A063956 Sum of unitary prime divisors of n.
0, 2, 3, 0, 5, 5, 7, 0, 0, 7, 11, 3, 13, 9, 8, 0, 17, 2, 19, 5, 10, 13, 23, 3, 0, 15, 0, 7, 29, 10, 31, 0, 14, 19, 12, 0, 37, 21, 16, 5, 41, 12, 43, 11, 5, 25, 47, 3, 0, 2, 20, 13, 53, 2, 16, 7, 22, 31, 59, 8, 61, 33, 7, 0, 18, 16, 67, 17, 26, 14, 71, 0, 73, 39, 3, 19, 18, 18, 79, 5, 0
Offset: 1
Examples
The prime divisors of 420 = 2^2 * 3 * 5 * 7. Among them, those that have exponent 1 (i.e., unitary prime divisors) are {3, 5, 7}, so a(420) = 3 + 5 + 7 = 15.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
Programs
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Mathematica
Table[DivisorSum[n, # &, And[PrimeQ@ #, GCD[#, n/#] == 1] &], {n, 81}] (* Michael De Vlieger, Feb 17 2019 *) f[p_, e_] := If[e == 1, p, 0]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 24 2024 *) Join[{0},Table[Total[Select[FactorInteger[n],#[[2]]==1&][[;;,1]]],{n,2,100}]] (* Harvey P. Dale, Jan 26 2025 *)
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PARI
{ for (n=1, 1000, f=factor(n)~; a=0; for (i=1, length(f), if (f[2, i]==1, a+=f[1, i])); write("b063956.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 04 2009
Formula
a(n*m) = a(n) + a(m) - a(gcd(n^2, m)) - a(gcd(n, m^2)) for all n and m > 0 (conjecture). - Velin Yanev, Feb 17 2019
From Amiram Eldar, Jul 24 2024: (Start)
Additive with a(p^e) = p is e = 1, and 0 otherwise. (End)
Extensions
Example clarified by Harvey P. Dale, Jan 26 2025