|
| |
|
|
A118240
|
|
The part of n in base phi left of the decimal, using a least-greedy algorithm representation.
|
|
1
| |
|
|
0, 0, 1, 10, 11, 101, 111, 1010, 1011, 1101, 1110, 1111, 10101, 10111, 11010, 11011, 11101, 11111, 101010, 101011, 101101, 101110, 101111, 110101, 110111, 111010, 111011, 111101, 111110, 111111, 1010101, 1010111, 1011010, 1011011, 1011101
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,4
|
|
|
COMMENTS
| Uses least-greedy algorithm (start with largest possible power of phi, writing a 1 only when required, then work downward)
constant (float): phi=(sqrt(5)+1)/2; variable (float): lphi=phi^floor[log(n)/log(phi)]; variable (float): rem=n; variable (integer): count=0; loop: while lphi>1 (count=count*10; lphi=lphi/phi; if(rem > lphi*phi) { rem=rem-lphi; count++;}}
|
|
|
LINKS
| R. Knott, Phigits and the Base Phi representation.
|
|
|
EXAMPLE
| 6 = 111.01101010... in base phi using the least-greedy
algorithm. The part to the left of the decimal is a(6) = 111.
|
|
|
CROSSREFS
| Cf. A055778, A104605, A118241, A105424.
Sequence in context: A205598 A037090 A171676 * A157845 A086084 A206073
Adjacent sequences: A118237 A118238 A118239 * A118241 A118242 A118243
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Graeme McRae (g_m(AT)mcraefamily.com), Apr 17, 2006
|
| |
|
|