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

1,6

Comments

Ordinal transform of A001414.

Links

Alois P. Heinz, Table of n, a(n) for n = 1..65536

Eric Weisstein's World of Mathematics, Sum of Prime Factors.

Formula

a(n) = |{j <= n : sopfr(j) = sopfr(n)}|.

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}]

Prog

(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

Crossrefs

Cf. A001414, A058933, A263025, A380654.

Sequence in context: A035170 A111949 A143323 * A368542 A344234 A338912

Adjacent sequences: A380650 A380651 A380652 * A380654 A380655 A380656

Keyword

nonn,changed

Author

Ilya Gutkovskiy, Jan 29 2025

Status

approved