OFFSET
1,1
COMMENTS
If k == 1 (mod 3) and 3 divides 2 + 3 + 5 + ... + prime(k) then k
is in the sequence. I conjecture that 3 is the only term of the sequence which is not of this form.
EXAMPLE
For each m, fp(1,m)=2 is prime so 1 is not in the sequence.
fp(2,2) = 2.2.3 = 223 is prime and 2 < 2*prime(2) so 2 isn't in the sequence. Also for each m, 5 divides fp(3,m) = 2.m.3.m.5 so fp(3,m) is composite and we deduce that 3 is in the sequence.
PROG
(PARI) is(k) = for(m=0, k*prime(k), if(ispseudoprime(eval(concat(concat([""], vector(2*k-1, i, if(i%2, prime(1+i\2), m)))))), return(0))); 1; \\ Jinyuan Wang, Apr 10 2020
CROSSREFS
KEYWORD
nonn,base,more
AUTHOR
Farideh Firoozbakht, Jun 15 2003
EXTENSIONS
Corrected and edited by Farideh Firoozbakht, Nov 04 2013
STATUS
approved