OFFSET
1,4
REFERENCES
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. 2nd Edition. Addison-Wesley, Reading, MA, 1994, p. 294.
LINKS
Freddy Barrera, Antidiagonals n = 1..50, flattened
FORMULA
For prime p, the formula holds: Fibonacci(k*p) = Fibonacci(p) * Sum_{i=0..floor((k-1)/2)} C(k-i-1, i)*(-1)^(i*p+i)*Lucas(p)^(k-2i-1).
A(n, k) = F((n-1)*k)*F(k+1) + F((n-1)*k-1)*F(k), where F(n) = A000045(n). - Freddy Barrera, Jun 24 2019
EXAMPLE
1, 1, 2, 3, 5, ...
1, 3, 8, 21, 55, ...
2, 8, 34, 144, 610, ...
3, 21, 144, 987, 6765, ...
5, 55, 610, 6765, 75025, ...
MATHEMATICA
Table[Fibonacci[k*(n-k+1)], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 10 2017 *)
PROG
(Sage)
F = fibonacci # A000045
def A(n, k):
return F((n-1)*k)*F(k+1) + F((n-1)*k - 1)*F(k)
[A(n, k) for d in (1..10) for n, k in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jun 24 2019
(Magma) /* As triangle */ [[Fibonacci(k*(n-k+1)): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Jul 04 2019
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Ralf Stephan, Jan 06 2005
STATUS
approved