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G.f.: A(x) = exp( Sum_{n>=1} 2^(n^2) * x^n/n ), a power series in x with integer coefficients.
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%I #32 Dec 27 2019 09:57:58

%S 1,2,10,188,16774,6745436,11466849412,80444398636280,

%T 2306003967992402758,268654794629082985019564,

%U 126765597346260977505891041836,241678070948246232010898235031930952,1858395916567787793818891330877931472153500,57560683587056536617649234722821582390470430186648

%N G.f.: A(x) = exp( Sum_{n>=1} 2^(n^2) * x^n/n ), a power series in x with integer coefficients.

%C More generally, it appears that for m integer, exp( Sum_{n >= 1} m^(n^2) * x^n/n ) is a power series in x with integer coefficients.

%C This is correct: if b(n) = m^(n^2) then by the little Fermat theorem the Gauss congruences hold: b(n*p^k) == b(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. Then apply Stanley, Ch. 5, Ex. 5.2(a). - _Peter Bala_, Dec 25 2019

%C Conjecture: highest exponent of 2 dividing a(n) = A000120(n) = number of 1's in binary expansion of n, so that a(n)/2^A000120(n) is odd for n >= 0. - _Paul D. Hanna_, Sep 01 2009

%D R. P. Stanley. Enumerative combinatorics, Vol. 2. Volume 62 of Cambridge Studies in Advanced Mathematics, Cambridge University Press, Cambridge, 1999.

%H Muniru A Asiru, <a href="/A155200/b155200.txt">Table of n, a(n) for n = 0..56</a>

%H Sawian Jaidee, Patrick Moss, Tom Ward, <a href="https://arxiv.org/abs/1809.09199">Time-changes preserving zeta functions</a>, arXiv:1809.09199 [math.DS], 2018.

%F Equals column 0 of triangle A155810.

%F G.f. satisfies: 2*A(x)*A(4x) + 8*x*A(x)*A'(4x) - A'(x)*A(4x) = 0. - _Paul D. Hanna_, Feb 24 2009

%F From _Paul D. Hanna_, Mar 11 2009: (Start)

%F The differential equation implies recurrence:

%F n*a(n) = 2*a(n-1) + sum(k = 1, n - 1, 4^k*a(k)*[2*(k+1)*a(n-1-k) - (n-k)*a(n-k)] for n > 0, with a(0) = 1.

%F G.f. A(x) generates A156631:

%F A156631(n) = [x^n] A(x)^(2^n) for n >= 0, where the g.f. of A156631 = Sum_{n >= 0} [Sum_{k >= 1} (2^n*2^k*x)^k/k]^n/n!. (End)

%F a(n) = (1/n)*Sum_{k = 1..n} 2^(k^2)*a(n-k), a(0) = 1. - _Vladeta Jovovic_, Feb 04 2009

%F Euler transform of A159034. - _Vladeta Jovovic_, Apr 02 2009

%F a(n) = B_n( 0!*2^(1^2), 1!*2^(2^2), 2!*2^(3^2), ..., (n-1)!*2^(n^2) ) / n!, where B_n() is the complete Bell polynomial. - _Max Alekseyev_, Oct 10 2014

%F a(n) ~ 2^(n^2) / n. - _Vaclav Kotesovec_, Oct 09 2019

%e G.f.: A(x) = 1 + 2*x + 10*x^2 + 188*x^3 + 16774*x^4 + 6745436*x^5 +...

%e log(A(x)) = 2*x + 2^4*x^2/2 + 2^9*x^3/3 + 2^16*x^4/4 + 2^25*x^5/5 +...

%p seq(coeff(series(exp(add(2^(k^2)*x^k/k,k=1..n)),x,n+1), x, n), n = 0 .. 15); # _Muniru A Asiru_, Dec 19 2018

%t nmax = 14; Exp[Sum[2^(n^2) x^n/n, {n, 1, nmax}]] + O[x]^nmax // CoefficientList[#, x]& (* _Jean-François Alcover_, Feb 14 2019 *)

%o (PARI) {a(n)=polcoeff(exp(sum(m=1,n+1,2^(m^2)*x^m/m)+x*O(x^n)),n)}

%o (PARI) {a(n)=if(n==0,1,(1/n)*(2*a(n-1) + sum(k=1,n-1,4^k*a(k)*(2*(k+1)*a(n-1-k) - (n-k)*a(n-k)))))} \\ _Paul D. Hanna_, Mar 11 2009

%o (PARI) {a(n)=if(n==0,1,(1/n)*sum(k=1,n,2^(k^2)*a(n-k)))} \\ _Paul D. Hanna_, Sep 01 2009

%Y Cf. A155201, A155202, A155810 (triangle), variants: A155203, A155207.

%Y Cf. A000120, A156631, A159034.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 04 2009