OFFSET
1,1
COMMENTS
These numbers are the products of two consecutive integers that are also Hardy-Ramanujan integers; that is, of the form 2^k1*3^k2*...*p_n^k_n, where k1 >= k2 >= ... >= k_n. This sequence is finite with last term a(14) = 2079*2080 = 4324320.
In this sequence, the greatest prime factor of all products has exponent 1, except for 8*9=72 (where the exponent of 3 equals 2). All the terms except 72 have the prime exponential form of factorials in that the exponents start at 1 with the greatest prime factor, and then are monotonic nondecreasing (with no gaps) up to the exponent of prime divisor 2. Given that only 2 and 6 are factorial numbers, this sequence limits the factorials that are the products of two consecutive integers to just those two products, giving 2! and 3!. - Ken Clements, Mar 13 2026
EXAMPLE
a(1) = 2 = 1*2 = 2^1.
a(2) = 6 = 2*3 = 2^1 * 3^1.
a(3) = 12 = 3*4 = 2^2 * 3^1.
a(4) = 30 = 5*6 = 2^1 * 3^1 * 5^1.
a(5) = 72 = 8*9 = 2^3 * 3^2.
a(6) = 210 = 14*15 = 2^1 * 3^1 * 5^1 * 7^1.
MATHEMATICA
Select[FactorialPower[Range[0, 3000], 2], (Max@Differences[(f = FactorInteger[#])[[;; , 2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]) &] (* Amiram Eldar, Aug 10 2025 *)
PROG
(Python)
from sympy import prime, factorint
def is_Hardy_Ramanujan(n):
factors = factorint(n)
p_idx = len(factors)
if list(factors.keys())[-1] != prime(p_idx):
return False
expos = list(factors.values())
e = expos[0]
for i in range(1, p_idx):
if expos[i] > e:
return False
e = expos[i]
return True
print([ n*(n+1) for n in range(1, 10_000) if is_Hardy_Ramanujan(n*(n+1))])
CROSSREFS
KEYWORD
nonn,fini,full
AUTHOR
Ken Clements, Aug 10 2025
STATUS
approved