%I #19 Jun 25 2020 15:00:37
%S 3,19,37,53,71,109,149,211,251,277,307,359,397,449,479,521,593,641,
%T 709,769,823,859,919,1009,1033,1087,1171,1217,1277,1321,1367,1399,
%U 1459,1549,1609,1637,1693,1747,1879,1973,2039,2099,2213,2341,2399,2437,2531,2663,2777,2879,2939,3061,3251,3433
%N Partition the sequence of consecutive primes into groups so that the absolute value of the alternating sum (1)^n (An) with n = 0,....m in each group is prime.
%C From _Robert Israel_, Jun 24 2020: (Start)
%C The alternating sum must consist of more than two terms, and a(n) is the absolute value of that alternating sum.
%C Is the sequence increasing? For k <= 99999, a(k+1) >= a(k)+14. (End)
%H Robert Israel, <a href="/A187828/b187828.txt">Table of n, a(n) for n = 1..10000</a>
%F a(x) = Sum_{(1)^n (An) with n = (0, 1, 2..m)}.
%e a(1)=3 because the absolute value of the alternating sum (1)^n (An) where An = (2, 3, 5, 7) with n = (0,1,2,3), is prime; a(2)=19 because the absolute value of the alternating sum (1)^n (An) where An = (11, 13, 17, 19, 23) with n = (0, 1, 2, 3), is prime; a(3)=37 because the absolute value of the alternating sum (1)^n (An) where An = (29, 31, 37, 41, 43) with n = (0, 1, 2, 3, 4) is prime.
%p p:= 1: R:= NULL:
%p for count from 1 to 50 do
%p q:= nextprime(p); p:= nextprime(q); t:= qp;
%p e:= 1;
%p do p:= nextprime(p);
%p t:= t + e*p;
%p e:= e;
%p until isprime(abs(t));
%p R:= R, abs(t);
%p od:
%p R; # _Robert Israel_, Jun 23 2020
%K nonn
%O 1,1
%A _Fabio Mercurio_, Dec 27 2012
%E More terms from _Robert Israel_, Jun 24 2020
