A274262 Number of positive integers possessing exactly n Fibonacci representations (A000121).

1, 2, 4, 6, 8, 12, 12, 18, 20, 24, 20, 44, 24, 36, 48, 54, 32, 76, 36, 88, 72, 60, 44, 156, 72, 72, 100, 132, 56, 208, 60, 162, 120, 96, 144, 316, 72, 108, 144, 312, 80, 312, 84, 220, 304, 132, 92, 540, 156, 280, 192, 264, 104, 460, 240, 468, 216, 168, 116, 116, 120, 180, 456, 486, 288, 520, 132, 352, 264, 624, 140

Offset: 1

Author

Steven Finch, Jun 16 2016

Keywords

• nonn

Examples

			Let phi denote the Euler totient.
The integer p^2*q has 8 multiplicative compositions:
  (p^2*q), p^2*q, q*p^2, p*(p*q), (p*q)*p, q*p*p, p*q*p, p*p*q
from which
  a(p^2*q) = 2*(3*phi(p^2)*phi(q) + 5*phi(p)^2*phi(q))
follows immediately.
		

Links

• Zai-Qiao Bai and Steven R. Finch, Fibonacci and Lucas Representations, Fibonacci Quart. 54 (2016), no. 4, 319-326.

Crossrefs

Cf. A000121, A067595.

Formula

Let p, q, r be distinct primes and k be a positive integer.

If n = p^k then a(n) = 2*(p-1)*(2*p-1)^(k-1).

If n = p*q then a(n) = 6*(p-1)*(q-1).

If n = p^2*q then a(n) = 2*(p-1)*(8*p-5)*(q-1).

If n = p^3*q then a(n) = 2*(p-1)*(2*p-1)*(10*p-7)*(q-1).

If n = p^4*q then a(n) = 6*(p-1)*(2*p-1)^2*(4*p-3)*(q-1).

If n = p^2*q^2 then a(n) = 2*(p-1)*(q-1)*(26*p*q-18*p-18*q+13).

If n = p*q*r then a(n) = 26*(p-1)*(q-1)*(r-1).