OFFSET
1,1
COMMENTS
Arrange the prime numbers into a triangle, with 2 at the top, 3 and 5 in the second row, 7, 11 and 13 in the third row, and so on:
2
3 5
7 11 13
17 19 23 29
31 37 41 43 47
...
The n-th term in the sequence is then the sum of the numbers in the upward diagonal beginning on the n-th row of this triangle.
LINKS
George Stagg, Table of n, a(n) for n = 1..1000
FORMULA
a(n) = Sum_{m=0..floor((n-1)/2)} prime((n-m)(n-m-1)/2+m+1).
EXAMPLE
a(1) = 2; a(2) = 3; a(3) = 7+5 = 12; a(4) = 17+11 = 28.
MAPLE
A249490:=n->add(ithprime((n-m)*(n-m-1)/2+m+1), m=0..floor((n-1)/2)): seq(A249490(n), n=1..50); # Wesley Ivan Hurt, Nov 07 2014
MATHEMATICA
Table[Sum[Prime[(n - m) (n - m - 1)/2 + m + 1], {m, 0, Floor[(n - 1)/2]}], {n, 50}] (* Wesley Ivan Hurt, Nov 07 2014 *)
PROG
(MATLAB/Octave)
p=primes(10000); a=[];
for n=1:30
m=0:floor((n-1)/2);
a=[a, sum(p((n-m).*(n-m-1)./2 + m + 1))];
end
a
(PARI) a(n) = sum(m=0, (n-1)\2, prime((n-m)*(n-m-1)/2+m+1)); \\ Michel Marcus, Nov 04 2014
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
George Stagg, Oct 30 2014
STATUS
approved