A020700 Numbers k such that k + sum of its prime factors = (k+1) + sum of its prime factors.

7, 14, 63, 80, 224, 285, 351, 363, 475, 860, 902, 1088, 1479, 2013, 2023, 3478, 3689, 3925, 5984, 6715, 8493, 9456, 13224, 15520, 17227, 18569, 19502, 20490, 21804, 24435, 24476, 27335, 31899, 32390, 35815, 37406, 37582, 41876, 49468, 50609, 54137, 57239

Offset: 1

Author

Enoch Haga

Keywords

• easy
• nonn

Comments

If p, (3/2)*(p+1), (3/2)*(p^2+p)+1 and (3/2)*(p^2+1)+2*p are all prime, then (3/2)*p*(3*p^2+4*p+3) is a term. The Generalized Bunyakovsky Conjecture implies that there are infinitely many of these. - Robert Israel, Apr 15 2022

Examples

			A075254(7) = 7+7 = 14 and A075254(8) = 8+2+2+2 = 14, so 7 is in the sequence.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..2500

Crossrefs

Cf. A020905, A228126.

Programs

Mathematica

SequencePosition[Table[n+Total[Times@@@FactorInteger[n]],{n,58000}],{x_,x_}][[;;,1]] (* Harvey P. Dale, Feb 26 2023 *)

PARI

A075254(n) = my(f = factor(n)); n + sum(i=1, #f~, f[i,1]*f[i,2]);
isok(n) = A075254(n) == A075254(n+1); \\ Michel Marcus, Jun 05 2014

Python

from sympy import factorint
from itertools import count, islice
def sopf(n): return sum(p*e for p, e in factorint(n).items())
def agen(): # generator of terms
    sopfkplus1 = 2
    for k in count(2):
        sopfk, sopfkplus1 = sopfkplus1, sopf(k+1)
        if k + sopfk == k + 1 + sopfkplus1: yield k
print(list(islice(agen(), 42))) # Michael S. Branicky, Apr 15 2022

Extensions

More terms from Michel Marcus, Jun 05 2014