A020700 - OEIS

%I #24 Feb 26 2023 17:54:55

%S 7,14,63,80,224,285,351,363,475,860,902,1088,1479,2013,2023,3478,3689,

%T 3925,5984,6715,8493,9456,13224,15520,17227,18569,19502,20490,21804,

%U 24435,24476,27335,31899,32390,35815,37406,37582,41876,49468,50609,54137,57239

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

%C 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

%H Robert Israel, <a href="/A020700/b020700.txt">Table of n, a(n) for n = 1..2500</a>

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

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

%o (PARI) A075254(n) = my(f = factor(n)); n + sum(i=1, #f~, f[i,1]*f[i,2]);

%o isok(n) = A075254(n) == A075254(n+1); \\ _Michel Marcus_, Jun 05 2014

%o (Python)

%o from sympy import factorint

%o from itertools import count, islice

%o def sopf(n): return sum(p*e for p, e in factorint(n).items())

%o def agen(): # generator of terms

%o     sopfkplus1 = 2

%o     for k in count(2):

%o         sopfk, sopfkplus1 = sopfkplus1, sopf(k+1)

%o         if k + sopfk == k + 1 + sopfkplus1: yield k

%o print(list(islice(agen(), 42))) # _Michael S. Branicky_, Apr 15 2022

%Y Cf. A020905, A228126.

%K easy,nonn

%O 1,1

%A _Enoch Haga_

%E More terms from _Michel Marcus_, Jun 05 2014