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