A316658 For any n >= 0 with base-5 expansion Sum_{k=0..w} d_k * 5^k, let f(n) = Sum_{k=0..w} [d_k > 0] * (2 + i)^k * i^(d_k - 1) (where [] is an Iverson bracket and i denotes the imaginary unit); a(n) equals the imaginary part of f(n).

0, 0, 1, 0, -1, 1, 1, 2, 1, 0, 2, 2, 3, 2, 1, -1, -1, 0, -1, -2, -2, -2, -1, -2, -3, 4, 4, 5, 4, 3, 5, 5, 6, 5, 4, 6, 6, 7, 6, 5, 3, 3, 4, 3, 2, 2, 2, 3, 2, 1, 3, 3, 4, 3, 2, 4, 4, 5, 4, 3, 5, 5, 6, 5, 4, 2, 2, 3, 2, 1, 1, 1, 2, 1, 0, -4, -4, -3, -4, -5, -3

Offset: 0

Author

Rémy Sigrist, Jul 09 2018

Keywords

• sign
• look
• base

Comments

See A316657 for the real part of f and additional comments.

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..15624

Crossrefs

Cf. A099456, A316657.

Programs

Mathematica

a[n_] := Module[{d, z}, d = IntegerDigits[n, 5] // Reverse; z = Sum[ If[d[[i]]>0, (2+I)^(i-1)*I^(d[[i]]-1), 0], {i, 1, Length[d]}]; Im[z]];
Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Nov 06 2021, after PARI code *)

PARI

a(n) = my (d=Vecrev(digits(n, 5)), z=sum(i=1, #d, if (d[i], (2+I)^(i-1) * I^(d[i]-1), 0))); imag(z)

Formula

a(5^n) = A099456(n-1) for any n > 0.