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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).

%I A289311 #28 Mar 14 2025 02:35:32
%S A289311 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,
%T A289311 15,-18,-17,29,-20,31,-38,14,19,12,-50,37,21,16,-57,41,-30,43,-29,-34,
%U A289311 25,47,-45,14,-38,20,-35,53,-70,16,-79,22,31,59,-80,61,33,-50
%N 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).
%C A289311 See A289310 for the real part of f and additional comments.
%C A289311 See A289320 for the square of the norm of f.
%C A289311 a(p) = p for any prime p.
%C A289311 The numbers 4 and 2700 are composite fixed points.
%C A289311 If a(n) = 0, then a(n^k) = 0 for any k > 0.
%C A289311 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).
%C A289311 a(n) = 0 for n = 1, 378, 1296, 142884, 489888, 639846, 1679616, 1873638, ...
%C A289311 As a(378) = 0 and 378 = 2 * 3^3 * 7, we have arctan(2) + arctan(3)*3 + arctan(7) = j * Pi (with j = 2).
%H A289311 Rémy Sigrist, <a href="/A289311/b289311.txt">Table of n, a(n) for n = 1..10000</a>
%e A289311 f(12) = f(2^2 * 3) = (1 + 2*I)^2 * (1 + 3*I) = -15 - 5*I, hence a(12) = -5.
%t A289311 Array[Im[Times @@ Map[(1 + #1 I)^#2 & @@ # &, FactorInteger@ #]] - Boole[# == 1] &, 63] (* _Michael De Vlieger_, Jul 03 2017 *)
%o A289311 (PARI) a(n) = my (f=factor(n)); imag (prod(i=1, #f~, (1 + f[i,1]*I) ^ f[i,2]))
%Y A289311 Cf. A289310, A289320.
%K A289311 sign
%O A289311 1,2
%A A289311 _Rémy Sigrist_, Jul 02 2017