A322382 a(n) = p*a(n/p) + 1, where p is the smallest prime divisor of n; a(1)=0.
0, 1, 1, 3, 1, 3, 1, 7, 4, 3, 1, 7, 1, 3, 4, 15, 1, 9, 1, 7, 4, 3, 1, 15, 6, 3, 13, 7, 1, 9, 1, 31, 4, 3, 6, 19, 1, 3, 4, 15, 1, 9, 1, 7, 13, 3, 1, 31, 8, 13, 4, 7, 1, 27, 6, 15, 4, 3, 1, 19, 1, 3, 13, 63, 6, 9, 1, 7, 4, 13, 1, 39, 1, 3, 19, 7, 8, 9, 1, 31, 40, 3, 1, 19, 6
Offset: 1
Keywords
Examples
For any prime p, a(p) = p*a(p/p)+1 = p*a(1)+1 = 1, because a(1) = 0. For n = 6, the least prime divisor is 2, so a(6) = 2*a(6/2)+1 = 2*a(3)+1 = 3. Using the equivalent definition we get w(6) = 1/3 + 1/6 = 1/2, so a(6) = 6*w(6) = 6*1/2 = 3. For n=32, a(32) = a(2^5) = 2^5 - 1 = 32 - 1 = 31.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16383
- Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Programs
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Mathematica
a[n_] := #*a[n/#] + 1 &[FactorInteger[n][[1, 1]] ]; a[1] = 0; Array[a, 1000] (* Michael De Vlieger, Jun 20 2025 *)
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PARI
a(n) = if (n==1, 0, my(p = vecmin(factor(n)[,1])); p*a(n/p)+1); \\ Michel Marcus, Jan 25 2019
Formula
From Antti Karttunen, Feb 28 2019 , Mar 04 2019: (Start)
If n is of the form p^k, k >= 1, then a(n) = A000203(A003557(n)). [Based on author's comments above] (End)
a(n) = Sum_{k=1..bigomega(n)} F^k(n), where F^k(n) is the k-th iterate of F(n) = A052126(n). - Ridouane Oudra, Aug 17 2024
Extensions
More terms from Michel Marcus, Jan 25 2019
Comments