A289310 Let f be the multiplicative function satisfying f(p^k) = (1 + p*I)^k for any prime p and k > 0 (where I^2 = -1); a(n) = the real part of f(n).

1, 1, 1, -3, 1, -5, 1, -11, -8, -9, 1, -15, 1, -13, -14, -7, 1, -20, 1, -23, -20, -21, 1, -5, -24, -25, -26, -31, 1, -30, 1, 41, -32, -33, -34, 0, 1, -37, -38, -1, 1, -40, 1, -47, -38, -45, 1, 65, -48, -44, -50, -55, 1, 10, -54, 3, -56, -57, 1, 10, 1, -61, -50

Offset: 1

Author

Rémy Sigrist, Jul 02 2017

Keywords

• sign

Comments

See A289311 for the imaginary part of f.

See A289320 for the square of the norm of f.

a(p) = 1 for any prime p.

If a(n) = 0, then a(n^(2*k-1)) = 0 and A289311(n^(2*k)) = 0 for any k > 0.

a(n) = 0 iff Sum_{i=1..k} ( arctan(p_i) * e_i ) = Pi/2 * (2*j + 1) for some integer j (where Product_{i=1..k} p_i^e_i is the prime factorization of n).

a(n) = 0 for n = 36, 3969, 13608, 46656, 1500282, 5143824, 6718383, ...

As a(36) = 0 and 36 = 2^2 * 3^3, we have arctan(2)*2 + arctan(3)*2 = Pi/2 * (2*j + 1) (with j = 1).

If |a(n)| = |A289311(n)|, then |a(n^(2k-1))| = |A289311(n^(2k-1))| for any k > 0.

|a(n)| = |A289311(n)| iff Sum_{i=1..k} ( arctan(p_i) * e_i ) = Pi/4 * (2*j + 1) for some integer j (where Product_{i=1..k} p_i^e_i is the prime factorization of n).

|a(n)| = |A289311(n)| for n = 6, 63, 216, 2268, 7776, 23814, 81648, 106641, 250047, 279936, 312273, 857304, ...

As |a(63)| = |A289311(63)| and 63 = 3^2 * 7, we have arctan(3)*2 + arctan(7) = Pi/4 * (2*j + 1) (with j=1).

The scatterplot of this sequence vs A289311 is interesting (see Links section).

Examples

			f(12) = f(2^2 * 3) = (1 + 2*I)^2 * (1 + 3*I) = -15 - 5*I, hence a(12) = -15.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Rémy Sigrist, Scatterplot of the first 1000000 terms of A289310 vs A289311

Crossrefs

Cf. A289311, A289320.

Programs

Mathematica

Array[Re[Times @@ Map[(1 + #1 I)^#2 & @@ # &, FactorInteger@ #]] &, 63] (* Michael De Vlieger, Jul 03 2017 *)

PARI

a(n) = my (f=factor(n)); real (prod(i=1, #f~, (1 + f[i,1]*I) ^ f[i,2]))