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

0,4

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

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

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 *)

Prog

(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 */

Crossrefs

Cf. A000045, A051782, A105796, A163626.

Sequence in context: A219703 A198947 A342717 * A178956 A319612 A247946

Adjacent sequences: A050943 A050944 A050945 * A050947 A050948 A050949

Keyword

nonn

Author

N. J. A. Sloane, Jan 02 2000

Status

approved