A104796 Triangle read by rows: T(n,k) = (n+1-k)*Fibonacci(n+2-k), for n>=1, 1<=k<=n.

1, 4, 1, 9, 4, 1, 20, 9, 4, 1, 40, 20, 9, 4, 1, 78, 40, 20, 9, 4, 1, 147, 78, 40, 20, 9, 4, 1, 272, 147, 78, 40, 20, 9, 4, 1, 495, 272, 147, 78, 40, 20, 9, 4, 1, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 1584, 890, 495, 272, 147, 78, 40, 20, 9, 4, 1, 2796, 1584, 890, 495, 272

Offset

1,2

Comments

The first column is A023607 (without the leading zero).

Links

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Example

Rows 1,2,3,4,5,6 and columns 1,2,3,4,5,6 of the triangle are:
1;
4, 1;
9, 4, 1;
20, 9, 4, 1;
40, 20, 9, 4, 1;
78, 40, 20, 9, 4, 1;
...
Row 3 for example is 3*F(4), 2*F(3), 1*F(2) = 3*3, 2*2, 1*1 = 9, 4, 1.
Row 4 is 4*F(5), 3*F(4), 2*F(3), 1*F(2) = 4*5, 3*3, 2*2, 1*1 = 20, 9, 4, 1.
Reading the rows backwards gives an initial segment of the terms of A023607 (but without the initial zero).

Mathematica

Table[(n+1-k)Fibonacci[n+2-k], {n, 20}, {k, n}]//Flatten (* Harvey P. Dale, Sep 24 2020 *)
Module[{nn=20, c}, c=LinearRecurrence[{2, 1, -2, -1}, {1, 4, 9, 20}, nn]; Table[ Reverse[ Take[c, n]], {n, nn}]]//Flatten (* Harvey P. Dale, Sep 25 2020 *)

Crossrefs

Row sums are in A094584.

Cf. A023607, A104762, A104765, A000045.

Sequence in context: A085691 A055461 A324999 * A132020 A175643 A143864

Adjacent sequences: A104793 A104794 A104795 * A104797 A104798 A104799

Keyword

nonn,tabl

Author

Gary W. Adamson, Mar 26 2005

Extensions

Edited by Ralf Stephan, Apr 05 2009

Entry revised by N. J. A. Sloane, Sep 23 2020

Status

approved