A130095 Inverse Möbius transform of odd-indexed Fibonacci numbers.

1, 3, 6, 16, 35, 97, 234, 626, 1603, 4218, 10947, 28767, 75026, 196654, 514269, 1346895, 3524579, 9229159, 24157818, 63250217, 165580380, 433505386, 1134903171, 2971244450, 7778742084, 20365086102, 53316292776, 139584059112, 365435296163, 956722544582

Offset: 1

Author

Gary W. Adamson, May 06 2007

Keywords

• nonn
• easy

Comments

Original name was: A051731 * A007436.

Conjecture: a(n)/a(n-1) tends to phi^2.

Examples

			The divisors of 6 are 1, 2, 3 and 6. Hence
a(6) = Fibonacci(1) + Fibonacci(3) + Fibonacci(5) + Fibonacci(11) = 97.
		

Crossrefs

Cf. A001519, A007435, A051731.

Programs

Maple

#A130095
with(combinat): with(numtheory):
f := n -> fibonacci(2*n-1):
g := proc (n) local div; div := divisors(n):
add(f(div[j]), j = 1 .. tau(n)) end proc:
seq(g(n), n = 1 .. 30); # Peter Bala, Mar 26 2015

Formula

From Peter Bala, Mar 26 2015: (Start)

a(n) = sum {d | n} Fibonacci(2*d - 1).

O.g.f. Sum_{n >= 1} Fibonacci(2*n - 1)*x^n/(1 - x^n) = Sum_{n >= 1} x^n*(1 - x^n)/(1 - 3*x^n + x^(2*n)).

Sum_{n >= 1} a(n)*x^(2*n) = Sum_{n >= 1} x^n/( 1/(x^n - 1/x^n) - (x^n - 1/x^n) ).

For p prime, a(p) == k (mod p) where k = 3 if p == 2, 3 (mod 5), k = 2 if p == 1, 4 (mod 5) and k = 0 if p = 5. (End)

Extensions

Incorrect original name removed and terms a(11) - a(30) added by Peter Bala, Mar 26 2015