A113177 Fully additive with a(p) = Fibonacci(p); If, for p prime, p^(m_{n,p}) is the highest power of p dividing n with m>=0, then a(n) = Sum_{p prime} F(p)*(m_{n,p}), where F(p) = p-th Fibonacci number.

0, 1, 2, 2, 5, 3, 13, 3, 4, 6, 89, 4, 233, 14, 7, 4, 1597, 5, 4181, 7, 15, 90, 28657, 5, 10, 234, 6, 15, 514229, 8, 1346269, 5, 91, 1598, 18, 6, 24157817, 4182, 235, 8, 165580141, 16, 433494437, 91, 9, 28658, 2971215073, 6, 26, 11, 1599, 235, 53316291173, 7, 94

Offset: 1

Author

Leroy Quet, Oct 16 2005

Keywords

• nonn
• easy

Examples

			12 = 2^2 * 3^1, so a(12) = F(2)*2 + F(3)*1 = 2 + 2 = 4.
		

Links

• Danny Rorabaugh, Table of n, a(n) for n = 1..4000

Crossrefs

Cf. A000045, A113178.

Cf. A373586 (indices of even terms), A373587 (of odd terms), A374052 (of multiples of 3), A374206 (2-adic valuation), A374207 (3-adic valuation), A374208 (5-adic valuation), A374209 [A007895(a(n))], A374124 [a(n) mod 360].

Cf. A374106 [gcd(a(n), A328845(n))], A374112 [gcd(a(n), A276085(n))].

For other completely additive sequences see the cross-references in A001414.

Programs

Mathematica

b[t_]:=Fibonacci[First[t]]Last[t] a[n_]:=Apply[Plus, Map[b, FactorInteger[n]]] (* Esa Peuha, Oct 26 2005 *)

PARI

{ for(n=1,100,f=factor(n);s=0;for(i=1,matsize(f)[1],s+=fibonacci(f[i,1])*f[i,2]);print1(s,",")) } \\ Lambert Klasen, Oct 26 2005

Sage

[0]+[sum([fibonacci(x[0])*x[1] for x in factor(n)]) for n in range(2,56)] # Danny Rorabaugh, Apr 03 2015

Formula

Totally additive with a(p) = A000045(p).

Extensions

More terms from Esa Peuha (esa.peuha(AT)helsinki.fi) and Lambert Klasen (lambert.klasen(AT)gmx.net), Oct 26 2005

Prefixed the name with a more succinct form of the definition given in comments. - Antti Karttunen, Jul 08 2024