A268386 a(n) = A193231(A268387(n)).
0, 1, 1, 3, 1, 0, 1, 2, 3, 0, 1, 2, 1, 0, 0, 5, 1, 2, 1, 2, 0, 0, 1, 3, 3, 0, 2, 2, 1, 1, 1, 4, 0, 0, 0, 0, 1, 0, 0, 3, 1, 1, 1, 2, 2, 0, 1, 4, 3, 2, 0, 2, 1, 3, 0, 3, 0, 0, 1, 3, 1, 0, 2, 6, 0, 1, 1, 2, 0, 1, 1, 1, 1, 0, 2, 2, 0, 1, 1, 4, 5, 0, 1, 3, 0, 0, 0, 3, 1, 3, 0, 2, 0, 0, 0, 5, 1, 2, 2, 0, 1, 1, 1, 3, 1, 0, 1, 1, 1, 1, 0, 4, 1, 1, 0, 2, 2, 0, 0, 2
Offset: 1
Keywords
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16384
Crossrefs
Programs
-
Mathematica
f[n_] := Which[0 <= # <= 1, #, EvenQ@ #, BitXor[2 #, #] &[f[#/2]], True, BitXor[#, 2 # + 1] &[f[(# - 1)/2]]] &@ Abs@ n; {0}~Join~Table[f[BitXor @@ Map[Last, FactorInteger@ n]], {n, 2, 120}] (* Michael De Vlieger, Feb 12 2016, after Robert G. Wilson v at A048724 and A065621 *)
-
PARI
a268387(n) = {my(f = factor(n), b = 0); for (k=1, #f~, b = bitxor(b, f[k, 2]); ); b; } a193231(n) = {my(x='x); subst(lift(Mod(1, 2)*subst(Pol(binary(n), x), x, 1+x)), x, 2)}; a(n) = a193231(a268387(n)); \\ Michel Marcus, May 09 2020
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Scheme
(define (A268386 n) (A193231 (A268387 n)))
Formula
a(2^k) = A193231(k). - Peter Munn, May 07 2020
From Peter Munn, Jun 02 2020: (Start)
Alternative definition, for n, k >= 1, where XOR denotes A003987:
a(prime(n)) = 1, where prime(n) = A000040(n);
a(n^2) = a(n) XOR (2 * a(n)) = A048720(a(n), 3);
a(A059897(n, k)) = a(n) XOR a(k).
(End)
Comments