A346690 Replace 6^k with (-1)^k in base-6 expansion of n.

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

Offset

0,3

Comments

If n has base-6 expansion abc..xyz with least significant digit z, a(n) = z - y + x - w + ...

Links

Robert Israel, Table of n, a(n) for n = 0..10000

Formula

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4) / (1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^6).

a(n) = n + 7 * Sum_{k>=1} (-1)^k * floor(n/6^k).

a(6*n+j) = j - a(n) for 0 <= j <= 5. - Robert Israel, Nov 21 2022

Example

59 = 135_6, 5 - 3 + 1 = 3, so a(59) = 3.

Maple

f:= proc(n) option remember; (n mod 6) - procname(floor(n/6)) end proc:
f(0):= 0:
map(f, [$1..100]); # Robert Israel, Nov 21 2022

Mathematica

nmax = 104; A[_] = 0; Do[A[x_] = x (1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4)/(1 - x^6) - (1 + x + x^2 + x^3 + x^4 + x^5) A[x^6] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
Table[n + 7 Sum[(-1)^k Floor[n/6^k], {k, 1, Floor[Log[6, n]]}], {n, 0, 104}]

Prog

(Python)
from sympy.ntheory.digits import digits
def a(n):
    return sum(bi*(-1)**k for k, bi in enumerate(digits(n, 6)[1:][::-1]))
print([a(n) for n in range(105)]) # Michael S. Branicky, Jul 29 2021
(PARI) a(n) = subst(Pol(digits(n, 6)), 'x, -1); \\ Michel Marcus, Nov 22 2022

Crossrefs

Cf. A007092, A053827, A055017, A065359, A065368, A346688, A346689, A346691.

Sequence in context: A256296 A256306 A030548 * A117724 A255826 A339256

Adjacent sequences: A346687 A346688 A346689 * A346691 A346692 A346693

Keyword

sign,base,easy

Author

Ilya Gutkovskiy, Jul 29 2021

Status

approved