OFFSET
1,1
COMMENTS
If we divide each a(n) by the two primes we get a sequence of the triangular numbers of (3 * A002820(n) - 1). If we take the differences between those triangular numbers we get A145061 + 1.
This is a subsequence of A011772, which is really the basic sequence here. - N. J. A. Sloane, Jul 06 2021
FORMULA
EXAMPLE
15 is the smallest triangular number that is a multiple of 3 and 5, so, a(1) = 15.
PROG
(PARI) a001359(n, p=3) = { while( p+2 < (p=nextprime( p+1 )) || n-->0, ); p-2};
a(n) = my(p=a001359(n), k = (p-1)*(p+2)/2); k*(k+1)/2; \\ Michel Marcus, Jun 10 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Ali Sada, Jun 01 2021
EXTENSIONS
a(22)-a(26) from Jon E. Schoenfield, Jun 01 2021
STATUS
approved