A289311 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 imaginary part of f(n).

0, 2, 3, 4, 5, 5, 7, -2, 6, 7, 11, -5, 13, 9, 8, -24, 17, -10, 19, -11, 10, 13, 23, -35, 10, 15, -18, -17, 29, -20, 31, -38, 14, 19, 12, -50, 37, 21, 16, -57, 41, -30, 43, -29, -34, 25, 47, -45, 14, -38, 20, -35, 53, -70, 16, -79, 22, 31, 59, -80, 61, 33, -50

Offset

1,2

Comments

See A289310 for the real part of f and additional comments.

See A289320 for the square of the norm of f.

a(p) = p for any prime p.

The numbers 4 and 2700 are composite fixed points.

If a(n) = 0, then a(n^k) = 0 for any k > 0.

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

a(n) = 0 for n = 1, 378, 1296, 142884, 489888, 639846, 1679616, 1873638, ...

As a(378) = 0 and 378 = 2 * 3^3 * 7, we have arctan(2) + arctan(3)*3 + arctan(7) = j * Pi (with j = 2).

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Example

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

Mathematica

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

Prog

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

Crossrefs

Cf. A289310, A289320.

Sequence in context: A051598 A086993 A238714 * A351926 A345303 A345310

Adjacent sequences: A289308 A289309 A289310 * A289312 A289313 A289314

Keyword

sign

Author

Rémy Sigrist, Jul 02 2017

Status

approved