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A126617 a(n) = Sum_{i=0..n} (-2)^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n). 15

%I #53 Jun 27 2022 03:22:18

%S 1,-1,2,-3,7,-10,31,-21,204,307,2811,12100,74053,432211,2768858,

%T 18473441,129941283,956187814,7351696139,58897405759,490681196604,

%U 4242903803727,38014084430983,352341755256348,3373662303816313,33326335433122711,339232538387804530

%N a(n) = Sum_{i=0..n} (-2)^(n-i)*B(i)*binomial(n,i) where B(n) = Bell numbers A000110(n).

%C a(n) is positive starting at n=8. - _Karol A. Penson_ and _Olivier Gérard_, Oct 22 2007

%C Hankel transform is A000178. - _Paul Barry_, Apr 23 2009

%H Vincenzo Librandi, <a href="/A126617/b126617.txt">Table of n, a(n) for n = 0..200</a>

%F E.g.f.: exp(exp(x)-2*x-1). - _Vladeta Jovovic_, Aug 04 2007

%F a(n) = e^(-1) * Sum_{k>=0} (k-2)^n / k!. This is a Dobinski-type formula. - _Karol A. Penson_ and _Olivier Gérard_, Oct 22 2007

%F G.f.: 1/(1+x-x^2/(1-2x^2/(1-x-3x^2/(1-2x-4x^2/(1-3x-5x^2/(1-.... (continued fraction). - _Paul Barry_, Apr 23 2009

%F Let A be the upper Hessenberg matrix of order n defined by: A[i,i-1]=-1, A[i,j]=binomial(j-1,i-1), (i<=j), and A[i,j]=0 otherwise. Then, for n>=1, a(n)=(-1)^(n)charpoly(A,2). - _Milan Janjic_, Jul 08 2010

%F G.f.: -1/U(0) where U(k) = x*k - 1 - x - x^2*(k+1)/U(k+1); (continued fraction, 1-step). - _Sergei N. Gladkovskii_, Sep 28 2012

%F G.f.: 1/G(0) where G(k) = 1 + 2*x/(1 + 1/(1 - 2*x*(k+1)/G(k+1))); (continued fraction, 3-step). - _Sergei N. Gladkovskii_, Nov 23 2012

%F G.f.: G(0)/(1+3*x) where G(k) = 1 - 2*x*(k+1)/((2*k+1)*(2*x*k-2*x-1) - x*(2*k+1)*(2*k+3)*(2*x*k-2*x-1)/(x*(2*k+3) - 2*(k+1)*(2*x*k-x-1)/G(k+1) )); (recursively defined continued fraction). - _Sergei N. Gladkovskii_, Dec 19 2012

%F From _Sergei N. Gladkovskii_, Feb 13 2013: (Start)

%F Conjecture: if the e.g.f. is E(x)= exp( exp(x) -1 + p*x) then

%F g.f.: (x+1-p*x)/x/(G(0)-x) - 1/x where G(k) = 2*x + 1 - p*x - x*k + x*(x*k - x - 1 + p*x)/G(k+1); (continued fraction).

%F So, for this sequence (p=-2), g.f.: (3*x+1)/x/( G(0)-x ) - 1/x where G(k) = 4*x + 1 - x*k + x*(x*k - 3*x - 1)/G(k+1);

%F (End)

%F G.f.: 1/Q(0), where Q(k) = 1 + 2*x - x/(1 - x*(k+1)/Q(k+1)); (continued fraction). - _Sergei N. Gladkovskii_, Apr 22 2013

%F a(0) = 1; a(n) = -2 * a(n-1) + Sum_{k=0..n-1} binomial(n-1,k) * a(k). - _Ilya Gutkovskiy_, Jul 30 2021

%F a(n) ~ n^(n-2) * exp(n/LambertW(n) - n - 1) / (sqrt(1 + LambertW(n)) * LambertW(n)^(n-2)). - _Vaclav Kotesovec_, Jun 27 2022

%e G.f.: 1 - 1*x + 2*x^2 - 3*x^3 + 7*x^4 - 10*x^5 + 31*x^6 - 21*x^7 + 204*x^8 + 307*x^9 + 2811*x^10 + 12100*x^11 + 74053*x^12 + 432211*x^13 + ...

%t Table[ Sum[ (-2)^(n - k) Binomial[n, k] BellB[k], {k, 0, n}], {n, 0, 50}] (* _Karol A. Penson_ and _Olivier Gérard_, Oct 22 2007 *)

%Y Cf. A000110, A000296, A005493, A124311, A126390, A153732.

%K sign

%O 0,3

%A _N. J. A. Sloane_, Aug 04 2007

%E More terms from _Karol A. Penson_ and _Olivier Gérard_, Oct 22 2007

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Last modified March 28 05:02 EDT 2024. Contains 371235 sequences. (Running on oeis4.)