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%I #23 Sep 05 2018 07:58:29
%S 1,19,27,1293,41419,41493,41519,41521,41601,1629899,3142492081,
%T 3142492231
%N Integers k such that A008347(k) is divisible by k.
%C Inspired by A008347.
%C a(n) - a(n-1) = 2 for n = 8.
%C a(11) > 10^9. - _David A. Corneth_, Aug 29 2018
%C a(13) > 7.35*10^12, if it exists. - _Giovanni Resta_, Sep 05 2018
%H Romeo Meštrović, <a href="https://arxiv.org/abs/1805.11657">On the distribution of primes in the alternating sums of concecutive primes</a> [sic], arXiv:1805.11657 [math.NT], 2018.
%F a(n) mod 2 = 1.
%e For first 19 prime numbers, (2 - 3 + 5 - 7 + 11 - 13 + 17 - 19 + 23 - 29 + 31 - 37 + 41 - 43 + 47 - 53 + 59 - 61 + 67) mod 19 = 38 mod 19 = 0. So a(2) = 19.
%t Select[Range@ 1500, Mod[Abs@ Sum[(-1)^k Prime[# - k], {k, 0, # - 1}], #] == 0 &] (* _Michael De Vlieger_, Nov 27 2015, after _Robert G. Wilson v_ at A008347 *)
%o (PARI) for(n=1, 1e5, if(sum(k=0, n-1, (-1)^(k)*prime(n-k)) % n == 0, print1(n,", ")))
%o (PARI) upto(n) = my(s = t = 0, u=prime(n), res=List()); forprime(p=2, u, t++; s = (p - s); if(s % t == 0, listput(res, i))); res \\ _David A. Corneth_, Aug 29 2018
%Y Cf. A008347.
%K nonn,more
%O 1,2
%A _Altug Alkan_, Nov 26 2015
%E a(10) from _David A. Corneth_, Aug 29 2018
%E a(11)-a(12) from _Giovanni Resta_, Sep 03 2018
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