^3)*2^9 + 2^16;
a(5) = 2^5 + 5*(1+2)^4*2 + 10*(1+2^2)^3*2^4 + 10*(1+2^3)^2*2^9 + 5*(1+2^4)*2^16 + 2^25; ...
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Table[Sum[Binomial[n,k]*(1+2^k)^k,{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 18 2014 *)
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{a(n)=sum(k=0, n, binomial(n, k)*(1+2^k)^k)}
for(n=0, 20, print1(a(n), ", "))
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{a(n)=sum(k=0, n, binomial(n, k)*(1+2^k)^(n-k)*2^(k^2))}
for(n=0, 20, print1(a(n), ", "))
Showing 1-2 of 2 results.
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