A160385 -id:A160385 - cp's OEIS frontend

A276134 a(5n) = a(n), a(5n+1) = a(5n+2) = a(5n+3) = a(5n+4) = a(n) + 1, a(0) = 0.

0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2

Offset: 0

Ilya Gutkovskiy, Aug 21 2016

Number of nonzero digits in the base 5 representation of n.
Fixed point of the mapping 0 -> 01111, 1 -> 12222, 2 -> 23333, ...
Self-similar or fractal sequence (underlining every fifth term, reproduce the original sequence).

			The evolution starting with 0 is: 0 -> 01111 -> 0111112222122221222212222 -> ...
...
a(0) = 0;
a(1) = a(5*0+1) = a(0) + 1 = 1;
a(2) = a(5*0+2) = a(0) + 1 = 1;
a(3) = a(5*0+3) = a(0) + 1 = 1;
a(4) = a(5*0+4) = a(0) + 1 = 1;
a(5) = a(5*1+0) = a(1) = 1;
a(6) = a(5*1+1) = a(1) + 1 = 2, etc.
...
Also a(10) = 1, because 10 (base 10) = 20 (base 5) and 20 has 1 nonzero digit.

Robert Israel, Table of n, a(n) for n = 0..10000
Michael Gilleland, Some Self-Similar Integer Sequences
Ilya Gutkovskiy, Illustration (mapping 0 -> 01111, 1 -> 12222, 2 -> 23333, ...)

Cf. A000034, A000120, A093178, A160384, A160385.

f:= n -> nops(subs(0=NULL,convert(n,base,5))):
map(f, [$0..100]); # Robert Israel, Sep 07 2016

Join[{0}, Table[IntegerLength[n, 5] - DigitCount[n, 5, 0], {n, 120}]]

a(5^k) = 1.
a(5^k-1) = k.
a(5^k-m) = k, k>0, m = 2,3,4.
a(5^k+m) = 2, k>0, m = 1,2,3,4.
a(5^k-a(5^k)) = k.
a(5^k+(-1)^k) = (k + (-1)^k*(k - 1) + 3)/2.
a(5^k+(-1)^k-1) = A093178(k).
a(5^k+(-1)^k+1) = A000034(k+1), k>0.
G.f. g(x) satisfies g(x) = (1+x+x^2+x^3+x^4)*g(x^5) + (x+x^2+x^3+x^4)/(1-x^5). - Robert Israel, Sep 07 2016

A336797 Numbers, not divisible by 3, whose squares have exactly 4 nonzero digits in base 3.

Original entry on oeis.org

7, 14, 16, 17, 26, 35, 47, 68, 350, 3788

Offset: 1

Michel Marcus, Jan 27 2021

Is this sequence infinite?

Next term, if it exists, is > 3^500. - James Rayman, Feb 05 2021

			7^2=49 in base 3 is 1211, so 7 is a term.
14^2=196 in base 3 is 21021, so 14 is a term.

Alessio Moscariello, On sparse perfect powers, arXiv:2101.10415 [math.NT], 2021. See Question 11 p. 9.

Cf. A007089 (numbers in base 3), A160385.

Select[Range[4000], Mod[#, 3] > 0 && Length @ Select[IntegerDigits[#^2, 3], #1 > 0 &] == 4 &] (* Amiram Eldar, Jan 27 2021 *)

isok(n) = (n%3) && #select(x->x, digits(n^2, 3)) == 4;

from gmpy2 import isqrt, is_square
import itertools
N = 1000
powers = [1]
a_list = []
while len(powers) < N: powers.append(3 * powers[-1])
def attempt(n):
    if is_square(n): a_list.append(int(isqrt(n)))
for A, B, C in itertools.combinations(powers[1:], 3):
    for a, b, c in itertools.product([1, 2], repeat=3):
            attempt(a*A + b*B + c*C + 1)
print(sorted(a_list)) # James Rayman, Feb 05 2021

cp's OEIS Frontend

A276134 a(5n) = a(n), a(5n+1) = a(5n+2) = a(5n+3) = a(5n+4) = a(n) + 1, a(0) = 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula