A322702 a(n) is the product of primes p such that p+1 divides n.
1, 1, 2, 3, 1, 10, 1, 21, 2, 1, 1, 330, 1, 13, 2, 21, 1, 170, 1, 57, 2, 1, 1, 53130, 1, 1, 2, 39, 1, 290, 1, 651, 2, 1, 1, 5610, 1, 37, 2, 399, 1, 5330, 1, 129, 2, 1, 1, 2497110, 1, 1, 2, 3, 1, 9010, 1, 273, 2, 1, 1, 10727970, 1, 61, 2, 651, 1, 10, 1, 201, 2
Offset: 1
Keywords
Examples
For n=12, the divisors of 12 are {1, 2, 3, 4, 6, 12}. The prime numbers p, such that p+1 is a divisor of 12, are {2, 3, 5, 11}, therefore a(12) = 2 * 3 * 5 * 11 = 330.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10080
Programs
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Maple
a:= n-> mul(`if`(isprime(d-1), d-1, 1), d=numtheory[divisors](n)): seq(a(n), n=1..100); # Alois P. Heinz, Dec 29 2018
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Mathematica
Array[Apply[Times, Select[Divisors@ #, PrimeQ[# - 1] &] - 1 /. {} -> {1}] &, 69] (* Michael De Vlieger, Jan 07 2019 *)
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PARI
a(n) = my(d=divisors(n)); prod(k=1, #d, if(isprime(d[k]-1), d[k]-1, 1));
Formula
a(n) = Product_{p prime, p+1 divides n} p.
a(n) = denominator of Sum_{p prime, p+1 divides n} 1/p.
a(n) = Product_{d|n, d-1 is prime} (d-1), where d runs over the divisors of n.
a(2*n + 1) = 2, iff n == 1 (mod 3), else a(2*n + 1) = 1.
Comments