A268379 Numbers having more prime factors of the form 4*k+1 than of the form 4*k+3, when counted with multiplicity.

5, 10, 13, 17, 20, 25, 26, 29, 34, 37, 40, 41, 50, 52, 53, 58, 61, 65, 68, 73, 74, 75, 80, 82, 85, 89, 97, 100, 101, 104, 106, 109, 113, 116, 122, 125, 130, 136, 137, 145, 146, 148, 149, 150, 157, 160, 164, 169, 170, 173, 175, 178, 181, 185, 193, 194, 195, 197, 200, 202, 205, 208, 212, 218, 221

Offset: 1

Author

Antti Karttunen, Feb 03 2016

Keywords

• nonn

Comments

Numbers n for which A083025(n) > A065339(n) or equally, for which A079635(n) > 0.

Closed under multiplication.

Examples

			75 = 3*5*5 is included as there are more prime factors of the form 4k+1 (here two 5's) than of the form 4k+3 (here just one 3).
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A065339, A079635, A083025.

Cf. also A001481, A072202, A268380.

Subsequence of A268381.

Differs from A221265 for the first time at n=22, as here a(22) = 75, a value missing from A221265.

Programs

Mathematica

Rest@ Position[Array[Map[Length, {Select[#, Mod[#, 4] == 1 &], Select[#, Mod[#, 4] == 3 &]}] &@ Flatten@ Apply[Table[#1, {#2}] &, FactorInteger@ #, 1] &, {221}], {a_, b_} /; a > b] // Flatten (* Michael De Vlieger, Feb 05 2016 *)

PARI

isok(n) = {my(f = factor(n), nb1 = 0, nb3 = 0); for (i=1, #f~, m = f[i,1] % 4; if (m == 1, nb1 += f[i,2], if (m == 3, nb3 += f[i,2]));); return (nb1 > nb3);} \\ Michel Marcus, Feb 04 2016

Python

from itertools import count, islice
from sympy import factorint
def A268379_gen(): # generator of terms
    return filter(lambda n:sum((f:=factorint(n)).values())-f.get(2,0) < 2*sum(f[p] for p in f if p & 3 == 1),count(1))
A268379_list = list(islice(A268379_gen(),30)) # Chai Wah Wu, Jun 28 2022