A213593 Stirling transform of the first kind of the Lucas numbers A000032.

2, 1, 2, -3, 10, -45, 250, -1645, 12490, -107415, 1031690, -10943955, 127058690, -1602400085, 21812913650, -318763741725, 4977247397650, -82695799908975, 1456703469048850, -27117356172328675, 531930264143933050

Offset: 0

Author

Jiaqiang Pan, Jun 15 2012

Keywords

• sign

Examples

			For n=4, a(4) = r*(r-1)*(r-2)*(r-3) + s*(s-1)*(s-2)*(s-3) = 10.
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..445
• Jiaqiang Pan, Matrix Decomposition of the Unified Generalized Stirling Numbers and Inversion of the Generalized Factorial Matrices, Journal of Integer Sequences 15 (2012) #12.6.6

Crossrefs

Cf. A000032 (Lucas numbers), the general-term formula of which is L(n) = r^n + s^n.

Cf. A008275, A323620.

Programs

GAP

Concatenation([2], List([1..25], n-> Sum([1..n], k-> (-1)^(n-k)* Stirling1(n, k)*Lucas(1,-1,k)[2] ))) # G. C. Greubel, Jul 06 2019

Magma

[2] cat [(&+[StirlingFirst(n,k)*Lucas(k): k in [1..n]]): n in [1..25]]; // G. C. Greubel, Jul 06 2019

Maple

A000032 := proc(n)
        combinat[fibonacci](n+1)+combinat[fibonacci](n-1) ;
end proc:
A213593 := proc(n)
        add(combinat[stirling1](n,i)*A000032(i),i=0..n) ;
end proc:
seq(A213593(n),n=0..20) ; # R. J. Mathar, Jun 26 2012

Mathematica

Expand@FunctionExpand@Table[Gamma[2 - GoldenRatio]/Gamma[2 - GoldenRatio - n] + Gamma[1 + GoldenRatio]/Gamma[1 - n + GoldenRatio], {n, 0, 20}] (* Vladimir Reshetnikov, Oct 20 2015 *)
Table[If[n==0, 2, Sum[StirlingS1[n, k]*LucasL[k], {k, n}]], {n, 0, 25}] (* G. C. Greubel, Jul 06 2019 *)

PARI

lucas(n) = fibonacci(n+1) + fibonacci(n-1);
vector(25, n, n--; if(n==0, 2, sum(k=1,n, stirling(n,k,1)*lucas(k)) )) \\ G. C. Greubel, Jul 06 2019

Sage

[2]+[sum((-1)^(n-k)*stirling_number1(n,k)*lucas_number2(k,1,-1) for k in (1..n)) for n in (1..25)] # G. C. Greubel, Jul 06 2019

Formula

a(0)=2; for n=1,2,3, ..., a(n) = r*(r-1)*(r-2)*...*(r-n+1) + s*(s-1)*(s-2)*...*(s-n+1), where r=(1+sqrt(5))/2 and s=(1-sqrt(5))/2.

From Vladimir Reshetnikov, Oct 20 2015: (Start)

Let phi=(1+sqrt(5))/2.

a(n) = Gamma(2-phi)/Gamma(2-phi-n)+Gamma(1+phi)/Gamma(1+phi-n).

Recurrence: a(0)=2, a(1)=1, a(n+2) = (1+n-n^2)*a(n) - 2*n*a(n+1).

E.g.f.: (1+(x+1)^sqrt(5))/(x+1)^(1/phi).

(End)

a(n) ~ (-1)^n * n! * n^((sqrt(5)-3)/2) / Gamma(2/(1+sqrt(5))). - Vaclav Kotesovec, Oct 21 2015

a(n) = Sum_{k=1..n} Stirling1(n,k)*Lucas(k). - G. C. Greubel, Jul 06 2019