A316707 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 square of the modulus of f(n).

0, 1, 1, 1, 1, 5, 10, 8, 2, 4, 5, 4, 10, 8, 2, 5, 2, 4, 10, 8, 5, 8, 2, 4, 10, 25, 32, 34, 20, 18, 50, 61, 61, 41, 41, 40, 45, 53, 37, 29, 10, 13, 17, 9, 5, 20, 29, 25, 13, 17, 25, 18, 32, 34, 20, 20, 17, 29, 25, 13, 50, 41, 61, 61, 41, 40, 29, 45, 53, 37, 10

Offset: 0

Author

Rémy Sigrist, Jul 11 2018

Keywords

• nonn
• base

Comments

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

Links

• Rémy Sigrist, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A316657, A316658.

Programs

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))); real(z)^2 + imag(z)^2

Formula

a(n) = A316657(n)^2 + A316658(n)^2.

a(5 * n) = 5 * a(n) for any n >= 0.

a(5^k) = 5^k for any k >= 0.