A380653 Number of positive integers less than or equal to n that have the same sum of prime factors (with repetition) as n.
1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 3, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 2, 1, 4, 2, 1, 3, 1, 4, 1, 2, 1, 5, 1, 1, 1, 3, 1, 2, 1, 2, 4, 1, 1, 5, 2, 3, 1, 2, 1, 6, 2, 3, 1, 2, 1, 4, 1, 1, 4, 5, 1, 3, 1, 2, 1, 3, 1, 6, 1, 1, 5, 2, 2, 3, 1, 6, 7, 2, 1, 4, 2, 1, 1, 3, 1, 7, 2, 1, 1, 1, 1, 8, 1, 4, 4, 5
Offset: 1
Keywords
Links
- Eric Weisstein's World of Mathematics, Sum of Prime Factors.
Programs
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Maple
b:= n-> add(i[1]*i[2], i=ifactors(n)[2]): p:= proc() 0 end: a:= proc(n) option remember; local t; t:= b(n); p(t):= p(t)+1 end: seq(a(n), n=1..100); # Alois P. Heinz, Jan 30 2025 -
Mathematica
sopfr[1] = 0; sopfr[n_] := Plus @@ Times @@@ FactorInteger@ n; Table[Length[Select[Range[n], sopfr[#] == sopfr[n] &]], {n, 1, 100}] -
Python
from sympy import factorint from collections import Counter from itertools import count, islice def agen(): # generator of terms sopfrcount = Counter() for n in count(1): key = sum(p*e for p, e in factorint(n).items()) sopfrcount[key] += 1 yield sopfrcount[key] print(list(islice(agen(), 100))) # Michael S. Branicky, Jan 30 2025
Formula
a(n) = |{j <= n : sopfr(j) = sopfr(n)}|.
Comments