A125127 Array L(k,n) read by antidiagonals: k-step Lucas numbers.

1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 1, 3, 7, 7, 1, 1, 3, 7, 11, 11, 1, 1, 3, 7, 15, 21, 18, 1, 1, 3, 7, 15, 26, 39, 29, 1, 1, 3, 7, 15, 31, 51, 71, 47, 1, 1, 3, 7, 15, 31, 57, 99, 131, 76, 1, 1, 3, 7, 15, 31, 63, 113, 191, 241, 123, 1

Offset: 1

Author

Jonathan Vos Post, Nov 21 2006

Keywords

• easy
• nonn
• tabl

Examples

			Table begins:
1 | 1  1  1   1   1   1    1    1    1    1
2 | 1  3  4   7  11  18   29   47   76  123
3 | 1  3  7  11  21  39   71  131  241  443
4 | 1  3  7  15  26  51   99  191  367  708
5 | 1  3  7  15  31  57  113  223  439  863
6 | 1  3  7  15  31  63  120  239  475  943
7 | 1  3  7  15  31  63  127  247  493  983
8 | 1  3  7  15  31  63  127  255  502 1003
9 | 1  3  7  15  31  63  127  255  511 1013
		

Links

• Freddy Barrera, Table of n, a(n) for n = 1..5050
• C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
• Eric Weisstein's World of Mathematics, Lucas n-Step Number

Crossrefs

n-step Lucas number analog of A092921 Array F(k, n) read by antidiagonals: k-generalized Fibonacci numbers (and see related A048887, A048888). L(1, n) = "1-step Lucas numbers" = A000012. L(2, n) = 2-step Lucas numbers = A000204. L(3, n) = 3-step Lucas numbers = A001644. L(4, n) = 4-step Lucas numbers = A001648 Tetranacci numbers A073817 without the leading term 4. L(5, n) = 5-step Lucas numbers = A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15. L(6, n) = 6-step Lucas numbers = A074584 Esanacci ("6-anacci") numbers. L(7, n) = 7-step Lucas numbers = A104621 Heptanacci-Lucas numbers. L(8, n) = 8-step Lucas numbers = A105754. L(9, n) = 9-step Lucas numbers = A105755. See A000295, A125129 for comments on partial sums of diagonals.

Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125129.

Programs

Sage

def L(k, n):
    if n < 0:
        return -1
    a = [-1]*(k-1) + [k] # [-1, -1, ..., -1, k]
    for i in range(1, n+1):
        a[:] = a[1:] + [sum(a)]
    return a[-1]
[L(k, n) for d in (1..12) for k, n in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jan 10 2019

Formula

L(k,n) = L(k,n-1) + L(k,n-2) + ... + L(k,n-k); L(k,n) = -1 for n < 0, and L(k,0) = k.

G.f. for row k: x*(dB(k,x)/dx)/(1-B(k,x)), where B(k,x) = x + x^2 + ... + x^k. - Petros Hadjicostas, Jan 24 2019

Extensions

Corrected by Freddy Barrera, Jan 10 2019