A371948 Numbers k such that k+1 is composite and A371699(k) != p^2 where p = A020639(k+1) is the smallest prime factor of k+1.

288, 298, 340, 360, 376, 516, 526, 550, 582, 736, 778, 792, 802, 816, 892, 988, 1002, 1006, 1072, 1138, 1146, 1198, 1206, 1246, 1270, 1338, 1342, 1348, 1356, 1390, 1402, 1456, 1500, 1516, 1536, 1576, 1632, 1642, 1702, 1726, 1738, 1750, 1768, 1816, 1828, 1842

Offset: 1

Author

Chai Wah Wu, Apr 13 2024

Keywords

• nonn

Comments

If k+1 is composite, then A371699(k) <= A020639(k+1)^2. This sequence lists numbers k where the inequality is strict.

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A020639, A027746, A259047, A322843, A248915, A371641, A371699, A371900.

Programs

Python

from itertools import count, islice
from sympy import isprime, primefactors, factorint, integer_log
def A371948_gen(startvalue=2): # generator of terms >= startvalue
    for n in count(max(startvalue,2)):
        if not isprime(n+1):
            q = min(primefactors(n+1))
            for m in range(4,q**2):
                f = factorint(m)
                if sum(f.values()) > 1:
                    c = 0
                    for p in sorted(f,reverse=True):
                        a = pow(n,integer_log(p,n)[0]+1,m)
                        for _ in range(f[p]):
                            c = (c*a+p)%m
                    if not c:
                        yield n
                        break
A371948_list = list(islice(A371948_gen(), 30))