A244041 -id:A244041 - cp's OEIS frontend

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

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

A024631 n written in fractional base 4/3.

Original entry on oeis.org

0, 1, 2, 3, 30, 31, 32, 33, 320, 321, 322, 323, 3210, 3211, 3212, 3213, 32100, 32101, 32102, 32103, 32130, 32131, 32132, 32133, 321020, 321021, 321022, 321023, 321310, 321311, 321312, 321313, 3210200, 3210201, 3210202, 3210203, 3210230, 3210231

Offset: 0

David W. Wilson

G. C. Greubel, Table of n, a(n) for n = 0..1000
K. Burde, Das Problem der Abzählreime und Zahlentwicklungen mit gebrochenen Basen [The problem of counting rhymes and number expansions with fractional bases], J. Number Theory 26(2) (1987), 192-209. [The author deals with the representation of n in fractional bases k/(k-1) and its relation to counting-off games (variations of Josephus problem). Here k = 4. See the review in MathSciNet (MR0889384) by R. G. Stoneham.]
Index entries for sequences related to the Josephus Problem

Cf. A244041 (sum of digits).

Cf. A024629 (base 3/2).

a:= proc(n) `if`(n<1, 0, irem(n, 4, 'q')+a(3*q)*10) end:
seq(a(n), n=0..45);  # Alois P. Heinz, Aug 20 2019

p:= 4; q:= 3; a[n_]:= a[n]= If[n==0, 0, 10*a[q*Floor[n/p]] + Mod[n, p]]; Table[a[n], {n,0,40}] (* G. C. Greubel, Aug 20 2019 *)

a(n) = my(p=4, q=3); if(n==0,0, 10*a(q*(n\p)) + (n%p));
vector(40, n, n--; a(n)) \\ G. C. Greubel, Aug 20 2019

def basepqExpansion(p, q, n):
    L, i = [n], 1
    while L[i-1] >= p:
        x=L[i-1]
        L[i-1]=x.mod(p)
        L.append(q*(x//p))
        i+=1
    return Integer(''.join(str(x) for x in reversed(L)))
[basepqExpansion(4,3,n) for n in [0..40]] # G. C. Greubel, Aug 20 2019

To represent a number in base b, if a digit is greater than or equal to b, subtract b and carry 1. In fractional base a/b, subtract a and carry b.

A357425 Smallest number for which the sum of digits in fractional base 4/3 is n.

Original entry on oeis.org

0, 1, 2, 3, 5, 6, 7, 10, 11, 15, 21, 22, 23, 31, 39, 43, 54, 55, 74, 75, 101, 102, 103, 138, 139, 183, 187, 246, 247, 330, 331, 439, 443, 587, 783, 790, 791, 1047, 1355, 1398, 1399, 1866, 1867, 2487, 2491, 3318, 3319, 4199, 4427, 5903, 5911, 7882, 7883, 9959

Offset: 0

Kevin Ryde, Sep 28 2022

The sum of digits is A244041 and k = a(n) is the smallest A244041(k) = n.
Terms are never multiples of 4, after a(0)=0, since a multiple of 4 is a final 0 digit in base 4/3 which can be removed for the same digit sum.
Terms are strictly increasing (and so are indices of record highs in A244041) since a(n) - 1 has sum of digits n-1 and so is an upper bound for a(n-1).
If a(n) != 3 (mod 4), then the next term is a(n+1) = a(n) + 1 by incrementing the least significant digit.
If a(n) == 3 (mod 4), then an upper bound on the next term is a(n+1) <= (a(n) - r)*4/3 + r+1, where r = a(n) mod 3, by reducing the last digit to reach a multiple of 3 then append a suitable additional digit.

			For n=10, a(10) = 21 = 32131 in base 4/3 is the smallest number with sum of digits = 10.
For n=11, a(11) = 22 = 32132 in base 4/3, and which differs from a(10) simply by increasing the least significant base 4/3 digit.

Kevin Ryde, Table of n, a(n) for n = 0..472
Kevin Ryde, C Code (revised)
Index entries for sequences related to fractional bases

Cf. A024631 (base 4/3), A244041 (sum of digits), A363758.

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A087165 a(n)=1 when n == 1 (mod 4), otherwise a(n) = a(n - ceiling(n/4)) + 1. Removing all the 1's results in the original sequence with every term incremented by 1.

Original entry on oeis.org

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

Offset: 1

Paul D. Hanna, Aug 24 2003

Indices of records are given by A087192: a(A087192(n))=n, where A087192(n) = ceiling(A087192(n-1)*4/3).
From Benoit Cloitre, Mar 07 2009: (Start)
To construct the sequence:
Step 1: start from a sequence of 1's, leaving 3 undefined places between 1's, giving 1,(),(),(),1,(),(),(),1,(),(),(),1,(),(),(),1,(),(),(),1,...
Step 2: replace the first undefined place with a 2 and leave 3 undefined places between 2's, giving 1,2,(),(),1,(),2,(),1,(),(),2,1,(),(),(),1,2,(),(),1,...
Step 3: replace the first undefined place with a 3 and leave 3 undefined places between 3's, giving 1,2,3,(),1,(),2,(),1,3,(),2,1,(),(),3,1,2,(),(),1,...
Step 4: replace the first undefined place with a 4 and leave 3 undefined places between 4's, giving 1,2,3,4,1,(),2,(),1,3,(),2,1,4,(),3,1,2,(),(),1,...
Iterating the process indefinitely yields the sequence: 1,2,3,4,1,5,2,6,1,3,7,2,1,4,8,3,1,2,5,9,1,... (End)

Cf. A001511, A087088, A087192, A244041.

a(n+1) - a(n) = 4*A018902(n-3), n > 2.

for n from 1 to 100 do
  if n mod 4 = 1 then A[n]:= 1
  else A[n]:= A[n - ceil(n/4)] + 1
  fi
od:
seq(A[n],n=1..100); # Robert Israel, Aug 05 2014

a(n)=my(s); while(n>4, if(n%4==1, return(s+1)); n=(n\4*3)+max(n%4 - 1,0); s++); s+n \\ Charles R Greathouse IV, Sep 22 2022

a(n) = 4 + A244041(4*(n-1)) - A244041(4*n). - Tom Edgar and James Van Alstine, Aug 05 2014
a(4*n) = a(3*n)+1.
a(4*n+1) = 1.
a(4*n+2) = a(3*n+1)+1.
a(4*n+3) = a(3*n+2)+1. - Robert Israel, Aug 05 2014
a(n) < k*log(n) + 4 for n > 1 where k = 1/log(4/3) < 3.5. - Charles R Greathouse IV, Sep 22 2022

A363758 Maximum sum of digits for any number with n digits in fractional base 4/3.

Original entry on oeis.org

0, 3, 6, 8, 9, 12, 13, 15, 17, 19, 22, 24, 26, 28, 30, 32, 33, 36, 37, 40, 42, 44, 46, 48, 50, 52, 54, 56, 57, 60, 62, 65, 67, 70, 71, 73, 75, 77, 80, 83, 84, 87, 90, 93, 94, 96, 98, 101, 104, 106, 108, 109, 112, 115, 117, 120, 122, 123, 126, 129, 131, 133, 134

Offset: 0

Kevin Ryde, Jun 20 2023

This sequence is strictly increasing since if a(n) is attained by the sum of digits of k, then the final digit of k is 3 and (k - (k mod 3))*4/3 + 3 is the same digits with a new second-least significant 1, 2 or 3 inserted, and so a(n+1) >= a(n) + 1.

Terms can be derived from A357425 by a(n) = s for the largest s where A357425(s) has n digits in base 4/3.

			For n=9, the numbers with 9 digits in base 4/3 are 60 to 79 and among them the maximum sum of digits is A244041(75) = 19 (those digits being 321023323), and so a(9) = 19.

Kevin Ryde, Table of n, a(n) for n = 0..207
Kevin Ryde, Mean Digit Plot (and LaTeX source).
Index entries for sequences related to fractional bases

Cf. A024631 (base 4/3), A244041 (sum of digits).

Cf. A357425 (smallest with sum s), A087192.

a(n) = Max_{4*A087192(n-1) <= i < 4*A087192(n)} A244041(i), for n>=2.

A364779 Largest integer with sum of digits n in fractional base 4/3.

Original entry on oeis.org

0, 1, 2, 4, 5, 8, 16, 17, 32, 44, 80, 256, 257, 344, 460, 464, 620, 1472, 1964, 2620, 2624, 3500, 6224, 8300, 11068, 11072, 26240, 34988, 46652, 262144, 262145, 349528, 349529, 466040, 621392, 828524, 1104700, 1532816, 3633344, 6459280, 6459281, 11483168, 19616912

Offset: 0

Kevin Ryde, Aug 13 2023

A largest integer exists since only a finite number of trailing 0 digits are possible, since each is a factor 4/3.
Each term k >= 3 has final digit d = k mod 4 which is always d < r where r = k mod 3 (and hence d = 0 or 1), since otherwise (k - r)*4/3 + r would split d into two final digits {d-r, r} for a larger number with the same sum of digits.
This sequence is strictly increasing since final digit d = 0 or 1 (and also a(2) = 2) can be incremented so that a(n)+1 is a candidate value for a(n+1).

Kevin Ryde, Table of n, a(n) for n = 0..150
Kevin Ryde, C Code
Index entries for sequences related to fractional bases

Cf. A024631 (base 4/3), A244041 (sum of digits).

Cf. A357425 (smallest of sum), A364780 (count by sum).

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A364780 Number of numbers with sum of digits n in fractional base 4/3.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 4, 3, 5, 6, 7, 14, 13, 15, 19, 19, 30, 39, 45, 56, 65, 75, 95, 124, 140, 174, 216, 268, 338, 417, 501, 627, 780, 974, 1203, 1454, 1825, 2266, 2769, 3427, 4268, 5188, 6433, 7930, 9671, 12000, 14738, 18265, 22642, 27961, 34528, 42523, 52325, 64425

Offset: 0

Kevin Ryde, Aug 13 2023

Only a finite number of numbers have sum of digits n (the largest is A364779(n)).

Kevin Ryde, Table of n, a(n) for n = 0..150
Kevin Ryde, C Code
Index entries for sequences related to fractional bases

Cf. A024631 (base 4/3), A244041 (sum of digits).
Cf. A357425 (smallest), A364779 (largest).
Cf. A245356 (count by length).

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A364751 Minimum sum of digits for any number of length n digits in fractional base 4/3.

Original entry on oeis.org

0, 3, 5, 6, 6, 8, 8, 9, 10, 10, 11, 11, 11, 11, 13, 14, 16, 17, 17, 17, 18, 19, 21, 22, 22, 23, 24, 26, 26, 26, 27, 28, 29, 29, 29, 29, 29, 29, 31, 33, 34, 35, 36, 37, 38, 38, 38, 39, 39, 41, 41, 42, 42, 43, 43, 45, 45, 46, 46, 48, 50, 50, 52, 52, 52, 52, 53, 55

Offset: 1

Kevin Ryde, Sep 07 2023

0 is taken to be 1 digit long so a(1) = 0.

Terms can be derived from A364779 by a(n) = s for the smallest s where k = A364779(s) is >= n digits long (noting that stripping trailing 0's from k suffices to show numbers with sum of digits s exist at each length down to where sum s-1 exists).

Kevin Ryde, Table of n, a(n) for n = 1..209
Kevin Ryde, Mean Digit Plot
Index entries for sequences related to fractional bases

Cf. A024631 (base 4/3), A244041 (sum of digits), A364779 (largest with sum).

Cf. A363758 (maximum sum).

a(n) = Min_{4*A087192(n-1) <= k < 4*A087192(n)} A244041(k), for n >= 2.

cp's OEIS Frontend

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

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A024631 n written in fractional base 4/3.

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A357425 Smallest number for which the sum of digits in fractional base 4/3 is n.

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A087165 a(n)=1 when n == 1 (mod 4), otherwise a(n) = a(n - ceiling(n/4)) + 1. Removing all the 1's results in the original sequence with every term incremented by 1.

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A363758 Maximum sum of digits for any number with n digits in fractional base 4/3.

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A364779 Largest integer with sum of digits n in fractional base 4/3.

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A364780 Number of numbers with sum of digits n in fractional base 4/3.

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A364751 Minimum sum of digits for any number of length n digits in fractional base 4/3.

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