A156170 G.f.: exp( Sum_{n>=1} [Sum_{k>=1} k^n*x^k]^n/n ), a power series in x with integer coefficients.

1, 1, 3, 10, 41, 219, 1602, 16635, 247171, 5242108, 157390565, 6663089873, 396778864166, 33200932308437, 3906922702271961, 646161881511137940, 150482521507292513413, 49318093291540113084965, 22790150225552744270503692, 14843990673285561887923674163, 13646527810852572644275538963207, 17710656073227095563348293151121448

Offset: 0

Author

Paul D. Hanna, Feb 05 2009

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 10*x^3 + 41*x^4 + 219*x^5 + 1602*x^6 +...
log(A(x)) = x + 5*x^2/2 + 22*x^3/3 + 117*x^4/4 + 821*x^5/5 + 7796*x^6/6 + 1810093*x^7/7 + 44561794*x^8/8 +...+ A276750(n)*x^n/n +...
The logarithm of g.f. A(x) equals the series:
log(A(x)) = Sum_{n>=1} (x + 2^n*x^2 + 3^n*x^3 +...+ k^n*x^k +...)^n/n,
or,
log(A(x)) = (x + 2*x^2 + 3*x^3 + 4*x^4 + 5*x^5 +...) +
(x + 2^2*x^2 + 3^2*x^3 + 4^2*x^4 + 5^2*x^5 +...)^2/2 +
(x + 2^3*x^2 + 3^3*x^3 + 4^3*x^4 + 5^3*x^5 +...)^3/3 +
(x + 2^4*x^2 + 3^4*x^3 + 4^4*x^4 + 5^4*x^5 +...)^4/4 + ...
This logarithmic series can be written using the Eulerian numbers like so:
log(A(x)) = x/(1-x)^2 + (x + x^2)^2/(1-x)^6/2 + (x + 4*x^2 + x^3)^3/(1-x)^12/3 + (x + 11*x^2 + 11*x^3 + x^4)^4/(1-x)^20/4 + (x + 26*x^2 + 66*x^3 + 26*x^4 + x^5)^5/(1-x)^30/5 + (x + 57*x^2 + 302*x^3 + 302*x^4 + 57*x^5 + x^6)^6/(1-x)^42/6 +...+ [ Sum_{k=1..n} A008292(n,k) * x^k ]^n / (1-x)^(n^2+n)/n +...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..150

Crossrefs

Cf. A276750, A276751, A276752, A156171, A155200, A008292, A292500.

Programs

PARI

{a(n) = polcoeff( exp( sum(m=1,n, sum(k=1,n, k^m*x^k +x*O(x^n))^m/m ) ),n)}
for(n=0,30,print1(a(n),", "))

PARI

{A008292(n,k) = sum(j=0,k, (-1)^j * (k-j)^n * binomial(n+1,j))}
{a(n) = my(A=1, Oxn=x*O(x^n)); A = exp( sum(m=1,n+1, sum(k=1,m, A008292(m,k)*x^k/(1-x +Oxn)^(m+1) )^m / m ) ); polcoeff(A,n)}
for(n=0,30,print1(a(n),", "))

Formula

G.f.: exp( Sum_{n>=1} [ Sum_{k=1..n} A008292(n,k) * x^k / (1-x)^(n+1) ]^n / n ), where A008292 are the Eulerian numbers. - Paul D. Hanna, Sep 13 2016

Conjecture: log(a(n)) ~ n^2 * log(2)/4. - Vaclav Kotesovec, Sep 02 2017