A086275 Number of distinct Gaussian primes in the factorization of n.

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

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*k-1 (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. Adams-Watters, 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: A112468 A207194 A349670 * A066855 A175685 A331182

Adjacent sequences: A086272 A086273 A086274 * A086276 A086277 A086278

Keyword

easy,nonn

Author

T. D. Noe, Jul 14 2003

Status

approved