|
| |
|
|
A158083
|
|
Possible answer to Marus du Satoy's puzzle sequence: a(n)=If[n <= 4, a[n - 1] + a[n - 2], If[n == 5, a[n - 1] + a[n - 2] + 9, a[n - 1] + 9]].
|
|
0
| |
|
|
2, 3, 5, 8, 13, 30, 39, 48, 57, 66, 75, 84, 93, 102, 111, 120, 129, 138, 147, 156, 165, 174, 183, 192, 201, 210, 219, 228, 237, 246, 255, 264, 273, 282, 291, 300, 309, 318, 327, 336, 345, 354, 363, 372, 381, 390, 399, 408, 417, 426, 435
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,1
|
|
|
COMMENTS
| The first four terms are Fibonacci
and the next is the fifth Fibonacci term plus 9 and the next is nine added to that...
There are probably other possible solutions,
like the fifth term as a sum of previous terms.
|
|
|
REFERENCES
| Marcus Du Sautoy, Symmetry: A Journey into the Patterns of Nature,Harper (March 11, 2008),page 96
|
|
|
FORMULA
| a(n)=If[n <= 4, a[n - 1] + a[n - 2], If[n == 5, a[n - 1] + a[n - 2] + 9, a[n - 1] + 9]].
G.f.: -(-2+x-x^2-x^3-2*x^4-12*x^5+8*x^6)/(-1+x)^2. a(n)=9n-15, n>4. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Apr 10 2009]
|
|
|
MATHEMATICA
| Clear[a, n];
a[0] = 2; a[1] = 3;
a[n_] := If[n <= 4, a[n - 1] + a[ n - 2], If[n == 5, a[n - 1] + a[n - 2] + 9, a[n - 1] + 9]];
Table[a[n], {n, 0, 50}]
|
|
|
CROSSREFS
| Sequence in context: A132915 A030036 A115212 * A042163 A014259 A001129
Adjacent sequences: A158080 A158081 A158082 * A158084 A158085 A158086
|
|
|
KEYWORD
| nonn,uned
|
|
|
AUTHOR
| Roger L. Bagula (rlbagulatftn(AT)yahoo.com), Mar 12 2009
|
| |
|
|
