A035327 -id:A035327 - cp's OEIS frontend

A171960 Values of the 2-complement of n in ternary representation.

2, 1, 0, 5, 4, 3, 2, 1, 0, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2

Offset: 0

Reinhard Zumkeller, Jan 20 2010

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A003462, A007089, A035327, A061601, A134025, A134026.

a[n_] := 3^(1 + Floor[Log[3, n]]) - n - 1; a[0] = 2; Array[a, 100] (* Amiram Eldar, Apr 03 2025 *)

a(n) = 3^(if(n,logint(n,3))+1) - 1 - n; \\ Kevin Ryde, Jul 16 2020

a(n) = if n < 3 then 2 - n else 3*a(floor(n/3)) + 2 - n mod 3.
a(A134026(n)) < A134026(n).
a(A003462(n)) = A003462(n).
a(A134025(n)) >= A134025(n).

A256293 Apply the transformation 0 -> 1 -> 2 -> 0 to the digits of n written in base 3, then convert back to base 10.

Original entry on oeis.org

1, 2, 0, 7, 8, 6, 1, 2, 0, 22, 23, 21, 25, 26, 24, 19, 20, 18, 4, 5, 3, 7, 8, 6, 1, 2, 0, 67, 68, 66, 70, 71, 69, 64, 65, 63, 76, 77, 75, 79, 80, 78, 73, 74, 72, 58, 59, 57, 61, 62, 60, 55, 56, 54, 13, 14, 12, 16, 17, 15

Offset: 0

M. F. Hasler, Mar 22 2015

Base 3 variant of A035327 (base 2) and A048379 (base 10).

See A256294 - A256299 for bases 4 through 9, and A256303 for the variant where the result is not converted back to base 10.

			a(3) = 7 because 3 = 10[3] becomes 21[3] = 7.
a(8) = 0 because 8 = 22[3] becomes 00[3] = 0.

Table[FromDigits[IntegerDigits[n,3]/.{0->1,1->2,2->0},3],{n,0,100}] (* Harvey P. Dale, Jan 21 2019 *)

A256293(n,b=3)=!n+apply(t->(t+1)%b,n=digits(n,b))*vector(#n,i,b^(#n-i))~

A256299 Apply the transformation 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 0 to the digits of n written in base 9, then convert back to base 10.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 0, 19, 20, 21, 22, 23, 24, 25, 26, 18, 28, 29, 30, 31, 32, 33, 34, 35, 27, 37, 38, 39, 40, 41, 42, 43, 44, 36, 46, 47, 48, 49, 50, 51, 52, 53, 45, 55, 56, 57, 58, 59, 60, 61, 62, 54, 64, 65, 66, 67, 68, 69

Offset: 0

M. F. Hasler, Mar 22 2015

Base 9 variant of A035327 (base 2) and A048379 (base 10). See A256293 - A256298 for bases 3 through 8, and A256289 for the variant where the result is not converted back to base 10.

			a(9) = 19 because 9 = 10[9] becomes 21[9] = 19.
a(80) = 0 because 80 = 88[9] becomes 00[9] = 0.

A256299(n,b=9)=!n+apply(t->(t+1)%b,n=digits(n,b))*vector(#n,i,b^(#n-i))~

A359208 Maximum value reached when starting from n during iteration of the map x->A359194(x) (binary complement of 3n), or -1 if infinite.

Original entry on oeis.org

0, 1, 2, 300, 300, 5, 300, 10, 10, 300, 10, 300, 328536, 300, 21, 300, 300, 328536, 300, 300, 300, 21, 72, 328536, 300, 328536, 661, 328536, 123130640068522377168864228132316865867184046004226894, 40, 300, 328536, 328536

Offset: 0

Joshua Searle, Dec 20 2022

It is unknown whether any terms are -1. The next term a(33) is equal to a(28), a 54-digit number. a(425720) is >= 2.09 * 10^114778, unresolved after 10^10 iterations.

a(425720) = 7.14... * 10^179246. - Joshua Searle, Jan 10 2023

			a(3) = 300 because the largest term in the iterated sequence: (3, 6, 13, 24, 55, 90, 241, 300, 123, 142, 85, 0) is 300.

Rémy Sigrist, Table of n, a(n) for n = 0..10000
Joshua Searle, Collatz-inspired sequences

Cf. A359194, A359207, A359209, A359215, A359218, A359219, A359220, A359221, A359222, A359255.

f[n_] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Table[Max@ NestWhileList[f, n, # != 0 &], {n, 0, 32}] (* Michael De Vlieger, Dec 21 2022 *)

f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
a(n) = my(x=n, m=n); while (m, m=f(3*m); if (m>x, x=m)); x; \\ Michel Marcus, Dec 21 2022

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def a(n):
    i, fi, m = 0, n, n
    while fi != 0: i, fi, m = i+1, f(fi), max(m, fi)
    return m
print([a(n) for n in range(33)]) # Michael S. Branicky, Dec 20 2022

A240857 Triangle read by rows: T(0,0) = 0; T(n+1,k) = T(n,k+1), 0 <= k < n; T(n+1,n) = T(n,0); T(n+1,n+1) = T(n,0)+1.

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, Apr 14 2014

Let h be the initial term of row n, to get row n+1, remove h and then append h and h+1.

			.   0:                                 0
.   1:                               0   1
.   2:                             1   0   1
.   3:                           0   1   1   2
.   4:                         1   1   2   0   1
.   5:                       1   2   0   1   1   2
.   6:                     2   0   1   1   2   1   2
.   7:                   0   1   1   2   1   2   2   3
.   8:                 1   1   2   1   2   2   3   0   1
.   9:               1   2   1   2   2   3   0   1   1   2
.  10:             2   1   2   2   3   0   1   1   2   1   2
.  11:           1   2   2   3   0   1   1   2   1   2   2   3
.  12:         2   2   3   0   1   1   2   1   2   2   3   1   2
.  13:       2   3   0   1   1   2   1   2   2   3   1   2   2   3
.  14:     3   0   1   1   2   1   2   2   3   1   2   2   3   2   3
.  15:   0   1   1   2   1   2   2   3   1   2   2   3   2   3   3   4 .

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened

Cf. A048881 (left edge), A000120 (right edge), A000788 (row sums), A000523 (row maxima), A240883 (central terms).

Cf. A035327.

a240857 n k = a240857_tabl !! n !! k
a240857_row n = a240857_tabl !! n
a240857_tabl = iterate (\(x:xs) -> xs ++ [x, x + 1]) [0]

T[n_, k_] := DigitCount[n + k + 1, 2, 1] - 1; Table[T[n, k], {n, 0, 12}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 01 2023 *)

from math import isqrt
def A240857(n): return (n-((r:=(m:=isqrt(k:=n+1<<1))+(k>m*(m+1)))*(r-3)>>1)).bit_count()-1 # Chai Wah Wu, Nov 11 2024

T(n,k) = A048881(n+k), 0 <= k <= n.

For n > 0: T(n,A035327(n)) = 0.

A326699 Numerator of the average position of a 1 in the reversed binary expansion of n.

Original entry on oeis.org

1, 2, 3, 3, 2, 5, 2, 4, 5, 3, 7, 7, 8, 3, 5, 5, 3, 7, 8, 4, 3, 10, 11, 9, 10, 11, 3, 4, 13, 7, 3, 6, 7, 4, 3, 9, 10, 11, 3, 5, 11, 4, 13, 13, 7, 15, 16, 11, 4, 13, 7, 14, 15, 4, 17, 5, 4, 17, 18, 9, 19, 4, 7, 7, 4, 9, 10, 5, 11, 4, 13, 11, 4, 13, 7, 14, 15, 4

Offset: 1

Gus Wiseman, Jul 20 2019

			The sequence of fractions begins: 1, 2, 3/2, 3, 2, 5/2, 2, 4, 5/2, 3, 7/3, 7/2, 8/3, 3, 5/2, 5, 3, 7/2, 8/3, 4.
For example, the reversed binary expansion of 18 is (0,1,0,0,1), and the average of {2,5} is 7/2, so a(18) = 7.

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

Cf. A000120, A029931, A035327, A048793, A070939, A291166, A295235, A326669, A326674, A326700 (denominators).

f:= proc(n) local L;
  L:= convert(n,base,2);
  L:= select(t -> L[t]=1, [$1..nops(L)]);
  numer(convert(L,`+`)/nops(L))
end proc:
map(f, [$1..100]); # Robert Israel, Oct 07 2019

Table[Numerator[Mean[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,100}]

A326700 Denominator of the average position of a 1 in the reversed binary expansion of n.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 2, 3, 1, 2, 1, 1, 2, 3, 1, 1, 3, 4, 2, 3, 3, 1, 1, 4, 2, 1, 1, 2, 1, 1, 2, 3, 3, 1, 1, 3, 1, 4, 3, 2, 4, 5, 2, 1, 3, 2, 3, 4, 1, 5, 1, 1, 4, 5, 2, 5, 1, 2, 1, 1, 2, 3, 1, 3, 1, 4, 2, 1, 3, 2, 3, 4, 1, 5, 1, 3, 3, 4, 1, 1, 4, 5

Offset: 1

Gus Wiseman, Jul 20 2019

The sequence of fractions begins: 1, 2, 3/2, 3, 2, 5/2, 2, 4, 5/2, 3, 7/3, 7/2, 8/3, 3, 5/2, 5, 3, 7/2, 8/3, 4.
For example, the reversed binary expansion of 18 is (0,1,0,0,1), and the average of {2,5} is 7/2, so a(18) = 2.
a(n) divides A000120(n). - Robert Israel, Oct 07 2019

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

Positions of 1's are A326669.

Cf. A000120, A029931, A035327, A048793, A070939, A291166, A295235, A326674, A326699 (numerators).

f:= proc(n) local L;
  L:= convert(n,base,2);
  L:= select(t -> L[t]=1, [$1..nops(L)]);
  denom(convert(L,`+`)/nops(L))
end proc:
map(f, [$1..100]); # Robert Israel, Oct 07 2019

Table[Denominator[Mean[Join@@Position[Reverse[IntegerDigits[n,2]],1]]],{n,100}]

A359209 Numbers that under iteration of the map x->A359194(x) (binary complement of 3n) until 0 is reached never exceed the initial term.

Original entry on oeis.org

0, 1, 2, 5, 10, 21, 39, 40, 42, 71, 72, 78, 85, 142, 150, 157, 163, 167, 168, 170, 285, 291, 300, 303, 311, 313, 315, 316, 317, 319, 320, 321, 322, 327, 328, 329, 331, 333, 334, 335, 336, 338, 339, 340, 341, 569, 571, 572, 573, 575, 576, 577, 578, 579

Offset: 1

Joshua Searle, Dec 20 2022

If it can be shown that all iterations of A359194 eventually reach one of the terms of this sequence, it would prove that all such trajectories are finite.
From M. F. Hasler, Dec 26 2022: (Start)
(1) If A359194(x) = a(k) for some a(k) < x, then x is also in the sequence.
(2) The orbit of any x under iterations of A359194 is finite if and only if it reaches a term of this sequence.
(3) No term has binary digits starting with '11...', i.e., all terms > 1 have binary digits starting '10...'.
(4) If the 3rd binary digit of a(n) is a '1', it cannot be followed by another bit 1, so all terms > 5 have a binary representation of the form '100...' or '1010...'
(5) A substring of '11', i.e., two consecutive bits 1, in a term of this sequence, is necessarily preceded by an earlier substring '00', i.e., two consecutive bits 0 to the left of the '11'. [This implies (3) and (4).] (End)

			40 is a term since its trajectory is 40 -> 7 -> 10 -> 1 -> 0, which never exceeds 40.

Cf. A035327, A359194, A359207, A359208, A359255.

bc:= n -> 2^(1+ilog2(n))-1-n: bc(0):= 1:
filter:= proc(n) local x;
  x:= n;
  while x <> 0 do
    x:= bc3(x);
    if x > n then return false fi;
  od;
  true
end proc:
select(filter, [$0..1000]); # Robert Israel, Dec 22 2022

f[n_] := BitXor[3 n, 2^IntegerPart[Log2[3 n] + 1] - 1]; Select[Range[0, 200], Function[n, AllTrue[NestWhileList[f, n, # != 0 &], # <= n &]]] (* Michael De Vlieger, Dec 21 2022 *)

f(n) = if(n, bitneg(n, exponent(n)+1), 1); \\ A035327
isok(m) = my(km=m); while (m, m=f(3*m); if (m>km, return(0))); return(1); \\ Michel Marcus, Dec 21 2022

def f(n): return 1 if n == 0 else (m:=3*n)^((1 << m.bit_length())-1)
def ok(n):
    i, fi, m = 0, n, n
    while fi != 0 and m <= n: i, fi, m = i+1, f(fi), max(m, fi)
    return m <= n
print([k for k in range(580) if ok(k)]) # Michael S. Branicky, Dec 20 2022

A167831 Largest m<=n such that no carry occurs when adding m to n in decimal arithmetic.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 3, 2, 1, 0, 10, 11, 12, 13, 14, 14, 13, 12, 11, 10, 20, 21, 22, 23, 24, 24, 23, 22, 21, 20, 30, 31, 32, 33, 34, 34, 33, 32, 31, 30, 40, 41, 42, 43, 44, 44, 43, 42, 41, 40, 49, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27

Offset: 0

Reinhard Zumkeller, Nov 14 2009

A167832(n) = a(n) + n.

R. Zumkeller, Table of n, a(n) for n = 0..9999

Cf. A167877, A035327 for the ternary and binary cases.

Cf. A031298.

a167831 n = head [x | let ds = a031298_row n, x <- [n, n-1 ..],
                      all (< 10) $ zipWith (+) ds (a031298_row x)]
-- Reinhard Zumkeller, Mar 15 2014

A167877 Largest m<=n such that no carry occurs when adding m to n in ternary arithmetic.

Original entry on oeis.org

0, 1, 0, 3, 4, 3, 2, 1, 0, 9, 10, 9, 12, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 27, 28, 27, 30, 31, 30, 29, 28, 27, 36, 37, 36, 39, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2

Offset: 0

Reinhard Zumkeller, Nov 14 2009

A167878(n) = a(n) + n.

R. Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A007089, A167831, A035327 for the decimal and binary cases.

Cf. A030341.

a167877 n = head [x | let ts = a030341_row n, x <- [n, n-1 ..],
                      all (< 3) $ zipWith (+) ts (a030341_row x)]
-- Reinhard Zumkeller, Mar 15 2014

cp's OEIS Frontend

A171960 Values of the 2-complement of n in ternary representation.

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A256293 Apply the transformation 0 -> 1 -> 2 -> 0 to the digits of n written in base 3, then convert back to base 10.

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A256299 Apply the transformation 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 0 to the digits of n written in base 9, then convert back to base 10.

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A359208 Maximum value reached when starting from n during iteration of the map x->A359194(x) (binary complement of 3n), or -1 if infinite.

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A240857 Triangle read by rows: T(0,0) = 0; T(n+1,k) = T(n,k+1), 0 <= k < n; T(n+1,n) = T(n,0); T(n+1,n+1) = T(n,0)+1.

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A326699 Numerator of the average position of a 1 in the reversed binary expansion of n.

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Maple

Mathematica

A326700 Denominator of the average position of a 1 in the reversed binary expansion of n.

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Maple

Mathematica

A359209 Numbers that under iteration of the map x->A359194(x) (binary complement of 3n) until 0 is reached never exceed the initial term.

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