|
|
A301852
|
|
Integers k such that the remainder of the sum of the first k primes divided by the k-th prime is equal to k.
|
|
1
|
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Integers k such that A071089(k) = k.
No more terms below 10^7.
Heuristically, the probability that k is a term is 1/prime(k) ~ 1/(k log k).
Since Sum_{k>=2} 1/(k log(k)) diverges, there should be infinitely many terms. However, the sum diverges very slowly, so terms may be very sparse: approximately log(log(k)) terms <= k. (End)
|
|
LINKS
|
|
|
EXAMPLE
|
2 is a term because prime(1)+prime(2) = 5 = 2 mod prime(2).
|
|
MAPLE
|
res:= NULL: p:= 1: s:= 0:
for m from 1 to 10^6 do
p:= nextprime(p);
s:= s+p;
if s mod p = m then res:= res, m fi
od:
|
|
PROG
|
(PARI) lista(nn)= my(p = 2, s = 2); for (n=1, nn, if ((s % p) == n, print1(n, ", ")); q = nextprime(p+1); s += q; p = q; ); \\ Michel Marcus, Mar 27 2018
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,more
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|