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%I #18 Aug 30 2018 08:21:17
%S 0,1,1,1,2,2,1,1,1,3,1,2,2,2,3,1,2,2,1,3,2,2,1,2,2,3,1,2,2,4,1,1,2,3,
%T 3,2,2,2,3,3,2,3,1,2,3,2,1,2,1,3,3,3,2,2,3,2,2,3,1,4,2,2,2,1,4,3,1,3,
%U 2,4,1,2,2,3,3,2,2,4,1,3,1,3,1,3,4,2,3,2,2,4,3,2,2,2,3,2,2,2,2,3
%N Number of distinct Gaussian primes in the factorization of n.
%C As shown in the formula, a(n) depends on the number of distinct primes of the forms 4*k+1 (A005089) and 4*k-1 (A005091) and whether n is divisible by 2 (A059841).
%C Note that associated divisors are counted only once. - _Jianing Song_, Aug 30 2018
%H T. D. Noe, <a href="/A086275/b086275.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/GaussianPrime.html">MathWorld: Gaussian Prime</a>
%F a(n) = A059841(n) + 2*A005089(n) + A005091(n).
%F Additive with a(p^e) = 2 if p = 1 (mod 4), 1 otherwise. - _Franklin T. Adams-Watters_, Oct 18 2006
%e a(1006655265000) = a(2^3*3^2*5^4*7^5*11^3) = 1 + 2*1 + 3 = 6 because n is divisible by 2, has 1 prime factor of the form 4*k+1 and 3 primes of the form 4*k+3. Over the Gaussian integers, 1006655265000 is factored as i*(1 + i)^6*(2 + i)^4*(2 - i)^4*3^2*7^5*11^3, the 6 distinct Gaussian factors are 1 + i, 2 + i, 2 - i, 3, 7 and 11.
%t Join[{0}, Table[f=FactorInteger[n, GaussianIntegers->True]; cnt=Length[f]; If[MemberQ[{-1, I, -I}, f[[1, 1]]], cnt-- ]; cnt, {n, 2, 100}]]
%o (PARI) a(n)=my(f=factor(n)[,1]); sum(i=1,#f,if(f[i]%4==1,2,1)) \\ _Charles R Greathouse IV_, Sep 14 2015
%Y Cf. A005089, A005091, A059841, A078458 (number of Gaussian primes, with multiplicity).
%K easy,nonn
%O 1,5
%A _T. D. Noe_, Jul 14 2003
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