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A391970
Numbers m such that the product m*(m+1) has a set of prime divisors, from greatest down to 2, that is missing exactly two prime divisors.
5
7, 10, 11, 32, 39, 84, 119, 143, 153, 168, 209, 220, 242, 272, 285, 324, 351, 374, 441, 455, 494, 624, 675, 728, 1224, 1539, 1700, 2057, 2184, 2499, 2600, 3059, 3135, 4913, 5082, 5775, 5928, 6174, 6655, 7865, 10625, 11270, 12635, 13376, 14079, 19227, 19550, 21504
OFFSET
1,1
COMMENTS
The number of distinct prime divisors, omega(m*(m+1)) is exactly two less than the index of the greatest prime divisor, pi(gpf(m*(m+1))): A252489(m) - A059957(m) = 2.
Although 1109496723125 may be the last term in this sequence, if not, subsequent terms are greater than 2*10^14 by computational verification.
The sequence must terminate because the growth rate of A252489 exceeds A059957 enough to produce a continuing difference of more than two eventually, so thereafter the factorization of the product will have more than two gaps.
LINKS
EXAMPLE
a(1) = 7 because 7*8 = 56 with gpf 7 and pi(7) = 4 while omega(56) = 2 because it has only 2 and 7 for prime divisors, missing prime divisors 3 and 5.
a(2) = 10 because 10*11 = 110 with gpf 11 and pi(11) = 5 while omega(110) = 3 because it has only 2, 5 and 11 for prime divisors, missing prime divisors 3 and 7.
a(3) = 11 because 11*12 = 132 with gpf 11 and pi(11) = 5 while omega(132) = 3 because it has only 2, 3, and 11 for prime divisors, missing prime divisors 5 and 7.
MATHEMATICA
q[m_] := Module[{p = Union[Flatten[FactorInteger[#][[;; , 1]] & /@ {m, m + 1}]]}, PrimePi[p[[-1]]] == Length[p] + 2]; Select[Range[22000], q] (* Amiram Eldar, Dec 25 2025 *)
PROG
(Python)
from sympy import primepi, primefactors
def ok(m):
pf = primefactors(m*(m+1))
omega, pidx = len(pf), primepi(pf[-1])
return pidx - omega == 2
print([m for m in range(2, 10**6) if ok(m)])
(PARI) isok(m) = my(f=factor(m*(m+1))); omega(f) == primepi(vecmax(f)) - 2; \\ Michel Marcus, Dec 26 2025
CROSSREFS
Sequence in context: A105140 A107018 A247868 * A048792 A024585 A190482
KEYWORD
nonn
AUTHOR
Ken Clements, Dec 24 2025
STATUS
approved