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%I #16 Oct 17 2020 07:39:10
%S 1,7,43,399,6091,151255,6550307,465127199,58293976795,12191724780647,
%T 4471204259257363,2799295142330495151,3026340345288168023883,
%U 5704756586858875194533367,18287793731664040419412785283,103736521111190203113027053903423,990788254951454647260121962687606203,16859931481746848392491523248553253264263,481447154976629475966161111465088882379644147
%N L.g.f.: Sum_{n>=1} [ Sum_{k>=1} (2*k-1)^n * x^k ]^n / n.
%C A060187(n,k) = Sum_{j=1..k} (-1)^(k-j) * binomial(n,k-j) * (2*j-1)^(n-1).
%C L.g.f. equals the logarithm of the g.f. of A292500.
%C Conjecture: a(n)^(1/n^2) tends to 3^(1/4). - _Vaclav Kotesovec_, Oct 17 2020
%H Paul D. Hanna, <a href="/A292502/b292502.txt">Table of n, a(n) for n = 1..130</a>
%F L.g.f.: 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 L.g.f: 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 + ...
%e The l.g.f. A(x) equals the series:
%e A(x) = Sum_{n>=1} (x + 3^n*x^2 + 5^n*x^3 + ... + (2*k-1)^n*x^k + ...)^n/n,
%e or,
%e A(x) = (x + 3*x^2 + 5*x^3 + 7*x^4 + 9*x^5 + ...) +
%e (x + 3^2*x^2 + 5^2*x^3 + 7^2*x^4 + 9^2*x^5 + ...)^2/2 +
%e (x + 3^3*x^2 + 5^3*x^3 + 7^3*x^4 + 9^3*x^5 + ...)^3/3 +
%e (x + 3^4*x^2 + 5^4*x^3 + 7^4*x^4 + 9^4*x^5 + ...)^4/4 + ...
%e This logarithmic series can be written using the Eulerian numbers of type B like so:
%e A(x) = (x + x^2) / (1-x)^2 +
%e (x + 6*x^2 + x^3)^2 / (1-x)^6/2 +
%e (x + 23*x^2 + 23*x^3 + x^4)^3 / (1-x)^12/3 +
%e (x + 76*x^2 + 230*x^3 + 76*x^4 + x^5)^4 / (1-x)^20/4 +
%e (x + 237*x^2 + 1682*x^3 + 1682*x^4 + 237*x^5 + x^6)^5 / (1-x)^30/5 +
%e (x + 722*x^2 + 10543*x^3 + 23548*x^4 + 10543*x^5 + 722*x^6 + x^7)^6 / (1-x)^42/6 +
%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)^56/7 + ... +
%e [ Sum_{k=0..n} A060187(n+1,k+1) * x^k ]^n / (1-x)^(n^2+n) * x^n/n + ...
%e Exponentiation yields the g.f. of A292500:
%e exp(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 + ... + A292500(n)*x^n + ...
%e which is an integer series.
%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 = 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 n*polcoeff(A, n)}
%o for(n=1, 30, print1(a(n), ", "))
%o (PARI) {a(n) = n*polcoeff( sum(m=1, n+1, sum(k=1, n, (2*k-1)^m * x^k +x*O(x^n))^m/m ), n)}
%o for(n=1, 30, print1(a(n), ", "))
%Y Cf. A292500, A276750, A060187.
%K nonn
%O 1,2
%A _Paul D. Hanna_, Sep 17 2017
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