A385415 Products of three consecutive integers whose prime divisors are consecutive primes starting at 2.

6, 24, 60, 120, 210, 720, 3360, 9240, 117600, 166320, 970200, 43243200, 85765680

Offset: 1

Author

Ken Clements, Jun 28 2025

Keywords

• nonn
• fini
• full

Comments

The final term is 440*441*442 = 85765680.

Let tau(a(n)) be the number of divisors of a(n), sigma(a(n)) be the sum of divisors of a(n), S be the strictly increasing sequence of divisors of a(n), and S(i) be the i-th element of S. Since sigma(a(n)) is even and for every i in the interval [1, tau(a(n))-1], 2*S(i) >= S(i+1), a(n) is a Zumkeller number (see Proposition 17 in Rao/Peng JNT paper at A083207). - Ivan N. Ianakiev, Jul 09 2025

Examples

			a(1) = 6 = 1*2*3 = 2^1 * 3^1.
a(2) = 24 = 2*3*4 = 2^3 * 3^1.
a(3) = 60 = 3*4*5 = 2^2 * 3^1 * 5^1.
a(4) = 120 = 4*5*6 = 2^3 * 3^1 * 5^1.
a(5) = 210 = 5*6*7 = 2^1 * 3^1 * 5^1 * 7^1.
a(6) = 720 = 8*9*10 = 2^4 * 3^2 * 5^1.
...
a(13) = 85765680 = 440*441*442 = 2^4 * 3^2 * 5^1 * 7^2 * 11^1 * 13^1 * 17^1.
		

Crossrefs

Intersection of A007531 and A055932.

Cf. A385189, A083207 (supersequence).

Programs

Mathematica

Select[(#*(# + 1)*(# + 2)) & /@ Range[500], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jun 28 2025 *)

Python

from sympy import prime, primefactors
def is_pi_complete(n): # Check for complete set of
    factors = primefactors(n) # prime factors
    return factors[-1] == prime(len(factors))
def aupto(limit):
    result = []
    for i in range(1, limit+1):
        n = i * (i+1) * (i+2)
        if is_pi_complete(n):
            result.append(n)
    return result
print(aupto(100_000))