%I
%S 260,3100,4699,29318,54760,82952,315657,380741,574883,873815,949282,
%T 1766959,2114033,3971361,5418111,6972931,7644772,9714402,15752927,
%U 30118112,91750525,129381240,130672476,395713882,450192670,523370293,553444691,833889991
%N Numbers equal to the sum of the prime factors, with multiplicity, of the previous and of the following k numbers, for some k.
%C Similar to A257367 but here the prime factors of a(n) are not considered.
%C Again, while in A257367 the prime numbers are not allowed because they would be just a trivial solution, here they are part of the terms of the sequence. The first one is 15752927.
%C Values of k are 2, 2, 3, 6, 4, 7, 5, 4, 7, 8, 11, 11, 5, 8, 11, 11, 9, 5, 9, 15, 14, 7, 9, 10, 12, 17, 19, 33, ...
%C Numbers tested up to 10^9.
%H Giovanni Resta, <a href="/A287923/b287923.txt">Table of n, a(n) for n = 1..40</a>
%F x = Sum_{i = k..k} A001414(i+x)  A001414(x), for some k.
%e 258 = 2*3*43, 259 = 7*37, 261 = 3*3*29, 262 = 2*131 and 2 + 3 + 43 + 7 + 37 + 3 + 3 + 29 + 2 + 131 = 260.
%p with(numtheory): P:= proc(q) local a,b,c,k,n;
%p for n from 1 to q do a:=0; k:=0; while a<n do k:=k+1;
%p b:=ifactors(nk)[2]; b:=add(b[j][1]*b[j][2], j=1..nops(b));
%p c:=ifactors(n+k)[2]; c:=add(c[j][1]*c[j][2], j=1..nops(c));
%p a:=a+b+c; od; if a=n then print(n); fi; od; end: P(10^9);
%Y Cf. A001414, A257367, A257524, A257525, A257929, A257930, A257976.
%K nonn
%O 1,1
%A _Paolo P. Lava_, Jun 15 2017
