A277611 Expansion of 1 / (1 - Sum_{k>=1} k^(k-2) * x^k ).

1, 1, 2, 6, 27, 180, 1678, 20388, 305331, 5423511, 111282445, 2587931469, 67239205808, 1929910531883, 60636166356164, 2069775112992573, 76268207153351225, 3017346008698599752, 127561003043924116908, 5738904556162964523209, 273764048456544759900846, 13802374108958236134168506, 733335098861491664742838394, 40953333749038944871704984923, 2398217239830108487402017089693, 146949291558772355319517897103987

Offset: 0

Author

Paul D. Hanna, Oct 23 2016

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 27*x^4 + 180*x^5 + 1678*x^6 + 20388*x^7 + 305331*x^8 + 5423511*x^9 + 111282445*x^10 + 2587931469*x^11 + 67239205808*x^12 +...
such that A(x) = 1 / (1 - Sum_{k>=1} k^(k-2) * x^k ).
The logarithm of the g.f. begins:
log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 83*x^4/4 + 746*x^5/5 + 8817*x^6/6 + 129340*x^7/7 + 2261195*x^8/8 + 45815431*x^9/9 + 1054594428*x^10/10 + 27167908186*x^11/11 + 774186515309*x^12/12 + 24174818590638*x^13/13 + 820795732075686*x^14/14 + 30104104733233598*x^15/15 +...
which equals the sum
log(A(x)) = (x + x^2 + 3*x^3 + 16*x^4 + 125*x^5 + 1296*x^6 +...) +
(x^2 + 2*x^3 + 7*x^4 + 38*x^5 + 291*x^6 + 2938*x^7 +...)/2 +
(x^3 + 3*x^4 + 12*x^5 + 67*x^6 + 507*x^7 + 5001*x^8 +...)/3 +
(x^4 + 4*x^5 + 18*x^6 + 104*x^7 + 783*x^8 + 7572*x^9 +...)/4 +
(x^5 + 5*x^6 + 25*x^7 + 150*x^8 + 1130*x^9 + 10751*x^10 +...)/5 +
(x^6 + 6*x^7 + 33*x^8 + 206*x^9 + 1560*x^10 + 14652*x^11 +...)/6 +
(x^7 + 7*x^8 + 42*x^9 + 273*x^10 + 2086*x^11 + 19404*x^12 +...)/7 +
... +
(x + 2^0*x^2 + 3^1*x^3 + 4^2*x^4 + 5^3*x^5 +...+ k^(k-2)*x^k +...)^n/n +
...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..388

Crossrefs

Cf. A088342, A277610.

Programs

Mathematica

CoefficientList[Series[1/(1 - Sum[k^(k-2) * x^k, {k, 1, 20}]), {x, 0, 20}], x] (* Vaclav Kotesovec, Nov 06 2016 *)

PARI

{a(n) = polcoeff( 1/(1 - sum(k=1, n+1, k^(k-2) * x^k +x*O(x^n)) ), n)}
for(n=0, 30, print1(a(n), ", "))

Formula

a(n) ~ n^(n-2) * (1 + 2*exp(-1)/n). - Vaclav Kotesovec, Nov 06 2016

a(0) = 1; a(n) = Sum_{k=1..n} k^(k-2) * a(n-k). - Ilya Gutkovskiy, Feb 07 2020