A175048 -id:A175048 - cp's OEIS frontend

A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

3, 12, 7, 24, 51, 28, 15, 48, 99, 204, 103, 56, 115, 60, 31, 96, 195, 396, 199, 408, 819, 412, 207, 112, 227, 460, 231, 120, 243, 124, 63, 192, 387, 780, 391, 792, 1587, 796, 399, 816, 1635, 3276, 1639, 824, 1651, 828, 415, 224, 451, 908, 455, 920, 1843, 924

Offset: 1

Leroy Quet, Dec 02 2009

A318921 expands the runs in a similar way, and A318921(a(n)) = A001477(n). - Andrew Weimholt, Sep 08 2018
From Chai Wah Wu, Nov 18 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 3 and f(1) = a(2) + a(3) = 19.
Then f(k) = 20*6^(k-1) - 2^(k-1) for k > 0.
Proof: by summing over the recurrence relations for a(n) (see formula section), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (6*a(2i) + 6*a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 19 shows that f(k) = 20*6^(k-1)-2^(k-1) for k > 0.
(End)

			6 in binary is 110. Increase each run by one digit to get 11100, which is 28 in decimal. So a(6) = 28.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175047, A175048, A324127 (partial sums).
Cf. also A030308, A007088, A182512, A318921.
For records see A319422, A319423, A319424.

import Data.List (group)
a175046 = foldr (\b v -> 2 * v + b) 0 .
          concatMap (\bs@(b:_) -> b : bs) . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

a[n_] := (Append[#, #[[1]]]& /@ Split[IntegerDigits[n, 2]]) // Flatten // FromDigits[#, 2]&;
Array[a, 60] (* Jean-François Alcover, Nov 12 2018 *)

A175046(n)={for(i=2,#n=binary (n*2+bittest (n,0)),n[i]!=n[i-1]&&n[i-1]*=[1,1]);fromdigits(concat(n),2)} \\ M. F. Hasler, Sep 08 2018

from re import split
def A175046(n):
    return int(''.join(d+'1' if '1' in d else d+'0' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2) # Chai Wah Wu, Sep 24 2018

def a(n):
    b = bin(n)[2:]
    return int(b.replace("01", "001").replace("10", "110") + b[-1], 2)
print([a(n) for n in range(1, 55)]) # Michael S. Branicky, Dec 07 2021

2n+1 <= a(n) < 2*(n+1/n)^2; a(n) mod 4 = 3*(n mod 2). - M. F. Hasler, Sep 08 2018
a(n) <= (9*n^2 + 12*n)/5, with equality iff n = (2/3)*(4^k-1) = A182512(k) for some k, i.e., n = 10101...10 in binary. - Conjectured by N. J. A. Sloane, Sep 09 2018, proved by M. F. Hasler, Sep 12 2018
From M. F. Hasler, Sep 12 2018: (Start)
Proof of N. J. A. Sloane's formula: For given (binary) length L(n) = floor(log_2(n)+1), the length of a(n) is maximal, L(a(n)) = 2*L(n), if and only if n's bits are alternating, i.e., n in A020988 (if even) or in A002450 (if odd).
For n = A020988(k) (= k times '10' in base 2) = (4^k - 1)*2/3, one has a(n) = A108020(k) (= k times '1100' in base 2) = (16^k - 1)*4/5. This yields a(n)/n = (4^k + 1)*6/5 = (n*9 + 12)/5, i.e., the given upper bound.
For n = A002450(k) = (4^k - 1)/3, one gets a(n) = A182512(k) = (16^k - 1)/5, whence a(n)/n = (4^k + 1)*3/5 = (n*9 + 6)/5, smaller than the bound.
If L(a(n)) < 2 L(n) - 1, then log_2(a(n)) < floor(log_2(a(n))+1) = L(a(n)) <= 2*L(n) - 2 = 2*floor(log_2(n)+1)-2 = 2*floor(log_2(n)) <= 2*log_2(n), whence a(n) < n^2.
It remains to consider the case L(a(n)) = 2 L(n) - 1. There are two possibilities:
If n = 10..._2, then n >= 2^(L(n)-1) and a(n) = 1100..._2 < 1101_2 * 2^(L(a(n))-4) = 13*2^(2*L(n)-5), so a(n)/n^2 < 13*2^(-5+2) = 13/8 = 1.625 < 9/5 = 1.8.
If n = 11..._2, then n >= 3*2^(L(n)-2) and a(n) = 111..._2 < 2^L(a(n)) = 2^(2*L(n)-1), so a(n)/n^2 < 2^(-1+4)/9 = 8/9 < 1 < 9/5.
This shows that a(n)/n^2 <= 9/5 + 12/(5*n) always holds, with equality iff n is in A020988; and a(n)/n^2 < 13/8 if n is not in A020988 or A002450. (End)
From M. F. Hasler, Sep 10 2018: (Start)
Right inverse of A318921: A318921 o A175046 = id (= A001477).
a(A020988(k)) = A108020(k); a(A002450(k)) = A182512(k); a(A000225(k)) = A000225(k+1) (achieves the lower bound a(n) >= 2n + 1) for all k >= 0. (End)
From David A. Corneth, Sep 20 2018: (Start)
a(4*k)   = 2*a(2*k).
a(4*k+1) = 4*a(2*k) + 3.
a(4*k+2) = 4*a(2*k+1).
a(4*k+3) = 2*a(2*k+1) + 1. (End)

Extended by Ray Chandler, Dec 18 2009

A175047 Write n in binary, then increase each run of 0's by one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

1, 4, 3, 8, 9, 12, 7, 16, 17, 36, 19, 24, 25, 28, 15, 32, 33, 68, 35, 72, 73, 76, 39, 48, 49, 100, 51, 56, 57, 60, 31, 64, 65, 132, 67, 136, 137, 140, 71, 144, 145, 292, 147, 152, 153, 156, 79, 96, 97, 196, 99, 200, 201, 204, 103, 112, 113, 228, 115, 120, 121, 124, 63, 128

Offset: 1

Leroy Quet, Dec 02 2009

From Reinhard Zumkeller, Dec 12 2009: (Start)
A070939(a(n)) = A070939(n) + A033264(n);
A171598 and A171599 give record values and where they occur. (End)

			12 in binary is 1100. Increase each run of 0 by one digit to get 11000, which is 24 in decimal. So a(12) = 24.

R. Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A175046, A175048.
Cf. A084471. - Robert G. Wilson v, Dec 11 2009
Cf. A030308, A106151, A007088.

import Data.List (group)
a175047 = foldr (\b v -> 2 * v + b) 0 . concatMap
   (\bs@(b:_) -> if b == 0 then 0 : bs else bs) . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

f[n_] := Block[{s = Split@ IntegerDigits[n, 2]}, FromDigits[ Flatten@ Insert[s, {0}, Table[{2 i}, {i, Floor[ Length@s/2]} ]], 2]]; Array[ f, 64] (* Robert G. Wilson v, Dec 11 2009 *)

from re import split
def A175047(n):
  return int(''.join(d+'0' if '0' in d else d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2) # Chai Wah Wu, Nov 21 2018

a(n) = if n<2 then n else 2*(1 + 0^((n+2) mod 4))*a([n/2]) + n mod 2. - Reinhard Zumkeller, Jan 20 2010

a(2^n) = 2^(n+1). - Chai Wah Wu, Nov 21 2018

Extended by Ray Chandler, Dec 18 2009

a(11) onwards from Robert G. Wilson v and Reinhard Zumkeller, Dec 11 2009

A348710 In the binary expansion of n, decrease the length of each run of 1-bits by one.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 0, 0, 0, 1, 0, 0, 2, 3, 8, 4, 4, 5, 12, 6, 14, 15, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 16, 8, 8, 9, 8, 4, 10, 11, 24, 12, 12, 13, 28, 14, 30, 31, 0, 0, 0, 1, 0, 0, 2, 3, 0, 0, 0, 1, 4, 2, 6, 7, 0, 0, 0

Offset: 0

Kevin Ryde, Oct 30 2021

Equivalently, change bits 01 -> 0, including a 0 reckoned above the most significant 1-bit of n so change there.
A single 1-bit run decreases to nothing.  The Fibbinary numbers (A003714) are those n with only single 1-bits so that a(n) = 0 iff n is in A003714.
a(n) = 1 iff n is in A213540 since those values end with bits 011 (which become 01) and otherwise have only single 1-bits, as do the Fibbinary numbers.
Decreasing each run is the inverse of the increase A175048 so that a(A175048(k)) = k.  This n = A175048(k) is the smallest n with a(n) = k and then other occurrences of k are by inserting single 1-bits into this n, including anywhere above the most significant bit.

			n    = 14551 = binary 111 000 11 0 1 0 111
a(n) =   787 = binary  11 000  1 0   0  11

Cf. A007088 (binary), A175048 (increase 1-bits), A090077 (decrease to single 1-bits).
Cf. A003714 (indices of 0's), A213540 (indices of 1's).
Cf. A106151 (decrease 0-bits), A318921 (decrease each run).

Table[FromDigits[Flatten[Split@IntegerDigits[n,2]/. {1,a___}:>{a}],2],{n,0,82}] (* Giorgos Kalogeropoulos, Nov 01 2021 *)

a(n) = my(v=binary(n),t=0); for(i=2,#v, if(v[i-1]||!v[i], v[t++]=v[i])); fromdigits(v[1..t],2);

def a(n): return int(bin(n).replace("b", "").replace("01", "0"), 2)
print([a(n) for n in range(83)]) # Michael S. Branicky, Oct 31 2021

A266150 Take the binary representation of n, increase each run of 0's by one 0 if the length of run is odd, otherwise, if length of run is even, remove one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

0, 1, 4, 3, 2, 9, 12, 7, 16, 5, 36, 19, 6, 25, 28, 15, 8, 33, 20, 11, 18, 73, 76, 39, 48, 13, 100, 51, 14, 57, 60, 31, 64, 17, 132, 67, 10, 41, 44, 23, 144, 37, 292, 147, 38, 153, 156, 79, 24, 97, 52, 27, 50, 201, 204, 103, 112, 29, 228, 115, 30, 121, 124, 63, 32

Offset: 0

Alex Ratushnyak, Dec 21 2015

This is a self-inverse permutation of the positive integers.

			a(4) = 2 since 4 = 100 binary -> 10 = 2 decimal.
a(5) = 9 since 5 = 101 binary -> 1001 = 9 decimal.
a(6) = 12 since 6 = 110 binary -> 1100 = 12 decimal.

Cf. A007088, A084483, A162853, A175046, A175047, A175048, A266151.

Table[FromDigits[#, 2] &@ Flatten[If[First@ # == 0, If[OddQ@ Length@ #, Append[IntegerDigits@ #, 0], Most@ IntegerDigits@ #], #] & /@ Split@ IntegerDigits[n, 2]], {n, 64}] (* Michael De Vlieger, Dec 22 2015 *)

a(n) = if (n==0, 0, my (b=n%2, r=valuation(n+b, 2), rr=if (b, r, r%2, r+1, r-1)); (a(n\2^r)+b)*2^rr-b) \\ Rémy Sigrist, Jan 20 2019

a(0) = 0 prepended by Rémy Sigrist, Jan 20 2019

A266151 Take the binary representation of n, increase each run of 1's by one 1 if the length of run is odd, otherwise, if length of run is even, remove one 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

0, 3, 6, 1, 12, 27, 2, 15, 24, 51, 54, 13, 4, 11, 30, 7, 48, 99, 102, 25, 108, 219, 26, 111, 8, 19, 22, 5, 60, 123, 14, 63, 96, 195, 198, 49, 204, 411, 50, 207, 216, 435, 438, 109, 52, 107, 222, 55, 16, 35, 38, 9, 44, 91, 10, 47, 120, 243, 246, 61, 28, 59, 126, 31

Offset: 0

Alex Ratushnyak, Dec 21 2015

This is a self-inverse permutation of the positive integers.

			a(4) = 12 since 4 = 100 binary -> 1100 = 12 decimal,
a(5) = 27 since 5 = 101 binary -> 110011 = 27 decimal,
a(6) = 2 since 6 = 110 binary -> 10 = 2 decimal.

Cf. A007088, A084483, A162853, A175046, A175047, A175048, A266150.

Table[FromDigits[#, 2] &@ Flatten[If[First@ # == 1, If[OddQ@ Length@ #, Append[IntegerDigits@ #, 1], Most@ IntegerDigits@ #], #] & /@ Split@ IntegerDigits[n, 2]], {n, 63}] (* Michael De Vlieger, Dec 22 2015 *)

a(n) = if (n==0, 0, my (b=n%2, r=valuation(n+b, 2), rr=if (b==0, r, r%2, r+1, r-1)); (a(n\2^r)+b)*2^rr-b) \\ Rémy Sigrist, Jan 20 2019

a(0) = 0 prepended by Rémy Sigrist, Jan 20 2019

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A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

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A175047 Write n in binary, then increase each run of 0's by one 0. a(n) is the decimal equivalent of the result.

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A348710 In the binary expansion of n, decrease the length of each run of 1-bits by one.

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A266150 Take the binary representation of n, increase each run of 0's by one 0 if the length of run is odd, otherwise, if length of run is even, remove one 0. a(n) is the decimal equivalent of the result.

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A266151 Take the binary representation of n, increase each run of 1's by one 1 if the length of run is odd, otherwise, if length of run is even, remove one 1. a(n) is the decimal equivalent of the result.

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