|
Search: id:A007502
|
|
|
| A007502 |
|
Les Marvin sequence: a(n) = F(n)+(n-1)F(n-1), F() = Fibonacci numbers. (Formerly M1170)
|
|
+0 6
|
|
| 1, 2, 4, 9, 17, 33, 61, 112, 202, 361, 639, 1123, 1961, 3406, 5888, 10137, 17389, 29733, 50693, 86204, 146246, 247577, 418299, 705479, 1187857, 1997018, 3352636, 5621097, 9412937, 15744681, 26307469, 43912648
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
Denominators of convergents of the continued fraction with the n partial quotients: [1;1,1,...(n-1 1's)...,1,n], starting with [1], [1;2], [1;1,3], [1;1,1,4], ... Numerators are A088209(n-1). - Paul D. Hanna (pauldhanna(AT)juno.com), Sep 23 2003
|
|
REFERENCES
|
A problem mentioned on page 230 of J. Rec. Math., Vol. 10, 1977.
|
|
LINKS
|
T. D. Noe, Table of n, a(n) for n=1..500
|
|
FORMULA
|
G.f.: (1-x^2+x^3)/(1-x-x^2)^2. - Paul D. Hanna (pauldhanna(AT)juno.com), Sep 23 2003
|
|
EXAMPLE
|
a(7)=F(7)+6*F(6)=13+6*8=61.
|
|
MATHEMATICA
|
F[0] = 0; F[1] = 1; F[n_] := F[n] = F[n - 1] + F[n - 2]; a[n_] := F[n] + (n - 1)F[n - 1]; Table[a[n], {n, 1, 40}]
|
|
CROSSREFS
|
Cf. A088209.
a(n+1) = A109754(n, n+1) = A101220(n, 0, n+1).
Adjacent sequences: A007499 A007500 A007501 this_sequence A007503 A007504 A007505
Sequence in context: A131095 A136379 A065026 this_sequence A088039 A115451 A077931
|
|
KEYWORD
|
nonn,easy,nice
|
|
AUTHOR
|
njas, Robert G. Wilson v (rgwv(AT)rgwv.com)
|
|
|
Search completed in 0.002 seconds
|