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
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.
Programs
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Mathematica
Select[(#*(# + 1)*(# + 2)*(# + 3)*(# + 4)) & /@ Range[12], PrimePi[(f = FactorInteger[#1])[[-1, 1]]] == Length[f] &] (* Amiram Eldar, Jul 05 2025 *)
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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))