A097706 Part of n composed of prime factors of form 4k+3.

1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 11, 3, 1, 7, 3, 1, 1, 9, 19, 1, 21, 11, 23, 3, 1, 1, 27, 7, 1, 3, 31, 1, 33, 1, 7, 9, 1, 19, 3, 1, 1, 21, 43, 11, 9, 23, 47, 3, 49, 1, 3, 1, 1, 27, 11, 7, 57, 1, 59, 3, 1, 31, 63, 1, 1, 33, 67, 1, 69, 7, 71, 9, 1, 1, 3, 19, 77, 3, 79, 1, 81, 1, 83, 21

Offset

1,3

Comments

Largest term of A004614 that divides n. - Peter Munn, Apr 15 2021

Links

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Formula

a(n) = n/A072436(n).

a(A004614(n)) = A004614(n).

a(A072437(n)) = 1.

a(n) = A000265(n)/A170818(n). - Peter Munn, Apr 15 2021

Maple

a:= n-> mul(`if`(irem(i[1], 4)=3, i[1]^i[2], 1), i=ifactors(n)[2]):
seq(a(n), n=1..100);  # Alois P. Heinz, Jun 09 2014

Mathematica

a[n_] := Product[{p, e} = pe; If[Mod[p, 4] == 3, p^e, 1], {pe, FactorInteger[n]}]; Array[a, 100] (* Jean-François Alcover, Jun 16 2015, updated May 29 2019 *)

Prog

(PARI) a(n)=local(f); f=factor(n); prod(k=1, matsize(f)[1], if(f[k, 1]%4<>3, 1, f[k, 1]^f[k, 2]))
(Python)
from sympy import factorint
from operator import mul
def a072436(n):
    f=factorint(n)
    return 1 if n == 1 else reduce(mul, [1 if i%4==3 else i**f[i] for i in f])
def a(n): return n/a072436(n) # Indranil Ghosh, May 08 2017
(Python)
from math import prod
from sympy import factorint
def A097706(n): return prod(p**e for p, e in factorint(n).items() if p & 3 == 3) # Chai Wah Wu, Jun 28 2022

Crossrefs

Equivalent sequence for distinct prime factors: A170819.

Equivalent sequences for prime factors of other forms: A000265 (2k+1), A170818 (4k+1), A072436 (not 4k+3), A248909 (6k+1), A343431 (6k+5).

Range of values: A004614.

Positions of 1's: A072437.

Cf. also A065338, A065339, A260728, A286363.

Sequence in context: A170819 A140211 A248101 * A132740 A106621 A011085

Adjacent sequences: A097703 A097704 A097705 * A097707 A097708 A097709

Keyword

nonn,mult,easy

Author

Ralf Stephan, Aug 30 2004

Status

approved