A277272 a(n+1) is the smallest number not already in the sequence whose sum of prime factors (with repetition) shares a factor with the sum of the prime factors of a(n); a(1)=2.

2, 4, 8, 3, 9, 14, 20, 24, 26, 5, 6, 21, 15, 16, 18, 25, 30, 32, 33, 7, 10, 12, 38, 27, 35, 36, 39, 42, 44, 46, 51, 49, 50, 55, 57, 11, 28, 40, 45, 48, 54, 62, 60, 64, 65, 66, 69, 13, 22, 56, 63, 74, 68, 70, 72, 77, 78, 81, 84, 85, 87, 91, 86, 92, 95, 93, 17, 52, 88, 99, 145, 98, 100, 94, 115, 102

Offset: 1

Author

David James Sycamore, Oct 08 2016

Keywords

• nonn

Comments

Inspired by A064413 (the EKG sequence). Conjecture: The sequence is a permutation of the natural numbers >1, in which composite terms appear once only (in irregular fashion) and the primes arise in their natural order. Immediately following the appearance of any prime p>3 the next term is q*(2^r) where q is the largest prime less than p, and r is the unique integer such that p=q+(2*r).

The above conjecture fails at a(95)=23, which appears before a(162)=19. (Also, the term appearing immediately after 37 is 218=2*109; the term immediately after 97 is 1335=3*5*89.) - Jon E. Schoenfield, Nov 05 2016

Examples

			a(2)=4 because sopf(4)=4 and is the smallest number (other than 2) to share a factor (2) with sopf(2)=2.
a(3)=8 since sopf(8)=6 and is the smallest number (other than 2 and 4) to share a factor (2) with sopf(4).
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000
• Jon E. Schoenfield, Magma program for generating a b-file

Crossrefs

Cf. A008472 (sopf), A064413.

Programs

Mathematica

f[n_] := Flatten@ Map[ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]; a = {2}; Do[k = 2; While[Nand[IntersectingQ[f[Total@ f@ k], f[Total@ f@ a[[n - 1]]]], ! MemberQ[a, k]], k++]; AppendTo[a, k], {n, 2, 76}]; a (* Michael De Vlieger, Oct 08 2016 *)

Python

from math import gcd
from sympy import factorint
from functools import cache
from itertools import count, islice
@cache
def sopfr(n): return sum(p*e for p,e in factorint(n).items())
def agen(): # generator of terms
    aset, an, mink = {2}, 2, 3
    while True:
        yield an
        s = sopfr(an)
        an = next(k for k in count(mink) if k not in aset and gcd(sopfr(k), s)!=1)
        aset.add(an)
        while mink in aset: mink += 1
print(list(islice(agen(), 76))) # Michael S. Branicky, Feb 21 2024