|
| |
|
|
A022105
|
|
Fibonacci sequence beginning 1 15.
|
|
2
| |
|
|
1, 15, 16, 31, 47, 78, 125, 203, 328, 531, 859, 1390, 2249, 3639, 5888, 9527, 15415, 24942, 40357, 65299, 105656, 170955, 276611, 447566, 724177, 1171743, 1895920, 3067663, 4963583, 8031246, 12994829
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| a(n-1)=sum(P(15;n-1-k,k),k=0..ceiling((n-1)/2)), n>=1, with a(-1)=14. These are the SW-NE diagonals in P(15;n,k), the (15,1) Pascal triangle. Cf. A093645 for the (10,1) Pascal triangle. Observation by Paul Barry (pbarry(AT)wit.ie, Apr 29 2004. Proof via recursion relations and comparison of inputs.
|
|
|
LINKS
| Tanya Khovanova, Recursive Sequences
|
|
|
FORMULA
| a(n)= a(n-1)+a(n-2), n>=2, a(0)=1, a(1)=15. a(-1):=14.
G.f.: (1+14*x)/(1-x-x^2).
a(n) = A101220(14,0,n+1). - Ross La Haye (rlahaye(AT)new.rr.com), May 02 2006
|
|
|
MATHEMATICA
| a={}; b=1; c=15; AppendTo[a, b]; AppendTo[a, c]; Do[b=b+c; AppendTo[a, b]; c=b+c; AppendTo[a, c], {n, 1, 12, 1}]; a (Vladimir Orlovsky, Jul 23 2008)
|
|
|
CROSSREFS
| a(n) = A109754(14, n+1).
a(k) = A118654(4, k).
Sequence in context: A193566 A079832 A037971 * A041456 A041458 A041454
Adjacent sequences: A022102 A022103 A022104 * A022106 A022107 A022108
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
|
| |
|
|
