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

1,1

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

Links

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

Example

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

Mathematica

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

Prog

(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

Crossrefs

Cf. A020905, A228126.

Sequence in context: A135536 A241201 A295388 * A110496 A117867 A291008

Adjacent sequences: A020697 A020698 A020699 * A020701 A020702 A020703

Keyword

easy,nonn

Author

Enoch Haga

Extensions

More terms from Michel Marcus, Jun 05 2014

Status

approved