A347038 Primes p such that there are no solutions to d(k+p) = sigma(k).

29, 37, 41, 53, 67, 89, 101, 109, 113, 127, 137, 151, 157, 173, 181, 197, 227, 229, 233, 257, 269, 277, 281, 293, 313, 349, 373, 389, 401, 409, 421, 439, 461, 557, 587, 593, 601, 613, 617, 641, 643, 653, 661, 673, 677, 701, 709, 739, 761, 773, 787, 821, 829

Offset: 1

Author

Angad Singh, Aug 12 2021

Keywords

• nonn

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

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000005, A000203, A247485.

Programs

Maple

filter:= proc(p) isprime(p) and not ormap(k -> numtheory:-tau(k+p) = numtheory:-sigma(k), [$1 .. 1 + 2*isqrt(p)]) end proc:
select(filter, [seq(i,i=3..1000,2)]); # Robert Israel, Aug 06 2025

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