A111075 a(n) = F(n) * Sum_{k|n} 1/F(k), where F(k) is the k-th Fibonacci number.

1, 2, 3, 7, 6, 21, 14, 50, 52, 122, 90, 427, 234, 784, 1038, 2351, 1598, 6860, 4182, 17262, 17262, 35622, 28658, 139703, 90031, 243308, 300405, 766850, 514230, 2367006, 1346270, 5188658, 5326470, 11409346, 11782764, 44717548, 24157818

Offset

1,2

Comments

a(n) = a(n+1) for n = 20, but for no other n < 25000. - Klaus Brockhaus, Oct 11 2005

If k|n then F(k)|F(n). Therefore A111075(n) = F(n) * sum{k|n} 1/F(k) = sum{k|n} F(n)/F(k) is a sum of integers. - Max Alekseyev, Oct 22 2005

Links

Alois P. Heinz, Table of n, a(n) for n = 1..4785

Formula

G.f.: Sum_{n>=1} x^n/(1 - Lucas(n)*x^n + (-1)^n*x^(2*n)) where Lucas(n) = A000204(n). [Paul D. Hanna, Jan 09 2012]

Example

a(6) = F(6) sum{k|6} 1/F(k) = F(6) * (1/F(1) + 1/F(2) + 1/F(3) + 1/F(6)) = 8 * (1/1 + 1/1 + 1/2 + 1/8) = 21.

Maple

with(combinat): with(numtheory): a:=proc(n) local div: div:=divisors(n): fibonacci(n)*sum(1/fibonacci(div[j]), j=1..tau(n)) end: seq(a(n), n=1..40); # Emeric Deutsch, Oct 11 2005
# second Maple program:
a:= n-> (F-> F(n)*add(1/F(d), d=numtheory[divisors(n)))(
         combinat[fibonacci]):
seq(a(n), n=1..42);  # Alois P. Heinz, Aug 20 2019

Mathematica

f[n_] := Fibonacci[n]*Plus @@ (1/Fibonacci /@ Divisors[n]); Table[ f[n], {n, 37}] (* Robert G. Wilson v, Oct 11 2005 *)

Prog

(PARI) {for(n=1, 37, d=divisors(n); print1(fibonacci(n)*sum(j=1, length(d), 1/fibonacci(d[j])), ", "))}
(PARI) {a(n)=fibonacci(n) * sumdiv(n, d, 1/fibonacci(d))} /* Paul D. Hanna, Oct 11 2005 */
(PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=polcoeff(sum(m=1, n, x^m/(1 - Lucas(m)*x^m + (-1)^m*x^(2*m) +x*O(x^n))), n)} /* Paul D. Hanna, Oct 11 2005 */

Crossrefs

Cf. A000045, A007435, A111159, A000204 (Lucas), A203318.

Sequence in context: A187015 A245467 A070964 * A011372 A104955 A011161

Adjacent sequences: A111072 A111073 A111074 * A111076 A111077 A111078

Keyword

nonn

Author

Leroy Quet, Oct 10 2005

Extensions

More terms from Robert G. Wilson v, Emeric Deutsch, Paul D. Hanna and Klaus Brockhaus, Oct 11 2005

Status

approved