A024631 -id:A024631 - cp's OEIS frontend

A245356 Number of numbers whose base-4/3 expansion (see A024631) has n digits.

4, 4, 4, 4, 8, 8, 12, 16, 20, 28, 36, 48, 64, 88, 116, 156, 208, 276, 368, 492, 656, 872, 1164, 1552, 2068, 2760, 3680, 4904, 6540, 8720, 11628, 15504, 20672, 27560, 36748, 48996, 65328, 87104, 116140, 154852, 206472, 275296, 367060, 489412, 652552, 870068

Offset: 1

Hailey R. Olafson, Jul 18 2014

			a(3) = 4 because 320, 321, 322, and 323 are the base-4/3 expansions for the numbers 9, 10, 11, and 12 respectively and these are the only numbers with 3 digits.

Cf. A072493, A024631, A081848.

A=[1]
for i in [1..60]:
    A.append(ceil(((4-3)/3)*sum(A)))
[4*x for x in A]

a(n) = 4*A072493(n).

A072493 a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 22, 29, 39, 52, 69, 92, 123, 164, 218, 291, 388, 517, 690, 920, 1226, 1635, 2180, 2907, 3876, 5168, 6890, 9187, 12249, 16332, 21776, 29035, 38713, 51618, 68824, 91765, 122353, 163138, 217517, 290023, 386697

Offset: 1

Benoit Cloitre, Nov 22 2002

nonn

Is this sequence, with its first 8 terms removed, the same as A005427? See also the similar conjecture with A005428/A073941. - Ralf Stephan, Apr 04 2003
Yes; the first 8 terms sum to 15, so upon dividing by 3 they are equivalent to the +5 in the formula for A005427. - Charlie Neder, Jan 10 2019
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture 1: a(n) equals the number of multiples of 3 whose representation in base 4/3 (see A024631) has n-1 digits. For example, a(8) = 4 because there are four multiples of 3 with n-1 = 7 digits in their representation in base 4/3: 33 = 3210201, 36 = 3210230, 39 = 3210233, and 42 = 3213122.
Conjecture 2: a(n) equals 1/4 times the number of nonnegative integers with the property that their 4/3-expansion has n digits (assuming that the 4/3-expansion of 0 has 1 digit). For example, a(7)*4 = 12 because the following 12 numbers have 4/3 expansions with n = 7 digits: 32 = 3210200, 33 = 3210201, 34 = 3210202, ..., 42 = 3213122, 43 = 3213123. (End)

Michel Marcus, Table of n, a(n) for n = 1..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.]
A. M. Odlyzko and H. S. Wilf, Functional iteration and the Josephus problem, Glasgow Math. J. 33(2) (1991), 235-240.
Index entries for sequences related to the Josephus Problem

Cf. A005427, A005428, A024631, A073941, A083286.

f[s_] := Append[s, Ceiling[Plus @@ s/3]]; Nest[f, {1}, 52] (* Robert G. Wilson v, Jul 07 2006 *)

lista(nn) = {va = vector(nn); va[1] = 1; for (n=2, nn, va[n] = ceil(sum(k=1, n-1, va[k])/3);); va;} \\ Michel Marcus, Apr 16 2015

a(n) = ceiling(c*(4/3)^n - 1/2) where c = 0.389324199524937508840138455...
From Petros Hadjicostas, Jul 21 2020: (Start)
Conjecture: The constant c above equals (3/16)*K(4), where K(q) = C(q/(q-1)) (q > 1) is described in Odlyzko and Wilf (1991).
For a > 1, the constant C(a) = limit_{n -> infinity} f_n(a)/a^n, where f_{n+1}(a) = ceiling(a*f_n(a)) for n >= 0 and f_0(a) = 1.
Thus, K(4) = limit_{n -> infinity} f_n(4/3)/(4/3)^n = 2.076395730799666... We have K(2) = 1 and K(3) = A083286 = 1.622270502884767315... (End)

A244041 Sum of digits of n written in fractional base 4/3.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 5, 6, 5, 6, 7, 8, 6, 7, 8, 9, 6, 7, 8, 9, 9, 10, 11, 12, 8, 9, 10, 11, 10, 11, 12, 13, 8, 9, 10, 11, 11, 12, 13, 14, 12, 13, 14, 15, 9, 10, 11, 12, 11, 12, 13, 14, 14, 15, 16, 17, 14, 15, 16, 17, 10, 11, 12, 13, 11, 12, 13, 14, 14, 15, 16, 17

Offset: 0

Hailey R. Olafson, Jun 17 2014

The base 4/3 expansion is unique and thus the sum of digits function is well-defined.

			In base 4/3 the number 14 is represented by 3212 and so a(14) = 3 + 2 + 1 + 2 = 8.

G. C. Greubel, Table of n, a(n) for n = 0..1000
F. M. Dekking, The Thue-Morse Sequence in Base 3/2, J. Int. Seq., Vol. 26 (2023), Article 23.2.3.
Kevin Ryde, Plot for Upper/Lower Bound Factors (and LaTeX source).
Index entries for sequences related to fractional bases

Cf. A000120, A007953, A024631, A053737, A244040.

p:=4; q:=3; a[n_]:= a[n]= If[n==0, 0, a[q*Floor[n/p]] + Mod[n, p]]; Table[a[n], {n,0,75}] (* G. C. Greubel, Aug 20 2019 *)

a(n) = p=4; q=3; if(n==0,0, a(q*(n\p)) + (n%p));
vector(75, n, n--; a(n)) \\ G. C. Greubel, Aug 20 2019

def base43sum(n):
    L, i = [n], 1
    while L[i-1]>3:
        x=L[i-1]
        L[i-1]=x.mod(4)
        L.append(3*floor(x/4))
        i+=1
    return sum(L)
[base43sum(n) for n in [0..75]]

a(n) = A007953(A024631(n)). - Michel Marcus, Jun 17 2014
a(n) < 3 log(n)/log(4/3) < 11 log(n) for n > 1. Possibly the constant factor can be replaced by 7 or 8. - Charles R Greathouse IV, Sep 22 2022
Conjecture: a(n) >> log(n), hence a(n) ≍ log(n). - Charles R Greathouse IV, Nov 03 2022

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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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).

C
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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

A245356 Number of numbers whose base-4/3 expansion (see A024631) has n digits.

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A072493 a(1) = 1 and a(n) = ceiling((Sum_{k=1..n-1} a(k))/3) for n >= 2.

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A244041 Sum of digits of 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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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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