A192936 Constant term of the reduction by x^2 -> x + 1 of the polynomial p(n,x) = Product_{k=1..n} (x+k).

1, 1, 3, 13, 71, 463, 3497, 29975, 287265, 3042545, 35284315, 444617525, 6048575335, 88347242335, 1378930649745, 22903345844335, 403342641729665, 7506843094993825, 147226845692229875, 3034786640911840925, 65592491119118514375

Offset: 0

Author

Clark Kimberling, Jul 13 2011

Keywords

• nonn

Comments

For an introduction to reductions of polynomials by substitutions such as x^2 -> x + 1, see A192232.

Essentially the same as A192239. - R. J. Mathar, Aug 10 2011

Examples

			The first four polynomials p(n,x) and their reductions are as follows:
p(0,x) = 1
p(1,x) = (x+1) -> 1 + x
p(2,x) = (x+1)*(x+2) -> 3 + 4*x
p(3,x) = (x+1)*(x+2)*(x+3) -> 13 + 19*x
From these, read
A192936=(1,1,3,13,...) and A192239=(0,1,3,13,...)
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..100

Crossrefs

Cf. A192232, A192744, A192239.

Programs

GAP

List([0..30], n-> (-1)^n*Sum([0..n+2], k-> (-1)^(n-k)* Stirling1(n+2, k)*Fibonacci(k+1)) ); # G. C. Greubel, Jul 27 2019

Magma

[(-1)^n*(&+[StirlingFirst(n+2,k)*Fibonacci(k+1): k in [0..n+2]]): n in [0..30]]; // G. C. Greubel, Feb 16 2019

Mathematica

(* First program *)
q = x^2; s = x + 1; z = 26;
p[0, x]:= 1; p[n_, x_]:= (x+n)*p[n-1, x];
Table[Expand[p[n, x]], {n, 0, 7}]
reduce[{p1_, q_, s_, x_}]:= FixedPoint[(s PolynomialQuotient @@ #1 + PolynomialRemainder @@ #1 &)[{#1, q, x}] &, p1];
t:= Table[reduce[{p[n, x], q, s, x}], {n, 0, z}];
Table[Coefficient[Part[t, n], x, 0], {n, 1, z}] (* A192936 *)
Table[Coefficient[Part[t, n], x, 1], {n, 1, z}] (* A306183 *)
(* Second program *)
Table[(-1)^n*Sum[StirlingS1[n+2, k]*Fibonacci[k+1], {k, 0, n+2}], {n, 0, 30}] (* G. C. Greubel, Feb 16 2019 *)

PARI

{a(n) = (-1)^n*sum(k=0,n+2, stirling(n+2,k,1)*fibonacci(k+1))};
vector(30, n, n--; a(n)) \\ G. C. Greubel, Feb 16 2019

Sage

[sum((-1)^k*stirling_number1(n+2,k)*fibonacci(k+1) for k in (0..n+2)) for n in (0..30)] # G. C. Greubel, Feb 16 2019

Formula

a(n) = 1/10*(5-sqrt(5))*Gamma(n+3/2+1/2*sqrt(5))/Gamma(3/2+1/2*sqrt(5)) - 1/10*(5+sqrt(5))*Gamma(1/2*sqrt(5)-1/2)*sin(1/2*Pi*(5+sqrt(5))) *Gamma(n+3/2-1/2*sqrt(5))/Pi. - Vaclav Kotesovec, Oct 26 2012

a(n) = (-1)^n*Sum_{k=0..n+2} Stirling1(n+2,k)*Fibonacci(k+1). - G. C. Greubel, Feb 16 2019