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G.f.: exp( Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n * (1-x)^n / n ).
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%I #17 Oct 17 2020 07:38:40

%S 1,1,3,13,91,1119,23235,879361,55447631,6274018595,1192773105789,

%T 400761393446831,231147252957096671,231434829013884972151,

%U 406000810484101907916927,1216355994930424625967455929,6474418584620388915674215696687,58229572245447428847208518694227279,936163501254507409972001699357677028097,25330794407893091120626418701416294765820223,1224635875718403110628189182372406488768960029317

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

%C A060187(n,k) = Sum_{j=1..k} (-1)^(k-j) * binomial(n,k-j) * (2*j-1)^(n-1).

%C Note that exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ] / (1-x)^n * x^n/ n ) does not yield an integer series.

%C Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - _Vaclav Kotesovec_, Oct 17 2020

%H Paul D. Hanna, <a href="/A292501/b292501.txt">Table of n, a(n) for n = 0..130</a>

%F G.f.: exp( Sum_{n>=1} [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2) * x^n/n ), where A060187 are the Eulerian numbers of type B.

%e G.f.: A(x) = 1 + x + 3*x^2 + 13*x^3 + 91*x^4 + 1119*x^5 + 23235*x^6 + 879361*x^7 + 55447631*x^8 + 6274018595*x^9 + 1192773105789*x^10 + 400761393446831*x^11 + 231147252957096671*x^12 + 231434829013884972151*x^13 + 406000810484101907916927*x^14 + 1216355994930424625967455929*x^15 +...

%e RELATED SERIES.

%e log(A(x)) = x + 5*x^2/2 + 31*x^3/3 + 305*x^4/4 + 5041*x^5/5 + 131477*x^6/6 + 5973311*x^7/7 + 436089793*x^8/8 + 55949083681*x^9/9 + 11863792842885*x^10/10 + 4395111080551775*x^11/11 + 2768928615166879025*x^12/12 + 3005637312940054635857*x^13/13 + 5680764740993004611483477*x^14/14 + 18239242940612856315412499071*x^15/15 +...

%e The logarithm of g.f. A(x) equals the series:

%e log(A(x)) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 +...+ (2*k-1)^n*x^k +...)^n * (1-x)^n/n,

%e or,

%e log(A(x)) = (x + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 +...) * (1-x) +

%e (x + 3^2*x^2 + 5^2*x^3 + 7^2*x^4 + 9^2*x^5 +...)^2 * (1-x)^2/2 +

%e (x + 3^3*x^2 + 5^3*x^3 + 7^3*x^4 + 9^3*x^5 +...)^3 * (1-x)^3/3 +

%e (x + 3^4*x^2 + 5^4*x^3 + 7^4*x^4 + 9^4*x^5 +...)^4 * (1-x)^4/4 + ...

%e This logarithmic series can be written using the Eulerian numbers of type B like so:

%e log(A(x)) = (x + x^2) / (1-x) +

%e (x + 6*x^2 + x^3)^2 / (1-x)^4/2 +

%e (x + 23*x^2 + 23*x^3 + x^4)^3 / (1-x)^9/3 +

%e (x + 76*x^2 + 230*x^3 + 76*x^4 + x^5)^4 / (1-x)^16/4 +

%e (x + 237*x^2 + 1682*x^3 + 1682*x^4 + 237*x^5 + x^6)^5 / (1-x)^25/5 +

%e (x + 722*x^2 + 10543*x^3 + 23548*x^4 + 10543*x^5 + 722*x^6 + x^7)^6 / (1-x)^36/6 +

%e (x + 2179*x^2 + 60657*x^3 + 259723*x^4 + 259723*x^5 + 60657*x^6 + 2179*x^7 + x^8)^7 / (1-x)^49/7 +...+

%e [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2) * x^n/n +...

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

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

%o (PARI) {A060187(n, k) = sum(j=1, k, (-1)^(k-j) * binomial(n, k-j) * (2*j-1)^(n-1))}

%o {a(n) = my(A=1, Oxn=x*O(x^n));

%o A = exp( sum(m=1,n+1, sum(k=0, m, A060187(m+1, k+1)*x^k)^m /(1-x +Oxn)^(m^2) * x^m/m ) ); polcoeff(A, n)}

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

%Y Cf. A292500, A156170, A276751, A276752, A156171, A155200, A060187.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Sep 17 2017