A385631 Products of five consecutive integers whose prime divisors are consecutive primes starting at 2.

120, 720, 2520, 6720, 15120, 30240, 55440, 240240, 360360

Offset: 1

Author

Ken Clements, Jul 05 2025

Keywords

• nonn
• fini
• full

Examples

			a(1) = 120 = 1*2*3*4*5 = 2^3 * 3^1 * 5^1.
a(2) = 720 = 2*3*4*5*6 = 2^4 * 3^2 * 5^1.
a(3) = 2520 = 3*4*5*6*7 = 2^3 * 3^2 * 5^1 * 7^1.
a(4) = 6720 = 4*5*6*7*8 = 2^6 * 3^1 * 5^1 * 7^1.
a(5) = 15120 = 5*6*7*8*9 = 2^4 * 3^3 * 5^1 * 7^1.
a(6) = 30240 = 6*7*8*9*10 = 2^5 * 3^3 * 5^1 * 7^1.
a(7) = 55440 = 7*8*9*10*11 = 2^4 * 3^2 * 5^1 * 7^1 * 11^1.
a(8) = 240240 = 10*11*12*13*14 = 2^4 * 3^1 * 5^1 * 7^1 * 11^1 * 13^1.
a(9) = 360360 = 11*12*13*14*15 = 2^3 * 3^2 * 5^1 * 7^1 * 11^1 * 13^1.
		

Crossrefs

Intersection of A052787 and A055932.

Cf. A217056, A385189, A385415.

Programs

Mathematica

Select[(#*(# + 1)*(# + 2)*(# + 3)*(# + 4)) & /@ Range[12], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jul 05 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) * (i+3) * (i+4)
        if is_pi_complete(n):
            result.append(n)
    return result
print(aupto(1000))