A050946 "Stirling-Bernoulli transform" of Fibonacci numbers.

0, 1, 1, 7, 13, 151, 421, 6847, 25453, 532231, 2473141, 63206287, 352444093, 10645162711, 69251478661, 2413453999327, 17943523153933, 708721089607591, 5927841361456981, 261679010699505967, 2431910546406522973, 118654880542567722871, 1212989379862721528101

Offset: 0

Author

N. J. A. Sloane, Jan 02 2000

Keywords

• nonn

Comments

From Paul Curtz, Oct 11 2013: (Start)

Differences table:

0, 1, 1, 7, 13, 151, 421, 6847, ...

1, 0, 6, 6, 138, 270, 6426, ...

-1, 6, 0, 132, 132, 6156, ...

7, -6, 132, 0, 6024, ...

-13, 138, -132, 6024, ...

151, -270, 6156, ...

-421, 6426, ...

6847, ... .

a(n) is an autosequence of first kind: the inverse binomial transform is the sequence signed, the main diagonal is A000004=0's.

The "Stirling-Bernoulli transform" applied to an autosequence of first kind is an autosequence of first kind.

Now consider the Akiyama-Tanigawa transform or algorithm applied to A000045(n):

0, 1, 1, 2, 3, 5, 8, ...

-1, 0, -3, -4, -10, -18, ... = -A006490

-1, 6, 3, 24, 40, ...

-7, 6, -63, -64, ...

-13, 138, 3, ...

-151, 270, ...

-421, ... .

Hence -a(n). The Akiyama-Tanigawa algorithm applied to an autosequence of first kind is an autosequence of first kind.

a(n+5) - a(n+1) = 150, 420, 6840, ... is divisible by 30.

For an autosequence of the second kind, the inverse binomial transform is the sequence signed with the main diagonal double of the first upper diagonal.

The Akiyama-Tanigawa algorithm applied to an autosequence leads to an autosequence of the same kind. Example: the A-T algorithm applied to the autosequence of second kind 1/n leads to the autosequence of the second kind A164555(n)/A027642(n).

Note that a2(n) = 2*a1(n+1) - a1(n) applied to the autosequence of the first kind a1(n) is a corresponding autosequence of the second kind. (End)

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..447
• C. J. Pita Ruiz V., Some Number Arrays Related to Pascal and Lucas Triangles, J. Int. Seq. 16 (2013) #13.5.7

Crossrefs

Cf. A000045, A051782, A105796, A163626.

Programs

Maple

with(combinat):
a:= n-> add((-1)^(k+1) *k! *stirling2(n+1, k+1)*fibonacci(k), k=0..n):
seq(a(n), n=0..30);  # Alois P. Heinz, May 17 2013

Mathematica

CoefficientList[Series[E^x*(1-E^x)/(1-3*E^x+E^(2*x)), {x, 0, 20}], x]* Range[0, 20]! (* Vaclav Kotesovec, Aug 13 2013 *)
t[0, k_] := Fibonacci[k]; t[n_, k_] := t[n, k] = (k+1)*(t[n-1, k] - t[n-1, k+1]); a[n_] := t[n, 0] // Abs; Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Oct 22 2013, after Paul Curtz *)

PARI

{a(n)=polcoeff(sum(m=0, n, fibonacci(m)*m!*x^m/prod(k=1, m, 1+k*x+x*O(x^n))), n)} /* Paul D. Hanna, Jul 20 2011 */

Formula

O.g.f.: Sum_{n>=1} Fibonacci(n) * n! * x^n / Product_{k=1..n} (1+k*x). - Paul D. Hanna, Jul 20 2011

A100872(n)=a(2*n) and A100868(n)=a(2*n-1).

From Paul Barry, Apr 20 2005: (Start)

E.g.f.: exp(x)*(1-exp(x))/(1-3*exp(x)+exp((2*x))).

a(n) = Sum_{k=0..n} (-1)^(n-k)*S2(n, k)*k!*Fibonacci(k). [corrected by Ilya Gutkovskiy, Apr 04 2019] (End)

a(n) ~ c * n! / (log((3+sqrt(5))/2))^(n+1), where c = 1/sqrt(5) if n is even and c = 1 if n is odd. - Vaclav Kotesovec, Aug 13 2013

a(n) = -1 * Sum_{k = 0..n} A163626(n,k)*A000045(k). - Philippe Deléham, May 29 2015