A053737 -id:A053737 - cp's OEIS frontend

A309954 Product of digits of (n written in base 4).

0, 1, 2, 3, 0, 1, 2, 3, 0, 2, 4, 6, 0, 3, 6, 9, 0, 0, 0, 0, 0, 1, 2, 3, 0, 2, 4, 6, 0, 3, 6, 9, 0, 0, 0, 0, 0, 2, 4, 6, 0, 4, 8, 12, 0, 6, 12, 18, 0, 0, 0, 0, 0, 3, 6, 9, 0, 6, 12, 18, 0, 9, 18, 27, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 3, 0, 2, 4, 6, 0, 3, 6, 9, 0, 0, 0, 0, 0

Offset: 0

Ilya Gutkovskiy, Aug 24 2019

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A007090, A007954, A053737.

Table[Times @@ IntegerDigits[n, 4], {n, 0, 100}]

G.f. A(x) satisfies: A(x) = x * (1 + 2*x + 3*x^2) * (1 + A(x^4)).

A216789 Table read by antidiagonals: T(n,k) is the digital sum of k in base n displayed in decimal.

Original entry on oeis.org

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

Offset: 2

Henry Bottomley & Robert G. Wilson v, Sep 16 2012

T(n,k) is the least number of powers of n that add up to k. - Mohammed Yaseen, Nov 12 2022

			A000120   0, 1, 1, 2, 1, 2, 2, 3, 1, 2,  2,  3,  2,  3,  3,  4, 1, 2, 2
A053735   0, 1, 2, 1, 2, 3, 2, 3, 4, 1,  2,  3,  2,  3,  4,  3, 4, 5, 2
A053737   0, 1, 2, 3, 1, 2, 3, 4, 2, 3,  4,  5,  3,  4,  5,  6, 1, 2, 3
A053824   0, 1, 2, 3, 4, 1, 2, 3, 4, 5,  2,  3,  4,  5,  6,  3, 4, 5, 6
A053827   0, 1, 2, 3, 4, 5, 1, 2, 3, 4,  5,  6,  2,  3,  4,  5, 6, 7, 3
A053828   0, 1, 2, 3, 4, 5, 6, 1, 2, 3,  4,  5,  6,  7,  2,  3, 4, 5, 6
A053829   0, 1, 2, 3, 4, 5, 6, 7, 1, 2,  3,  4,  5,  6,  7,  8, 2, 3, 4
A053830   0, 1, 2, 3, 4, 5, 6, 7, 8, 1,  2,  3,  4,  5,  6,  7, 8, 9, 2
A007953   0, 1, 2, 3, 4, 5, 6, 7, 8, 9,  1,  2,  3,  4,  5,  6, 7, 8, 9
A053831   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,  1,  2,  3,  4,  5, 6, 7, 8
A053832   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11,  1,  2,  3,  4, 5, 6, 7
A053833   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,  1,  2,  3, 4, 5, 6
A053834   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,  1,  2, 3, 4, 5
A053835   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14,  1, 2, 3, 4
A053836   0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 2, 3

Robert Israel, Table of n, a(n) for n = 2..10012 (indices corrected to start at 2 by Sidney Cadot, Jan 07 2023)
K. Atanassov, On Some of Smarandache's Problems
Eric Weisstein's World of Mathematics, Digit Sum
Wikipedia, Digit sum

Cf. A000120, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836.

[seq(seq(convert(convert(n-b,base,b),`+`),b=n..2,-1),n=1..15)]; # Robert Israel, Aug 02 2020

DigitSum[n_, b_: 10] := Total[IntegerDigits[n, b]]; Table[ DigitSum[n - b, b], {n, 2, 13}, {b, n, 2, -1}] // Flatten

Name and offset corrected by Mohammed Yaseen, Nov 12 2022

A260112 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 4.

Original entry on oeis.org

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

Offset: 0

Peter Kagey, Jul 16 2015

a(n) = (Weight of quaternary expansion of n) + (length of quaternary expansion of n) - 1.

			For a(308) = 9, the nine steps are: 308 => 77 => 76 => 19 => 18 => 17 => 16 => 4 => 1 => 0.

Peter Kagey, Table of n, a(n) for n = 0..10000

Analogous sequences with a different multiplier k: A056792 (k=2), A061282 (k=3).

Cf. A053737, A110591, A007090: base 4 sequences.

c i = if i `mod` 4 == 0 then i `div` 4 else i - 1
b 0 foldCount = foldCount
b sheetCount foldCount = b (c sheetCount) (foldCount + 1)
a260112 n = b n 0 -- Peter Kagey, Sep 02 2015

a:= n-> (l-> nops(l)+add(i, i=l)-1)(convert(n, base, 4)):
seq(a(n), n=0..105);  # Alois P. Heinz, Jul 16 2015

a(n)=sumdigits(n,4)+#digits(n,4)-1 \\ Charles R Greathouse IV, Jul 16 2015

def a(n); n.to_s(4).length + n.to_s(4).split('').map(&:to_i).reduce(:+) - 1 end

a(n) = A053737(n) + A110591(n) - 1. - Michel Marcus, Jul 17 2015

A334841 a(0) = 0; for n > 0, a(n) = (number of 1's and 3's in base 4 representation of n) - (number of 0's and 2's in base 4 representation of n).

Original entry on oeis.org

0, 1, -1, 1, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -3, -1, -3, -1, -1, 1, -1, 1, -3, -1, -3, -1, -1, 1, -1, 1, -1, 1, -1, 1, 1, 3, 1, 3, -1, 1, -1, 1, 1, 3, 1, 3, -2, 0, -2, 0, 0, 2, 0, 2, -2, 0, -2, 0, 0, 2, 0, 2, 0, 2, 0, 2, 2, 4, 2, 4, 0

Offset: 0

Alexander Van Plantinga, May 13 2020

Values are even for base 4 representations of n with an even number of digits, and odd for base 4 representations of n with an odd number of digits, except for a(0).

			      n in    #odd    #even
  n  base 4  digits - digits = a(n)
  =  ======  =======================
  0    0        0   -            0
  1    1        1   -    0   =   1
  2    2        0   -    1   =  -1
  3    3        1   -    0   =   1
  4   10        1   -    1   =   0
  5   11        2   -    0   =   2
  6   12        1   -    1   =   0
  7   13        2   -    0   =   2

Cf. A053737, A010065, A037863, A007090, A301336, A333596.

a:= n-> `if`(n=0, 0, add(`if`(i in [1, 3], 1, -1), i=convert(n, base, 4))):
seq(a(n), n=0..100);  # Alois P. Heinz, May 30 2020

a[0] = 0; a[n_] := Total[-(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; Array[a, 100, 0] (* Amiram Eldar, May 13 2020 *)
Join[{0},Table[Total[If[EvenQ[#],-1,1]&/@IntegerDigits[n,4]],{n,90}]] (* Harvey P. Dale, Sep 06 2020 *)

a(n) = my(ret=0); if(n,forstep(i=0,logint(n,2),2, if(bittest(n,i),ret++,ret--))); ret; \\ Kevin Ryde, May 24 2020

import numpy as np
def qnary(n):
    e = n//4
    q = n%4
    if n == 0 : return 0
    if e == 0 : return q
    if e != 0 : return np.append(qnary(e), q)
m = 400
v = [0]
for i in range(1, m+1) :
    t = np.array(qnary(i))
    t[t%2 != 0] = 1
    t[t%2 == 0] = -1
    v = np.append(v, np.sum(t))

def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0 # Chai Wah Wu, Sep 03 2020

qnary = function(n, e, q){
  e = floor(n/4)
  q = n%%4
  if(n == 0 ){return(0)}
  if(e == 0){return(q)}
  else{return(c(qnary(e), (q)))}
}
m = 400
s = seq(2, m)
v = c(0)
for(i in s){
  x = qnary(i-1)
  x[which(x%%2!=0)] = 1
  x[which(x%%2==0)] = -1
  v[i] = sum(x)
}

a(n) = 2*A139351(n) - A110591(n), n>0. - R. J. Mathar, Sep 02 2020

A355487 Bitwise XOR of the base-4 digits of n.

Original entry on oeis.org

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

Offset: 0

Kevin Ryde, Jul 04 2022

Equivalently, the parity of the odd position 1-bits of n and the parity of the even position 1-bits of n, combined as a(n) = 2*A269723(n) + A341389(n).

In GF(2)[x] polynomials encoded as bits of an integer (least significant bit for the constant term), a(n) is remainder n mod x^2 + 1.

			n=35 has base-4 digits 203 so a(35) = 2 XOR 0 XOR 3 = 1.

Cf. A030373 (base 4 digits), A003987 (XOR).
Cf. A341389, A269723, A010060.
Cf. A353167 (indices of 0's).
Other digit operations: A053737 (sum), A309954 (product).

a[n_] := BitXor @@ IntegerDigits[n, 4]; Array[a, 100, 0] (* Amiram Eldar, Jul 05 2022 *)

a(n) = if(n==0,0, fold(bitxor,digits(n,4)));

from operator import xor
from functools import reduce
from sympy.ntheory import digits
def a(n): return reduce(xor, digits(n, 4)[1:])
print([a(n) for n in range(87)]) # Michael S. Branicky, Jul 05 2022

Fixed point of the morphism 0 -> 0,1; 1 -> 2,3; 2 -> 1,0; 3 -> 3,2 starting from 0.

A037316 Numbers whose base-4 and base-5 expansions have the same digit sum.

Original entry on oeis.org

1, 2, 3, 28, 29, 40, 41, 42, 43, 52, 53, 54, 76, 77, 78, 79, 90, 91, 100, 101, 102, 103, 115, 136, 137, 138, 139, 160, 161, 162, 163, 188, 189, 210, 211, 236, 237, 238, 239, 270, 271, 280, 281, 282, 283, 295, 305, 306, 307, 330, 331

Offset: 1

Clark Kimberling

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A053737 (base-4 sum of digits), A053824 (base-5 sum of digits).

filter:= n -> convert(convert(n,base,4),`+`)=convert(convert(n,base,5),`+`):
select(filter, [$1..1000]); # Robert Israel, Mar 11 2018

Select[Range[400],Total[IntegerDigits[#,4]]==Total[IntegerDigits[#,5]]&] (* Harvey P. Dale, Sep 15 2018 *)

isok(k) = sumdigits(k, 4) == sumdigits(k, 5); \\ Michel Marcus, Jun 02 2021

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

A269224 Factorial of the sum of digits of n in base 4.

Original entry on oeis.org

1, 1, 2, 6, 1, 2, 6, 24, 2, 6, 24, 120, 6, 24, 120, 720, 1, 2, 6, 24, 2, 6, 24, 120, 6, 24, 120, 720, 24, 120, 720, 5040, 2, 6, 24, 120, 6, 24, 120, 720, 24, 120, 720, 5040, 120, 720, 5040, 40320, 6, 24, 120, 720, 24, 120, 720, 5040, 120, 720, 5040, 40320, 720, 5040, 40320

Offset: 0

M. F. Hasler, Mar 15 2016

See sequences A093659, A269223 and A269221 for the base 2, base 3 and base 10 analog.

Cf. A000142, A053737, A166350, A093659, A269223, A269221.

Table[Total[IntegerDigits[n, 4]]!, {n, 0, 62}] (* Michael De Vlieger, Mar 15 2016 *)

A269224(n)=sumdigits(n,4)! \\ sumdigits(.,4) requires version >= 2.7; see A053737 for a substitute.

a(n) = vecsum(digits(n,4))!; \\ Michel Marcus, Mar 15 2016

a(n) = A000142(A053737(n)).

A333596 a(0) = 0; for n > 0, a(n) = a(n-1) + (number of 1's and 3's in base-4 representation of n) - (number of 0's and 2's in base-4 representation of n).

Original entry on oeis.org

0, 1, 0, 1, 1, 3, 3, 5, 3, 3, 1, 1, 1, 3, 3, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 18, 17, 14, 13, 12, 13, 12, 13, 10, 9, 6, 5, 4, 5, 4, 5, 4, 5, 4, 5, 6, 9, 10, 13, 12, 13, 12, 13, 14, 17, 18, 21, 19, 19, 17, 17, 17, 19, 19, 21, 19, 19

Offset: 0

Alexander Van Plantinga, Mar 27 2020

Local maxima values minus 1 are divisible by 4.

For a digit-wise recurrence, it's convenient to sum n terms so b(n) = a(n-1) = Sum_{i=0..n-1} A334841(i). Then b(4n+r) = 4*b(n) + r*A334841(n) + (1 if r even), for 0 <= r <= 3 and 4n+r >= 1. This is 4 copies of terms 0..n-1 and r copies of the following n. The new lowest digits cancel when r is odd, or net +1 when r is even. Repeatedly expanding gives the PARI code below. - Kevin Ryde, Jun 02 2020

			      n in    #odd    #even
  n  base 4  digits - digits + a(n-1) = a(n)
  =  ======  ===============================
  0    0        0   -                     0
  1    1        1   -    0   +    0   =   1
  2    2        0   -    1   +    1   =   0
  3    3        1   -    0   +    0   =   1
  4   10        1   -    1   +    1   =   1
  5   11        2   -    0   +    1   =   3
  6   12        1   -    1   +    3   =   3
  7   13        2   -    0   +    3   =   5

Cf. A053737, A010065, A037863, A268289, A007090, A231664, A301336, A334841.

a:= proc(n) option remember; `if`(n=0, 0, a(n-1) +add(
     `if`(i in [1, 3], 1, -1), i=convert(n, base, 4)))
    end:
seq(a(n), n=0..80);  # Alois P. Heinz, May 30 2020

f[n_] := Total[(-1)^(r = Range[0, 3]) * DigitCount[n, 4, r]]; a[0] = 0; a[n_] := a[n] = a[n - 1] - f[n]; Array[a, 100, 0] (* Amiram Eldar, Apr 24 2020 *)

a(n) = my(v=digits(n+1,4),s=0); for(i=1,#v, my(t=v[i]); v[i]=t*s+!(t%2); s-=(-1)^t); fromdigits(v,4); \\ Kevin Ryde, May 30 2020

b(n)=my(d=digits(n,4)); -sum(i=1,#d,(-1)^d[i])
first(n)=my(s); concat(0,vector(n,k,s+=b(k))) \\ Charles R Greathouse IV, Jul 04 2020

import numpy as np
def qnary(n):
    e = n//4
    q = n%4
    if n == 0 : return 0
    if e == 0 : return q
    if e != 0 : return np.append(qnary(e), q)
m = 400
v = [0]
for i in range(1,m+1) :
    t = np.array(qnary(i))
    t[t%2 != 0] = 1
    t[t%2 == 0] = -1
    v = np.append(v, np.sum([np.sum(t), v[i-1]]))

from itertools import accumulate
def A334841(n):
    return 2*bin(n)[-1:1:-2].count('1')-(len(bin(n))-1)//2 if n > 0 else 0
A333596_list = list(accumulate(A334841(n) for n in range(10000))) # Chai Wah Wu, Sep 03 2020

qnary = function(n, e, q){
  e = floor(n/4)
  q = n%%4
  if(n == 0 ){return(0)}
  if(e == 0){return(q)}
  else{return(c(qnary(e), (q)))}
}
m = 400
s = seq(2,m)
v = c(0)
for(i in s){
  x = qnary(i-1)
  x[which(x%%2!=0)] = 1
  x[which(x%%2==0)] = -1
  v[i] = sum(x,v[i-1])
}

A037309 Numbers whose base-3 and base-4 expansions have the same digit sum.

Original entry on oeis.org

1, 2, 21, 22, 23, 33, 34, 35, 40, 41, 78, 79, 88, 89, 100, 101, 112, 113, 136, 137, 150, 151, 156, 157, 158, 188, 204, 205, 206, 210, 211, 228, 229, 230, 236, 261, 262, 263, 273, 274, 275, 280, 281, 294, 295, 312, 313, 314, 328

Offset: 1

Clark Kimberling

{n: A053735(n) = A053737(n).} - R. J. Mathar, Jun 30 2021

cp's OEIS Frontend

A309954 Product of digits of (n written in base 4).

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

Formula

A216789 Table read by antidiagonals: T(n,k) is the digital sum of k in base n displayed in decimal.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Extensions

A260112 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 4.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

PARI

Ruby

Formula

A334841 a(0) = 0; for n > 0, a(n) = (number of 1's and 3's in base 4 representation of n) - (number of 0's and 2's in base 4 representation of n).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Python

R

Formula

A355487 Bitwise XOR of the base-4 digits of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Python

Formula

A037316 Numbers whose base-4 and base-5 expansions have the same digit sum.

Views

Author

Keywords

Links

Crossrefs

Programs

Maple

Mathematica

PARI

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

Views

Author

Keywords

Comments

Examples