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If n is a prime power, then a(n) = lambda(tau(n)) = A014963(A000005(n)); otherwise, a(n) = 1.
2

%I #16 Jul 19 2017 20:09:16

%S 1,2,2,3,2,1,2,2,3,1,2,1,2,1,1,5,2,1,2,1,1,1,2,1,3,1,2,1,2,1,2,1,1,1,

%T 1,1,2,1,1,1,2,1,2,1,1,1,2,1,3,1,1,1,2,1,1,1,1,1,2,1,2,1,1,7,1,1,2,1,

%U 1,1,2,1,2,1,1,1,1,1,2,1,5,1,2,1,1,1,1

%N If n is a prime power, then a(n) = lambda(tau(n)) = A014963(A000005(n)); otherwise, a(n) = 1.

%C For any integer sequence a, the sequence b such that b(n) = Product_{d|n} a(d) is a divisibility sequence. Since A253139(n) = Product_{d|n} a(d), A253139 is a divisibility sequence.

%C a(n) depends only on prime signature of n (cf. A025487). So a(24) = a(375) since 24 = 2^3*3 and 375 = 3*5^3 both have prime signature (3,1).

%H Antti Karttunen, <a href="/A253141/b253141.txt">Table of n, a(n) for n = 1..10000</a>

%H Morgan Ward, <a href="https://doi.org/10.1090/S0002-9904-1939-06980-2">A note on divisibility sequences</a>, Bull. Amer. Math. Soc., 45 (1939), 334-336.

%H <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>

%e 2 is a prime number, i.e., a prime power with 2 divisors; a(2) = A014963(2) = 2.

%e 6 = 2*3 is not a prime power; a(6) = 1.

%e 8 = 2^3 is a prime power with 4 divisors; a(8) = A014963(4) = 2.

%e 32 = 2^5 is a prime power with 6 divisors; a(32) = A014963(6) = 1.

%t Table[If[PrimePowerQ[n], Exp[MangoldtLambda[DivisorSigma[0, n]]], 1], {n, 1, 100}] (* _Indranil Ghosh_, Jul 19 2017 *)

%o (PARI)

%o A014963(n) = ispower(n, , &n); if(isprime(n), n, 1); \\ This function from _Charles R Greathouse IV_, Jun 10 2011

%o A253141(n) = if(1==omega(n), A014963(numdiv(n)), 1); \\ _Antti Karttunen_, Jul 19 2017

%Y Cf. A000005, A014963, A253139.

%K nonn,easy

%O 1,2

%A _Matthew Vandermast_, Dec 27 2014