A187828 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.

3, 19, 37, 53, 71, 109, 149, 211, 251, 277, 307, 359, 397, 449, 479, 521, 593, 641, 709, 769, 823, 859, 919, 1009, 1033, 1087, 1171, 1217, 1277, 1321, 1367, 1399, 1459, 1549, 1609, 1637, 1693, 1747, 1879, 1973, 2039, 2099, 2213, 2341, 2399, 2437, 2531, 2663, 2777, 2879, 2939, 3061, 3251, 3433

Offset: 1

Author

Fabio Mercurio, Dec 27 2012

Keywords

• nonn

Comments

From Robert Israel, Jun 24 2020: (Start)

The alternating sum must consist of more than two terms, and a(n) is the absolute value of that alternating sum.

Is the sequence increasing? For k <= 99999, a(k+1) >= a(k)+14. (End)

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Programs

Maple

p:= 1: R:= NULL:
for count from 1 to 50 do
  q:= nextprime(p); p:= nextprime(q); t:= q-p;
  e:= 1;
  do p:= nextprime(p);
     t:= t + e*p;
     e:= -e;
  until isprime(abs(t));
  R:= R, abs(t);
od:
R; # Robert Israel, Jun 23 2020

Formula

a(x) = Sum_{(-1)^n (An) with n = (0, 1, 2..m)}.

Extensions

More terms from Robert Israel, Jun 24 2020