A346689 Replace 5^k with (-1)^k in base-5 expansion of n.

0, 1, 2, 3, 4, -1, 0, 1, 2, 3, -2, -1, 0, 1, 2, -3, -2, -1, 0, 1, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, -2, -1, 0, 1, 2, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8

Offset: 0

Author

Ilya Gutkovskiy, Jul 29 2021

Keywords

• sign
• base

Comments

If n has base-5 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

Examples

			48 = 143_5, 3 - 4 + 1 = 0, so a(48) = 0.
		

Crossrefs

Cf. A007091, A053824, A055017, A065359, A065368, A346688, A346690, A346691.

Programs

Mathematica

nmax = 104; A[] = 0; Do[A[x] = x (1 + 2 x + 3 x^2 + 4 x^3)/(1 - x^5) - (1 + x + x^2 + x^3 + x^4) A[x^5] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 6 Sum[(-1)^k Floor[n/5^k], {k, 1, Floor[Log[5, n]]}], {n, 0, 104}]

Python

from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 5)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021

Formula

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3) / (1 - x^5) - (1 + x + x^2 + x^3 + x^4) * A(x^5).

a(n) = n + 6 * Sum_{k>=1} (-1)^k * floor(n/5^k).