A079635 Sum of (2 - p mod 4) for all prime factors p of n (with repetition).
0, 0, -1, 0, 1, -1, -1, 0, -2, 1, -1, -1, 1, -1, 0, 0, 1, -2, -1, 1, -2, -1, -1, -1, 2, 1, -3, -1, 1, 0, -1, 0, -2, 1, 0, -2, 1, -1, 0, 1, 1, -2, -1, -1, -1, -1, -1, -1, -2, 2, 0, 1, 1, -3, 0, -1, -2, 1, -1, 0, 1, -1, -3, 0, 2, -2, -1, 1, -2, 0, -1, -2, 1, 1, 1, -1, -2, 0, -1, 1, -4, 1, -1, -2, 2, -1, 0, -1, 1
Offset: 1
Keywords
Examples
a(55) = a(5*11) = (2 - 5 mod 4)+(2 - 11 mod 4) = (2-1)+(2-3) = (1)+(-1) = 0.
Links
- Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
-
Haskell
a079635 1 = 0 a079635 n = sum $ map ((2 - ) . (`mod` 4)) $ a027746_row n -- Reinhard Zumkeller, Jan 10 2012
-
Maple
f:= proc(n) local t; add(t[2]*(2-(t[1] mod 4)), t=ifactors(n)[2]) end proc: map(f, [$1..100]); # Robert Israel, Feb 05 2016
-
Mathematica
f[n_]:=Plus@@((2-Mod[#[[1]],4])*#[[2]]&/@If[n==1,{},FactorInteger[n]]); Table[f[n],{n,100}] (* Ray Chandler, Dec 20 2011 *)
-
Scheme
(define (A079635 n) (- (A083025 n) (A065339 n))) ;; Antti Karttunen, Feb 03 2016
Formula
Other identities. For all n >= 1:
a(A267099(n)) = -a(n). - Antti Karttunen, Feb 03 2016
Totally additive with a(2) = 0, a(p) = 1 if p == 1 (mod 4), and a(p) = -1 if p == 3 (mod 4). - Amiram Eldar, Jun 17 2024
Extensions
Edited by Ray Chandler, Dec 20 2011
Comments