A053827 -id:A053827 - cp's OEIS frontend

A245321 Sum of digits of n written in fractional base 6/5.

0, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 9, 10, 11, 12, 13, 14, 12, 13, 14, 15, 16, 17, 14, 15, 16, 17, 18, 19, 15, 16, 17, 18, 19, 20, 15, 16, 17, 18, 19, 20, 20, 21, 22, 23, 24, 25, 19, 20, 21, 22, 23, 24, 23, 24, 25, 26, 27, 28, 21, 22, 23, 24, 25, 26, 24, 25

Offset: 0

Tom Edgar, Jul 18 2014

The base 6/5 expansion is unique and thus the sum of digits function is well-defined.

			In base 6/5 the number 15 is represented by 543 and so a(15) = 5 + 4 + 3 = 12.

Amiram Eldar, Table of n, a(n) for n = 0..10000 (terms 0..1000 from G. C. Greubel)
Index entries for sequences related to fractional bases.

Cf. A000120, A007953, A024638, A053827, A244040.

a:= proc(n) `if`(n<1, 0, irem(n, 6, 'q')+a(5*q)) end:
seq(a(n), n=0..100);  # Alois P. Heinz, Aug 19 2019

a[n_]:= a[n] = If[n==0, 0, a[5*Floor[n/6]] + Mod[n,6]]; Table[a[n], {n, 0, 70}] (* G. C. Greubel, Aug 19 2019 *)

a(n) = if(n == 0, 0, a(n\6 * 5) + n % 6); \\ Amiram Eldar, Jul 31 2025

def basepqsum(p,q,n):
    L=[n]
    i=1
    while L[i-1]>=p:
        x=L[i-1]
        L[i-1]=x.mod(p)
        L.append(q*(x//p))
        i+=1
    return sum(L)
[basepqsum(6,5,i) for i in [0..70]]

a(n) = A007953(A024638(n)).

A037322 Numbers whose base-5 and base-6 expansions have the same digit sum.

Original entry on oeis.org

1, 2, 3, 4, 45, 46, 47, 66, 67, 68, 69, 85, 86, 87, 88, 89, 145, 146, 147, 148, 149, 168, 169, 186, 187, 188, 189, 225, 226, 227, 265, 266, 267, 268, 269, 306, 307, 308, 309, 325, 326, 327, 328, 329, 370, 371, 408, 409, 490, 491

Offset: 1

Clark Kimberling

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Select[Range[500],Total[IntegerDigits[#,5]]==Total[IntegerDigits[#,6]]&] (* Harvey P. Dale, Mar 27 2022 *)

{n: A053824(n) = A053827(n).} - R. J. Mathar, Jun 30 2021

A089293 Sum of digits in the mixed-base enumeration system n=...d(4)d(3)d(2)d(1), where the digits satisfy 0<=d(i)<=1 if i is odd, 0<=d(i)<=2 if i is even.

Original entry on oeis.org

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

Offset: 0

John W. Layman, Jan 15 2004

Counting 0,1,2,3,... (base 10) in this mixed-base system proceeds as follows: 0,1,10,11,20,21,100,101,110,111,120,121,1000,.. = A109827.

			11(base 10) = 121(mixed-base), so a(11)=4.

Cf. A109827 (mixed-base).

Sums of digits in other bases: A000120 (binary), A053735 (ternary), A053827 (base 6).

a(n) = vecsum(digits(n,6)\/2); \\ Kevin Ryde, Aug 03 2021

a(n)=a(n-1)+1 if n=1, 3, 5 mod 6; a(n)=a(n-1) if n=2, 4 mod 5; a(n)=a(n/6) if n=0 mod 6.

A135738 Least positive integer with even digit sum in bases 2..n.

Original entry on oeis.org

3, 6, 10, 10, 54, 54, 54, 54, 130, 130, 130, 130, 390, 390, 2000, 2000, 3238, 3238, 4080, 4080, 7326, 7326, 16584, 16584, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 17310, 231000, 231000, 231000, 231000, 466352, 466352, 466352, 466352, 3020830

Offset: 2

M. F. Hasler, Dec 06 2007

The sequence is obviously increasing. It seems that a(2n+1) = a(2n) for n > 1. Is there a simple proof? Is there a simple way to construct a(n)? Notice the pattern in base N, e.g., 130 = 10000010_2 = 11211_3 = 2002_4 = 1010_5 = 334_6 = 244_7 = 202_8 = 154_9 = 109_11 = {10}{10}_12 = {10}0_13.

			a(2)=3 since 1=1_2, 2=10_2, so 3=11_2 is the number > 0 with even digit sum (1+1) in base 2.
a(3)=6 since 4=100_2, 5=12_3, so 6=20_3=110_2 is the least N > 0 with even digit sum in base 2 and in base 3.
a(4)=a(5)=10=1010_2=101_3=22_4=20_5 is the least N > 0 having even digit sum in bases 2 through 4 and has so also in base 5.

"Davar55" on mersenneforum.org, Puzzles / "Sum of digits".

Cf. A000120, A053735, A053737, A053824, A053827-A053836, A007953.

digitsum(n,b=10,s)={n=[n];while(n=divrem(n[1],b),s+=n[2]);s}
A135738(Bmax,n=1)={until(!n++,for(b=2,Bmax,digitsum(n,b)%2&next(2));return(n))} /* n-th element of the sequence */
t=1;for(b=2,100,print(b,":",t=A135738(b,t))) /* display the list */

Corrected example a(3)=5 to a(3)=6 David Yablon (davar55(AT)yahoo.com), Mar 19 2010

A037327 Numbers whose base-6 and base-7 expansions have the same digit sum.

Original entry on oeis.org

1, 2, 3, 4, 5, 66, 67, 68, 69, 126, 127, 128, 129, 130, 131, 156, 157, 158, 159, 160, 189, 190, 191, 246, 247, 248, 249, 250, 251, 280, 281, 308, 309, 310, 311, 366, 367, 368, 369, 370, 396, 397, 398, 456, 457, 458, 459, 460, 461, 518, 519, 520, 521, 546, 547

Offset: 1

Clark Kimberling

Cf. A053827, A053828.

isok(k) = sumdigits(k, 6) == sumdigits(k, 7); \\ Michel Marcus, Mar 18 2023

from numpy import base_repr
def ok(n):
    return sum(map(int, base_repr(n, 6))) == sum(map(int, base_repr(n, 7)))
print([n for n in range(1, 10**5) if ok(n)])
# Christoph B. Kassir, Apr 05 2023

{n: A053827(n) = A053828(n)}. - R. J. Mathar, Jun 30 2021

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A245321 Sum of digits of n written in fractional base 6/5.

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A037322 Numbers whose base-5 and base-6 expansions have the same digit sum.

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A089293 Sum of digits in the mixed-base enumeration system n=...d(4)d(3)d(2)d(1), where the digits satisfy 0<=d(i)<=1 if i is odd, 0<=d(i)<=2 if i is even.

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A135738 Least positive integer with even digit sum in bases 2..n.

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A037327 Numbers whose base-6 and base-7 expansions have the same digit sum.

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