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A136516
a(n) = (2^n+1)^n.
16
1, 3, 25, 729, 83521, 39135393, 75418890625, 594467302491009, 19031147999601100801, 2460686496619787545743873, 1280084544196357822418212890625, 2672769719437237714909813214827010049, 22366167213460480200139104627873703828439041
OFFSET
0,2
COMMENTS
More generally, Sum_{n>=0} m^n * q^(n^2) * exp(b*q^n*x) * x^n / n! = Sum_{n>=0} (m*q^n + b)^n * x^n / n! for all q, m, b.
Main diagonal of A264871. - Omar E. Pol, Nov 27 2015
LINKS
FORMULA
E.g.f.: A(x) = Sum_{n>=0} 2^(n^2) * exp(2^n*x) * x^n/n!.
O.g.f.: Sum_{n>=0} 2^(n^2)*x^n/(1 - 2^n*x)^(n+1) = Sum_{n>=0} (2^n+1)^n*x^n. [Paul D. Hanna, Sep 15 2009]
a(n) = 2^(n^2) + n 2^(n^2-n) + O(n^2 2^(n^2-2n)). - Robert Israel, Nov 27 2015
EXAMPLE
A(x) = 1 + 3x + 5^2*x^2/2! + 9^3*x^3/3! + 17^4*x^4/4! +... + (2^n+1)^n*x^n/n! +...
A(x) = exp(x) + 2*exp(2x) + 2^4*exp(4x)*x^2/2! + 2^9*exp(8x)*x^3/3! +...+ 2^(n^2)*exp(2^n*x)*x^n/n! +...
This is a special case of the more general statement:
Sum_{n>=0} m^n * F(q^n*x)^b * log( F(q^n*x) )^n / n! = Sum_{n>=0} x^n * [y^n] F(y)^(m*q^n + b) where F(x) = exp(x), q=2, m=1, b=1.
MAPLE
seq((2^n+1)^n, n=0..30); # Robert Israel, Nov 27 2015
MATHEMATICA
Table[(2^n+1)^n, {n, 0, 16}] (* Vladimir Joseph Stephan Orlovsky, Feb 14 2011*)
PROG
(PARI) a(n)=polcoeff(sum(k=0, n, 2^(k^2)*exp(2^k*x)*x^k/k!), n)
(PARI) {a(n)=polcoeff(sum(k=0, n, 2^(k^2)*x^k/(1-2^k*x +x*O(x^n))^(k+1)), n)} \\ Paul D. Hanna, Sep 15 2009
(Magma) [(2^n+1)^n: n in [0..45]]; // Vincenzo Librandi, Apr 21 2011
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Jan 02 2008
STATUS
approved