

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).


4



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
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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



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



KEYWORD

sign


AUTHOR



STATUS

approved



