

A086275


Number of distinct Gaussian primes in the factorization of n.


11



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, 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, 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
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OFFSET

1,5


COMMENTS

As shown in the formula, a(n) depends on the number of distinct primes of the forms 4*k+1 (A005089) and 4*k1 (A005091) and whether n is divisible by 2 (A059841).
Note that associated divisors are counted only once.  Jianing Song, Aug 30 2018


LINKS

T. D. Noe, Table of n, a(n) for n = 1..10000
Eric W. Weisstein, MathWorld: Gaussian Prime


FORMULA

a(n) = A059841(n) + 2*A005089(n) + A005091(n).
Additive with a(p^e) = 2 if p = 1 (mod 4), 1 otherwise.  Franklin T. AdamsWatters, Oct 18 2006


EXAMPLE

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.


MATHEMATICA

Join[{0}, Table[f=FactorInteger[n, GaussianIntegers>True]; cnt=Length[f]; If[MemberQ[{1, I, I}, f[[1, 1]]], cnt ]; cnt, {n, 2, 100}]]


PROG

(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


CROSSREFS

Cf. A005089, A005091, A059841, A078458 (number of Gaussian primes, with multiplicity).
Sequence in context: A112465 A112468 A207194 * A066855 A175685 A331182
Adjacent sequences: A086272 A086273 A086274 * A086276 A086277 A086278


KEYWORD

easy,nonn


AUTHOR

T. D. Noe, Jul 14 2003


STATUS

approved



