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A332413 a(n) is the imaginary part of f(n) = Sum_{d_k > 0} 3^k * i^(d_k-1) where Sum_{k >= 0} 5^k * d_k is the base 5 representation of n and i denotes the imaginary unit. Sequence A332412 gives real parts.

%I A332413 #12 Feb 15 2020 10:44:04
%S A332413 0,0,1,0,-1,0,0,1,0,-1,3,3,4,3,2,0,0,1,0,-1,-3,-3,-2,-3,-4,0,0,1,0,-1,
%T A332413 0,0,1,0,-1,3,3,4,3,2,0,0,1,0,-1,-3,-3,-2,-3,-4,9,9,10,9,8,9,9,10,9,8,
%U A332413 12,12,13,12,11,9,9,10,9,8,6,6,7,6,5,0,0,1,0
%N A332413 a(n) is the imaginary part of f(n) = Sum_{d_k > 0} 3^k * i^(d_k-1) where Sum_{k >= 0} 5^k * d_k is the base 5 representation of n and i denotes the imaginary unit. Sequence A332412 gives real parts.
%H A332413 Rémy Sigrist, <a href="/A332413/b332413.txt">Table of n, a(n) for n = 0..15624</a>
%F A332413 a(n) = 0 iff the n-th row of A031219 has neither 2's nor 4's.
%F A332413 a(5*n)   = 3*a(n).
%F A332413 a(5*n+1) = 3*a(n).
%F A332413 a(5*n+2) = 3*a(n) + 1.
%F A332413 a(5*n+3) = 3*a(n).
%F A332413 a(5*n+4) = 3*a(n) - 1.
%e A332413 For n = 103:
%e A332413 - 103 = 4*5^2 + 3*5^0,
%e A332413 - so f(123) = 3^2 * i^(4-1) + 3^0 * i^(3-1) = -1 - 9*i,
%e A332413 - and a(n) = -9.
%o A332413 (PARI) a(n) = { my (d=Vecrev(digits(n,5))); imag(sum (k=1, #d, if (d[k], 3^(k-1)*I^(d[k]-1), 0))) }
%Y A332413 Cf. A031219, A289814, A332412 (real parts and additional comments).
%K A332413 sign,base
%O A332413 0,11
%A A332413 _Rémy Sigrist_, Feb 12 2020