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%I #35 Jan 12 2019 20:46:17
%S 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,
%T 39,1,290,1,651,2,1,1,5610,1,37,2,399,1,5330,1,129,2,1,1,2497110,1,1,
%U 2,3,1,9010,1,273,2,1,1,10727970,1,61,2,651,1,10,1,201,2
%N a(n) is the product of primes p such that p+1 divides n.
%C In general, a(n) is the product of A072627(n) distinct prime factors, with a(n) = 1 iff A072627(n) = 0.
%H Antti Karttunen, <a href="/A322702/b322702.txt">Table of n, a(n) for n = 1..10080</a>
%F a(n) = Product_{p prime, p+1 divides n} p.
%F a(n) = denominator of Sum_{p prime, p+1 divides n} 1/p.
%F a(n) = Product_{d|n, d-1 is prime} (d-1), where d runs over the divisors of n.
%F a(2*n + 1) = 2, iff n == 1 (mod 3), else a(2*n + 1) = 1.
%F A001221(a(n)) = A072627(n). - _Antti Karttunen_, Jan 12 2019
%e 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.
%p a:= n-> mul(`if`(isprime(d-1), d-1, 1), d=numtheory[divisors](n)):
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Dec 29 2018
%t Array[Apply[Times, Select[Divisors@ #, PrimeQ[# - 1] &] - 1 /. {} -> {1}] &, 69] (* _Michael De Vlieger_, Jan 07 2019 *)
%o (PARI) a(n) = my(d=divisors(n)); prod(k=1, #d, if(isprime(d[k]-1), d[k]-1, 1));
%Y Cf. A072627, A027760, A322356, A323156.
%K nonn
%O 1,3
%A _Daniel Suteu_, Dec 23 2018
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