OFFSET
1,1
COMMENTS
If p is not in the sequence and d(k+p) = sigma(k), then k <= 1+2*sqrt(p). Proof: We have d(m) <= 2*sqrt(m) (see formula in A000005), so 2*sqrt(k+p) >= d(k+p) = sigma(k) >= k+1 (if k > 1). After squaring and simplifying, we get k <= 1+2*sqrt(p). - Pontus von Brömssen, Aug 20 2021
PROG
(Python)
from sympy import divisor_count as d, divisor_sigma as sigma, primerange
from math import isqrt
def A347038_list(pmax):
a = []
for p in primerange(2, pmax + 1):
if not any(d(k + p) == sigma(k) for k in range(1, 2 + isqrt(4 * p))):
a.append(p)
return a # Pontus von Brömssen, Aug 20 2021
CROSSREFS
KEYWORD
nonn
AUTHOR
Angad Singh, Aug 12 2021
STATUS
approved