A053735 -id:A053735 - cp's OEIS frontend

A134451 Ternary digital root of n.

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

Offset: 0

Reinhard Zumkeller, Oct 27 2007

Continued fraction expansion of sqrt(3) - 1. - N. J. A. Sloane, Dec 17 2007. Cf. A040001, A048878/A002530.
Minimum number of terms required to express n as a sum of odd numbers.
Shadow transform of even numbers A005843. - Michel Marcus, Jun 06 2013
From Jianing Song, Nov 01 2022: (Start)
For n > 0, a(n) is the minimal gap of distinct numbers coprime to n. Proof: denote the minimal gap by b(n). For odd n we have A058026(n) > 0, hence b(n) = 1. For even n, since 1 and -1 are both coprime to n we have b(n) <= 2, and that b(n) >= 2 is obvious.
The maximal gap is given by A048669. (End)

			n=42: A007089(42) = '1120', A053735(42) = 1+1+2+0 = 4,
A007089(4)='11', A053735(4)=1+1=2: therefore a(42) = 2.
0.732050807568877293527446341... = 0 + 1/(1 + 1/(2 + 1/(1 + 1/(2 + ...)))). - _Harry J. Smith_, May 31 2009

Harry J. Smith, Table of n, a(n) for n = 0..20000
Lorenz Halbeisen and Norbert Hungerbuehler, Number theoretic aspects of a combinatorial function, Notes on Number Theory and Discrete Mathematics 5(4) (1999), 138-150; see Definition 7 for the shadow transform.
N. J. A. Sloane, Transforms.
Eric Weisstein's World of Mathematics, Digital Root.
Eric Weisstein's World of Mathematics, Ternary.

Cf. A000010, A055034, A134452, A160390 (decimal expansion).
Apart from a(0) the same as A040001.
Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1).

a134451 = until (< 3) a053735
-- Reinhard Zumkeller, May 12 2011

A134451:=n->((n+1) mod 2)+2*signum(n)-1; seq(A134451(n), n=0..100); # Wesley Ivan Hurt, Dec 06 2013

Table[Mod[n + 1, 2] + 2 Sign[n] - 1, {n, 0, 100}] (* Wesley Ivan Hurt, Dec 06 2013 *)

a(n) = if(!n, 0, 2 - n%2) \\ Andrew Howroyd, Dec 08 2024

a(n) = n if n <= 2, otherwise a(A053735(n)).
a(A005408(n)) = 1; a(A005843(n)) = 2 for n>0;
a(n) = 0 if n=0, otherwise A000034(n-1).
a(n) = ((n+1) mod 2) + 2*sign(n) - 1. - Wesley Ivan Hurt, Dec 06 2013
Multiplicative with a(2^e) = 2, a(p^e) = 1 for odd prime p. - Andrew Howroyd, Aug 06 2018
a(0) = A055034(1) / A000010(1), a(n) = A000010(n+1) / A055034(n+1), n>1. - Torlach Rush, Oct 29 2019
Dirichlet g.f.: zeta(s)*(1+1/2^s). - Amiram Eldar, Jan 01 2023

A356867 For n >= 1, write n = 3^m + k, where m >= 0 is the greatest power of 3 <= n, and k is in the range 0 <= k < 3^(m+1) - 3^m, then for n such that k=0, a(n)=n, and for n such that k > 0, a(n) is the smallest prime multiple p*a(k), p != 3, that is not already a term.

Original entry on oeis.org

1, 2, 3, 5, 4, 6, 10, 8, 9, 7, 14, 15, 25, 20, 12, 50, 16, 18, 35, 28, 30, 125, 40, 24, 100, 32, 27, 11, 22, 21, 55, 44, 42, 70, 56, 45, 49, 98, 75, 175, 140, 60, 250, 80, 36, 245, 196, 150, 625, 200, 48, 500, 64, 54, 77, 110, 105, 275, 88, 84, 350, 112, 90, 343

Offset: 1

David James Sycamore, Sep 01 2022

Any prime p may be used to generate a sequence D(p) of this kind. The present sequence is D(3), and D(2) is the Doudna sequence, A005940.
Conjectured to be a permutation of the positive integers in which the primes appear in order.
From Antti Karttunen, Sep 16 2023: (Start)
The conjecture is true: Sequence is a permutation of natural numbers. By definition it is injective, and the surjectivity is guaranteed by the fact that there are infinitely many such n > k encountered by the greedy algorithm that a(n) will be a multiple of a(k), and "the smallest prime multiple" condition guarantees that all multiples of a(k) will eventually appear. That the primes and A100484 appear in order follows from the formulas a(3^m + 1) = prime(m+2), and a(3^m + 2) = 2*prime(m+2).
If the base-3 representation of n-1 has the base-3 representation of k-1 as its suffix, then a(n) is a multiple of a(k). For example, A007089(16-1) = 120, and A007089(43-1) = 1120, thus the former is the suffix of the latter, and a(16) = 50 indeed divides a(43) = 250.
(End)

			n=1=3^0+0 so a(1)=1. n=2=3^0+1 so k=1 and a(2)=2. Similarly a(3)=3 and a(9)=9.
n=10=3^2+1, therefore k=1 and a(1)=1 so a(10)=1*7=7 (since 2 and 5 have already occurred).

Michael De Vlieger, Table of n, a(n) for n = 1..19683 (19683 = 3^9)
Michael De Vlieger, Fan style ternary tree showing a(n) for n = 1..3^9, with a heat map color function for level m where 3^m is blue, smaller values are bluer, and larger are yellow-green. The smallest value in level m is shown in purple and largest is shown in red.
Index entries for sequences that are permutations of the natural numbers

Cf. A007089, A007949, A011655, A048473, A100484, A053735, A364958 (fixed points), A365390 (inverse permutation), A365424, A365459, A365462 [= a(n)-n], A365463 [= gcd(a(n),n)], A365464, A365465, A365717 [= A348717(a(1+n))], A365719 [= A046523(a(1+n))], A365721 [= omega(a(1+n))], A365722 [= bigomega(a(1+n))].

Cf. also A005940, A364611, A364628 for variants D(2), D(5) and D(7).

Block[{a, c, i, j, k, m, t, nn}, nn = 64; m = 1; i = 2; p = Prime[i]; c[] = False; Do[Set[{m, k}, {1, n - p^Floor[Log[p, n]]}]; If[k == 0, Set[{a[n], c[n]}, {n, True}], While[Set[t, Prime[m] a[k]]; Or[m == i, c[t]], m++]; Set[{a[n], c[t]}, {t, True}]], {n, nn}]; Array[a, nn] ] (* _Michael De Vlieger, Sep 01 2022 *)

up_to = 19683;
A356867list(up_to) = { my(v=vector(up_to),met=Map(),h=0,ak); for(i=1,#v,if(1==vecsum(digits(i,3)), v[i] = i; h = i, ak = v[i-h]; forprime(p=2,,if(3!=p && !mapisdefined(met,p*ak), v[i] = p*ak; break))); mapput(met,v[i],i)); (v); };
v356867 = A356867list(up_to);
A356867(n) = v356867[n]; \\ Antti Karttunen, Sep 15 2023

from sympy import nextprime
from sympy.ntheory import digits
from itertools import count, islice
def b(n): return n - 3**(len(digits(n,3)) - 2)
def agen():
    aset, alst = set(), [None]
    for n in count(1):
        k = b(n)
        if k == 0: an = n
        else:
            ak, p = alst[k], 2
            while p == 3 or p*ak in aset: p = nextprime(p)
            an = p*ak
        yield an; aset.add(an); alst.append(an)
print(list(islice(agen(), 64))) # Michael S. Branicky, Sep 02 2022

a(3^m + 1) = prime(m+2) for m >= 1.
Conjectures from Jianing Song, Nov 23 2022: (Start)
(1) a(3^m+2) = 2*prime(m+2) for m >= 2. - [The conjecture is true because a(2) = 2 and 3^m + 2 < 3^(1+m) + (3^m) + 1 for all m - Antti Karttunen, Sep 16 2023]
(2) For n > m >= 1, a(3^n+3^m+1) = prime(m+2)^2 for n = m+1; prime(n+2)*prime(m+2)^2 for n >= m+2.
(3) For n > m >= 1, a(3^n+3^m+2) = 4*prime(n+2) for n >= 3, m = 1; 2*prime(m+2)^2 for n = m+1, m >= 2; 2*prime(m+2)*prime(m+3) for n = m+2, m >= 2; 2*prime(n+2)*prime(m+2)^2 for n >= m+3, m >= 2. (End)
From Antti Karttunen, Sep 17 2023: (Start)
If A053735(n) = 1, then a(n) = n, otherwise a(n) = A365424(n) * a(A365459(n)).
For all n >= 1, A007949(a(n)) = A007949(n) and a(3*n) = 3*a(n).
For n >= 1, a(3^n - 1) = 2^(2n - 1), a(A048473(n)) = 2^(2*(n-1)).
These are conjectures so far:
For n >= 1, a(3^n - 2) = 10^(n-1).
For n >= 2, a(3^n - 3) = A002023(n-2) = 6*4^(n-2).
(End)

More terms from Michael De Vlieger, Sep 01 2022

A064150 Numbers divisible by the sum of their ternary digits.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 10, 12, 15, 16, 18, 20, 21, 24, 25, 27, 28, 30, 32, 33, 35, 36, 39, 40, 45, 48, 54, 56, 57, 60, 63, 64, 65, 72, 75, 77, 78, 80, 81, 82, 84, 87, 88, 90, 92, 93, 95, 96, 99, 100, 105, 108, 111, 112, 115, 117, 120, 132, 133, 135, 136, 144, 145, 150, 152

Offset: 1

Len Smiley, Sep 11 2001

a(n) mod A053735(a(n)) = 0. - Reinhard Zumkeller, Nov 25 2009

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000 (first 1000 terms from Harry J. Smith).
Paul Dahlenberg and Tom Edgar, Consecutive factorial base Niven numbers, Fibonacci Quarterly, Vol. 56, No. 2 (2018), pp. 163-166.

Cf. A005349 (Decimal), A049445 (Binary).

a064150 n = a064150_list !! (n-1)
a064150_list = filter (\x -> x `mod` a053735 x == 0) [1..]
-- Reinhard Zumkeller, Oct 28 2012

Select[Range[200], IntegerQ[#/(Plus@@IntegerDigits[#, 3])] &] (* Alonso del Arte, May 27 2011 *)

isok(m)={m % sumdigits(m, 3) == 0} \\ Harry J. Smith, Sep 09 2009

import numpy as np
def gen():
    for dec_num in range(1,153):
        tern_num = np.base_repr(dec_num, 3)
        sum_tern_digits = 0
        for i in tern_num:
            sum_tern_digits += int(i)
        if dec_num % sum_tern_digits == 0:
            yield dec_num
print(list((gen()))) # Adrienne Leonardo, Dec 28 2024

Corrected and extended by Vladeta Jovovic, Sep 22 2001

Offset corrected by Reinhard Zumkeller, Oct 28 2012

A138530 Triangle read by rows: T(n,k) = sum of digits of n in base k representation, 1<=k<=n.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 2, 1, 5, 2, 3, 2, 1, 6, 2, 2, 3, 2, 1, 7, 3, 3, 4, 3, 2, 1, 8, 1, 4, 2, 4, 3, 2, 1, 9, 2, 1, 3, 5, 4, 3, 2, 1, 10, 2, 2, 4, 2, 5, 4, 3, 2, 1, 11, 3, 3, 5, 3, 6, 5, 4, 3, 2, 1, 12, 2, 2, 3, 4, 2, 6, 5, 4, 3, 2, 1, 13, 3, 3, 4, 5, 3, 7, 6, 5, 4, 3, 2, 1, 14, 3, 4, 5, 6, 4, 2, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Reinhard Zumkeller, Mar 26 2008

A131383(n) = sum of n-th row;
A000027(n) = T(n,1);
A000120(n) = T(n,2) for n>1;
A053735(n) = T(n,3) for n>2;
A053737(n) = T(n,4) for n>3;
A053824(n) = T(n,5) for n>4;
A053827(n) = T(n,6) for n>5;
A053828(n) = T(n,7) for n>6;
A053829(n) = T(n,8) for n>7;
A053830(n) = T(n,9) for n>8;
A007953(n) = T(n,10) for n>9;
A053831(n) = T(n,11) for n>10;
A053832(n) = T(n,12) for n>11;
A053833(n) = T(n,13) for n>12;
A053834(n) = T(n,14) for n>13;
A053835(n) = T(n,15) for n>14;
A053836(n) = T(n,16) for n>15;
A007395(n) = T(n,n-1) for n>1;
A000012(n) = T(n,n).

			Start of the triangle for n in base k representation:
......................1
....................11....10
......... ........111....11...10
................1111...100...11..10
..............11111...101...12..11..10
............111111...110...20..12..11..10
..........1111111...111...21..13..12..11..10
........11111111..1000...22..20..13..12..11..10
......111111111..1001..100..21..14..13..12..11..10
....1111111111..1010..101..22..20..14..13..12..11..10
..11111111111..1011..102..23..21..15..14..13..12..11..10
111111111111..1100..110..30..22..20..15..14..13..12..11..10,
and the triangle of sums of digits starts:
......................1
.....................2...1
......... ..........3...2...1
...................4...1...2...1
..................5...2...3...2...1
.................6...2...2...3...2...1
................7...3...3...4...3...2...1
...............8...1...4...2...4...3...2...1
..............9...2...1...3...5...4...3...2...1
............10...2...2...4...2...5...4...3...2...1
...........11...3...3...5...3...6...5...4...3...2...1
..........12...2...2...3...4...2...6...5...4...3...2...1.

Alois P. Heinz, Rows n = 1..141, flattened
Eric Weisstein's World of Mathematics, Digit Sum

Cf. A007953. See A240236 for another version.

Cf. A002260.

a138530 n k = a138530_tabl !! (n-1) !! (k-1)
a138530_row n = a138530_tabl !! (n-1)
a138530_tabl = zipWith (map . flip q) [1..] a002260_tabl where
   q 1 n = n
   q b n = if n < b then n else q b n' + d where (n', d) = divMod n b
-- Reinhard Zumkeller, Apr 29 2015

T[n_, k_] := If[k == 1, n, Total[IntegerDigits[n, k]]];
Table[T[n, k], {n, 1, 14}, {k, 1, n}] // Flatten (* Jean-François Alcover, Oct 25 2021 *)

A054635 Champernowne sequence: write n in base 3 and juxtapose.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Apr 16 2000

Essentially the same as A003137. - R. J. Mathar, Aug 29 2009
An irregular table in which the n-th row lists the base-3 digits of n. - Jason Kimberley, Dec 07 2012
The base-3 Champernowne constant (A077771): it is normal in base 3. - Jason Kimberley, Dec 07 2012

Reinhard Zumkeller, Rows n = 0..1000 of triangle, flattened
Eric Weisstein's World of Mathematics, Ternary Champernowne Constant
Wikipedia, Ternary numeral system

Cf. A054637 (partial sums).
Cf. A081604 (row lengths), A053735 (row sums), A030341 (rows reversed), A007089, A077771.
Table in which the n-th row lists the base b digits of n: A030190 and A030302 (b=2), A003137 and this sequence (b=3), A030373 (b=4), A031219 (b=5), A030548 (b=6), A030998 (b=7), A031035 and A054634 (b=8), A031076 (b=9), A007376 and A033307 (b=10). - Jason Kimberley, Dec 06 2012

a054635 n k = a054635_tabf !! n !! k
a054635_row n = a054635_tabf !! n
a054635_tabf = map reverse a030341_tabf
a054635_list = concat a054635_tabf
-- Reinhard Zumkeller, Feb 21 2013

[0]cat &cat[Reverse(IntegerToSequence(n,3)):n in[1..31]]; // Jason Kimberley, Dec 07 2012

almostNatural[n_, b_] := Block[{m = 0, d = n, i = 1, l, p}, While[m <= d, l = m; m = (b - 1) i*b^(i - 1) + l; i++]; i--; p = Mod[d - l, i]; q = Floor[(d - l)/i] + b^(i - 1); If[p != 0, IntegerDigits[q, b][[p]], Mod[q - 1, b]]]; Array[ almostNatural[#, 3] &, 105, 0] (* Robert G. Wilson v, Jun 29 2014 *)
First[RealDigits[ChampernowneNumber[3], 3, 100, 0]] (* Paolo Xausa, Jun 19 2024 *)

from sympy.ntheory.digits import digits
def agen(limit):
    for n in range(limit):
        yield from digits(n, 3)[1:]
print([an for an in agen(35)]) # Michael S. Branicky, Sep 01 2021

A134599 Sum of digital sums (base 3) of the prime factors of n.

Original entry on oeis.org

0, 2, 1, 4, 3, 3, 3, 6, 2, 5, 3, 5, 3, 5, 4, 8, 5, 4, 3, 7, 4, 5, 5, 7, 6, 5, 3, 7, 3, 6, 3, 10, 4, 7, 6, 6, 3, 5, 4, 9, 5, 6, 5, 7, 5, 7, 5, 9, 6, 8, 6, 7, 7, 5, 6, 9, 4, 5, 5, 8, 5, 5, 5, 12, 6, 6, 5, 9, 6, 8, 7, 8, 5, 5, 7, 7, 6, 6, 7, 11, 4, 7, 3, 8, 8, 7, 4, 9, 5, 7, 6, 9, 4, 7, 6, 11, 5, 8, 5, 10, 5, 8, 5, 9, 7

Offset: 1

Hieronymus Fischer, Nov 11 2007

			a(6) = 3, since 6 = 2*3 and so a(6) = ds_3(2) + ds_3(3) = 2 + 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A053735, A080773 (base 2), A118503.

f[p_, e_] := e * DigitSum[p, 3]; a[1] = 0; a[n_] := Plus @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jul 30 2025 *)

a(n) = {my(f = factor(n)); sum(i = 1, #f~, f[i, 2] * sumdigits(f[i, 1], 3));} \\ Amiram Eldar, Jul 30 2025

If p_1*p_2*p_3* ... *p_m = n is the unique prime factorization of n, then a(n) = Sum_{k=1..m} ds_3(p_k), where ds_3 is the digital sum base 3.

Totally additive with a(p) = A053735(p). - Amiram Eldar, Jul 30 2025

a(1) = 0 prepended by Amiram Eldar, Jul 30 2025

A230640 Let M(1)=0 and for n>1, B(n)=(M(ceiling(n/2))+M(floor(n/2))+2)/2, M(n)=3^B(n)+M(floor(n/2))+1. This sequence gives M(n).

Original entry on oeis.org

0, 4, 28, 248, 129140168, 68630377364912, 2088595827392656793085408064780643444068898148936888424953199350296

Offset: 1

N. J. A. Sloane, Oct 31 2013

Max A. Alekseyev and N. J. A. Sloane, On Kaprekar's Junction Numbers, arXiv:2112.14365, 2021; Journal of Combinatorics and Number Theory 12:3 (2022), 115-155.

Cf. A230639.
Related base-3 sequences: A053735, A134451, A230641, A230642, A230643, A230853, A230854, A230855, A230856, A230639, A230640, A010063 (trajectory of 1)
Smallest number m such that u + (sum of base-b digits of u) = m has exactly n solutions, for bases 2 through 10: A230303, A230640, A230638, A230867, A238840, A238841, A238842, A238843, A006064.

f:=proc(n) option remember; local B, M;
if n<=1 then RETURN([0, 0]);
else
B:=(f(ceil(n/2))[2] + f(floor(n/2))[2] + 2)/2;
M:=3^B+f(floor(n/2))[2]+1; RETURN([B, M]); fi;
end proc;
[seq(f(n)[2], n=1..7)];

A053838 a(n) = (sum of digits of n written in base 3) modulo 3.

Original entry on oeis.org

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

Offset: 0

Henry Bottomley, Mar 28 2000

Start with 0, repeatedly apply the morphism 0->012, 1->120, 2->201. This is a ternary version of the Thue-Morse sequence A010060. See Brlek (1989). - N. J. A. Sloane, Jul 10 2012

A090193 is generated by the same mapping starting with 1. A090239 is generated by the same mapping starting with 2. - Andrey Zabolotskiy, May 04 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..2000
S. Brlek, Enumeration of factors in the Thue-Morse word, Discrete Applied Math. 24 (1989), 83-96.
Arthur Dolgopolov, Equitable Sequencing and Allocation Under Uncertainty, Preprint, 2016.
Glen Joyce C. Dulatre, Jamilah V. Alarcon, Vhenedict M. Florida, and Daisy Ann A. Disu, On Fractal Sequences, DMMMSU-CAS Science Monitor (2016-2017) Vol. 15 No. 2, 109-113.
Michael Gilleland, Some Self-Similar Integer Sequences
Michel Rigo, Relations on words, arXiv preprint arXiv:1602.03364 [cs.FL], 2016. See Example 17.
Robert Walker, Self Similar Sloth Canon Number Sequences
Index entries for sequences that are fixed points of mappings

Cf. A004128, A010060, A053837, A053839-A053844.

Equals A026600(n+1) - 1.

A053838 := proc(n)
    add(d,d=convert(n,base,3)) ;
    modp(%,3) ;
end proc:
seq(A053838(n),n=0..100) ; # R. J. Mathar, Nov 04 2017

Nest[ Flatten[ # /. {0 -> {0, 1, 2}, 1 -> {1, 2, 0}, 2 -> {2, 0, 1}}] &, {0}, 7] (* Robert G. Wilson v, Mar 08 2005 *)

a(n) = vecsum(digits(n, 3)) % 3; \\ Michel Marcus, May 04 2016

from sympy.ntheory import digits
def A053838(n): return sum(digits(n,3)[1:])%3 # Chai Wah Wu, Feb 28 2025

a(n) = A010872(A053735(n)) =(n+a(floor[n/3])) mod 3. So one can construct sequence by starting with 0 and mapping 0->012, 1->120 and 2->201 (e.g. 0, 012, 012120201, 012120201120201012201012120, ...) and looking at n-th digit of a term with sufficient digits.

a(n) = A004128(n) mod 3. [Gary W. Adamson, Aug 24 2008]

A244040 Sum of digits of n in fractional base 3/2.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 3, 4, 5, 3, 4, 5, 5, 6, 7, 4, 5, 6, 5, 6, 7, 7, 8, 9, 5, 6, 7, 5, 6, 7, 7, 8, 9, 8, 9, 10, 5, 6, 7, 7, 8, 9, 6, 7, 8, 7, 8, 9, 9, 10, 11, 9, 10, 11, 5, 6, 7, 7, 8, 9, 8, 9, 10, 6, 7, 8, 8, 9, 10, 8, 9, 10, 9, 10, 11, 11, 12, 13, 10, 11, 12, 5

Offset: 0

James Van Alstine, Jun 17 2014

The base 3/2 expansion is unique, and thus the sum of digits function is well-defined.

Fixed point starting with 0 of the two-block substitution a,b -> a,a+1,a+2 for a = 0,1,2,... and b = 0,1,2,.... - Michel Dekking, Sep 29 2022

			In base 3/2 the number 7 is represented by 211 and so a(7) = 2 + 1 + 1 = 4.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Michel Dekking, The Thue-Morse sequence in base 3/2, J. Int. Seq., Vol. 26 (2023), Article 23.2.3; arXiv preprint, arXiv:2301.13563 [math.CO], 2023.
Index entries for sequences related to fractional bases.

Cf. A024629, A007953, A000120, A053735, A244041, A246435.

a244040 0 = 0
a244040 n = a244040 (2 * n') + t where (n', t) = divMod n 3
-- Reinhard Zumkeller, Sep 05 2014

a[n_]:= a[n]= If[n==0, 0, a[2*Floor[n/3]] + Mod[n,3]]; Table[a[n], {n, 0, 85}] (* G. C. Greubel, Aug 20 2019 *)

a(n) = if(n == 0, 0, a(n\3 * 2) + n % 3); \\ Amiram Eldar, Jul 30 2025

a244040 = lambda n: a244040((n // 3) * 2) + (n % 3) if n else 0 # David Radcliffe, Aug 21 2021

def base32sum(n):
    L, i = [n], 1
    while L[i-1]>2:
        x=L[i-1]
        L[i-1]=x.mod(3)
        L.append(2*floor(x/3))
        i+=1
    return sum(L)
[base32sum(n) for n in [0..85]]

a(0) = 0, a(3n+r) = a(2n)+r for n >= 0 and r = 0, 1, 2. - David Radcliffe, Aug 21 2021

a(n) = A007953(A024629(n)). - Amiram Eldar, Jul 30 2025

A037301 Numbers whose base-2 and base-3 expansions have the same digit sum.

Original entry on oeis.org

0, 1, 6, 7, 10, 11, 12, 13, 18, 19, 21, 36, 37, 46, 47, 58, 59, 60, 61, 86, 92, 102, 103, 114, 115, 120, 121, 166, 167, 172, 173, 180, 181, 198, 199, 216, 217, 222, 223, 261, 273, 282, 283, 285, 298, 299, 300, 301, 306, 307, 309, 318

Offset: 1

Clark Kimberling

If Sum_{i=0..k} (binomial(k,i) mod 2) == Sum_{i=0..k} (binomial(k,i) mod 3) then k is in the sequence. (The converse does not hold.) - Benoit Cloitre, Nov 16 2003
Problem: To prove that the sequence is infinite. A generalization: Let s_m(k) denote the sum of digits of k in base m; does the Diophantine equation s_p(k) = s_q(k), where p,q are fixed distinct primes, have infinitely many solutions? - Vladimir Shevelev, Jul 30 2009
Also, numbers k such that the exponent of the largest power of 2 dividing k! is exactly twice the exponent of the largest power of 3 dividing k!. - Ivan Neretin, Mar 08 2015
a(5) = 10, a(6) = 11, a(7) = 12 and a(8) = 13 is the first time that four consecutive terms appear in this sequence. Conjecture: There is no occurrence of five or more consecutive terms of a(n). Tested by exhaustive search up to a(n) = 3^29. - Thomas König, Aug 15 2020

Stanislav Sykora, Table of n, a(n) for n = 1..10000
Vladimir Shevelev, Compact integers and factorials, Acta Arith. 126 (2007), no. 3, 195-236.
Vladimir Shevelev, Binomial predictors, arXiv:0907.3302 [math.NT], 2009.
Lukas Spiegelhofer, Collisions of the binary and ternary sum-of-digits functions, arXiv:2105.11173 [math.NT], 2021. Proves that sequence is infinite.

Cf. A000120, A053735, A180017.

Cf. A001316, A051638, A212222, A330904, A334765.

Select[ Range@ 320, Total@ IntegerDigits[#, 2] == Total@ IntegerDigits[#, 3] &] (* Robert G. Wilson v, Oct 24 2014 *)

is(n)=sumdigits(n,3)==hammingweight(n) \\ Charles R Greathouse IV, May 21 2015

A053735(a(n)) = A000120(a(n)); A180017(a(n)) = 0. - Reinhard Zumkeller, Aug 06 2010

Zero prepended by Zak Seidov, May 31 2010

cp's OEIS Frontend

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