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A072438
Remove prime factors of form 4*k+1.
5
1, 2, 3, 4, 1, 6, 7, 8, 9, 2, 11, 12, 1, 14, 3, 16, 1, 18, 19, 4, 21, 22, 23, 24, 1, 2, 27, 28, 1, 6, 31, 32, 33, 2, 7, 36, 1, 38, 3, 8, 1, 42, 43, 44, 9, 46, 47, 48, 49, 2, 3, 4, 1, 54, 11, 56, 57, 2, 59, 12, 1, 62, 63, 64, 1, 66, 67, 4, 69, 14, 71, 72, 1, 2, 3, 76, 77, 6, 79, 16, 81, 2
OFFSET
1,2
COMMENTS
a(n) <= n; a(a(n)) = a(n).
All factors p^m of a(n) are of the form p=2 or p=4*k+3.
LINKS
FORMULA
Multiplicative with a(p)=(if p==1 (mod 4) then 1 else p).
EXAMPLE
a(90) = a(2*3*3*5) = a(2*(4*0+3)^2*(4*1+1)^1) = 2*3^2*1 = 18.
MAPLE
a:= n-> mul(`if`(irem(i[1], 4)=1, 1, i[1]^i[2]), i=ifactors(n)[2]):
seq(a(n), n=1..100); # Alois P. Heinz, Jun 09 2014
MATHEMATICA
a[n_] := n/Product[{p, e} = pe; If[Mod[p, 4] == 1, p^e, 1], {pe, FactorInteger[n]}];
Array[a, 100] (* Jean-François Alcover, May 29 2019 *)
PROG
(PARI) a(n) = my(f=factor(n)); for (i=1, #f~, if ((f[i, 1] % 4) == 1, f[i, 1] = 1)); factorback(f); \\ Michel Marcus, Jun 09 2014
(Python)
from sympy import factorint, prod
def a(n):
f = factorint(n)
return 1 if n == 1 else prod(i**f[i] for i in f if i%4 != 1) # Indranil Ghosh, May 08 2017
CROSSREFS
KEYWORD
nonn,mult,easy,changed
AUTHOR
Reinhard Zumkeller, Jun 17 2002
STATUS
approved