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A292500 G.f.: exp( Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n / n ).

%I A292500 #28 Oct 17 2020 04:54:57
%S A292500 1,1,4,18,122,1382,26992,967860,59207134,6539607238,1225903048760,
%T A292500 407719392472476,233686070341415140,233030334505100451484,
%U A292500 407716349332865096406960,1219594666823043463552070760,6484753389847998264537623184230,58288150472645787928029816422705798,936721167715228772497787011017302901192,25340260842241991639562678352357479545874188
%N A292500 G.f.: exp( Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n / n ).
%C A292500 A060187(n,k) = Sum_{j=1..k} (-1)^(k-j) * binomial(n,k-j) * (2*j-1)^(n-1).
%C A292500 Note that exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]  / (1-x)^(n+1) * x^n/n ) does not yield an integer series.
%C A292500 Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - _Vaclav Kotesovec_, Oct 17 2020
%H A292500 Paul D. Hanna, <a href="/A292500/b292500.txt">Table of n, a(n) for n = 0..130</a>
%F A292500 G.f.: exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n  / (1-x)^(n^2+n) * x^n/n ), where A060187 are the Eulerian numbers of type B.
%e A292500 G.f.: A(x) = 1 + x + 4*x^2 + 18*x^3 + 122*x^4 + 1382*x^5 + 26992*x^6 + 967860*x^7 + 59207134*x^8 + 6539607238*x^9 + 1225903048760*x^10 + 407719392472476*x^11 + 233686070341415140*x^12 + 233030334505100451484*x^13 + 407716349332865096406960*x^14 + 1219594666823043463552070760*x^15 +...
%e A292500 RELATED SERIES.
%e A292500 log(A(x)) = x + 7*x^2/2 + 43*x^3/3 + 399*x^4/4 + 6091*x^5/5 + 151255*x^6/6 + 6550307*x^7/7 + 465127199*x^8/8 + 58293976795*x^9/9 + 12191724780647*x^10/10 + 4471204259257363*x^11/11 + 2799295142330495151*x^12/12 + 3026340345288168023883*x^13/13 + 5704756586858875194533367*x^14/14 +...+ A292502(n)*x^n/n +...
%e A292500 The logarithm of g.f. A(x) equals the series:
%e A292500 log(A(x)) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 +...+ (2*k-1)^n*x^k +...)^n/n,
%e A292500 or,
%e A292500 log(A(x)) = (x + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 +...) +
%e A292500 (x + 3^2*x^2 + 5^2*x^3 + 7^2*x^4 + 9^2*x^5 +...)^2/2 +
%e A292500 (x + 3^3*x^2 + 5^3*x^3 + 7^3*x^4 + 9^3*x^5 +...)^3/3 +
%e A292500 (x + 3^4*x^2 + 5^4*x^3 + 7^4*x^4 + 9^4*x^5 +...)^4/4 + ...
%e A292500 This logarithmic series can be written using the Eulerian numbers of type B like so:
%e A292500 log(A(x)) = (x + x^2) / (1-x)^2 +
%e A292500 (x + 6*x^2 + x^3)^2 / (1-x)^6/2 +
%e A292500 (x + 23*x^2 + 23*x^3 + x^4)^3 / (1-x)^12/3 +
%e A292500 (x + 76*x^2 + 230*x^3 + 76*x^4 + x^5)^4 / (1-x)^20/4 +
%e A292500 (x + 237*x^2 + 1682*x^3 + 1682*x^4 + 237*x^5 + x^6)^5 / (1-x)^30/5 +
%e A292500 (x + 722*x^2 + 10543*x^3 + 23548*x^4 + 10543*x^5 + 722*x^6 + x^7)^6 / (1-x)^42/6 +
%e A292500 (x + 2179*x^2 + 60657*x^3 + 259723*x^4 + 259723*x^5 + 60657*x^6 + 2179*x^7 + x^8)^7 / (1-x)^56/7 +...+
%e A292500 [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2+n) * x^n/n +...
%t A292500 nmax = 20; CoefficientList[Series[Exp[Sum[2^(k^2) * x^k * LerchPhi[x, -k, 1/2]^k/k, {k, 1, nmax}]], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Oct 17 2020 *)
%o A292500 (PARI) {a(n) = polcoeff( exp( sum(m=1, n+1, sum(k=1, n+1, (2*k-1)^m * x^k +x*O(x^n))^m/m ) ), n)}
%o A292500 for(n=0, 30, print1(a(n), ", "))
%o A292500 (PARI) {A060187(n, k) = sum(j=1, k, (-1)^(k-j) * binomial(n, k-j) * (2*j-1)^(n-1))}
%o A292500 {a(n) = my(A=1, Oxn=x*O(x^n));
%o A292500 A = exp( sum(m=1,n+1, sum(k=0, m, A060187(m+1, k+1)*x^k)^m /(1-x +Oxn)^(m^2+m) * x^m/m ) );
%o A292500 polcoeff(A, n)}
%o A292500 for(n=0, 30, print1(a(n), ", "))
%Y A292500 Cf. A292501, A292502 (log), A156170, A276751, A276752, A156171, A155200, A060187.
%K A292500 nonn
%O A292500 0,3
%A A292500 _Paul D. Hanna_, Sep 17 2017