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

0,6

Comments

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

Links

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

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.

Prog

(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

Crossrefs

Cf. A316657, A316658.

Sequence in context: A067843 A245942 A280943 * A109360 A141622 A144136

Adjacent sequences: A316704 A316705 A316706 * A316708 A316709 A316710

Keyword

nonn,base

Author

Rémy Sigrist, Jul 11 2018

Status

approved