A319423 -id:A319423 - 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

A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

2, 9, 6, 17, 38, 25, 14, 33, 70, 153, 78, 49, 102, 57, 30, 65, 134, 281, 142, 305, 614, 313, 158, 97, 198, 409, 206, 113, 230, 121, 62, 129, 262, 537, 270, 561, 1126, 569, 286, 609, 1222, 2457, 1230, 625, 1254, 633, 318, 193, 390, 793, 398, 817, 1638, 825, 414

Offset: 1

Chai Wah Wu, Oct 04 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 21 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) = 2 and f(1) = a(2) + a(3) = 15.
Then f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. By summing over the recurrence relations for a(n), 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} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first order recurrence relation with the initial condition f(1) = 15 shows that f(k) = 16*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by appending the opposite digit to get 11001, which is 25 in decimal. So a(6) = 25.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320038.

from re import split
def A320037(n):
    return int(''.join(d+'0' if '1' in d else d+'1' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2a(2n)-1, a(4n+1) = 4a(2n)+2, a(4n+2) = 4a(2n+1)+1 and a(4n+3) = 2a(2n+1)+2. - Chai Wah Wu, Nov 21 2018

A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

1, 6, 3, 12, 25, 14, 7, 24, 49, 102, 51, 28, 57, 30, 15, 48, 97, 198, 99, 204, 409, 206, 103, 56, 113, 230, 115, 60, 121, 62, 31, 96, 193, 390, 195, 396, 793, 398, 199, 408, 817, 1638, 819, 412, 825, 414, 207, 112, 225, 454, 227, 460, 921, 462, 231, 120, 241

Offset: 1

Chai Wah Wu, Oct 04 2018

nonn

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 25 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) = 1 and f(1) = a(2) + a(3) = 9.
Then f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
Proof: looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), 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} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 9 shows that f(k) = 10*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by prepending the opposite digit to get 01110, which is 14 in decimal. So a(6) = 14.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037.

a[n_] := Split[IntegerDigits[n, 2]] /. {a0:{(0)...} :> Prepend[a0, 1], a1:{(1)...} :> Prepend[a1, 0]} // Flatten // FromDigits[#, 2]&;
Array[a, 60] (* Jean-François Alcover, Dec 02 2018 *)

from re import split
def A320038(n):
    return int(''.join('0'+d if '1' in d else '1'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(n) = floor(A175046(n)/2).

a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+1, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018

A319422 Records in A175046.

Original entry on oeis.org

3, 12, 24, 51, 99, 204, 396, 408, 819, 1587, 1635, 3276, 6348, 6540, 6552, 13107, 25395, 26163, 26211, 52428, 101580, 104652, 104844, 104856, 209715, 406323, 418611, 419379, 419427, 838860, 1625292, 1674444, 1677516, 1677708, 1677720, 3355443, 6501171

Offset: 1

N. J. A. Sloane, Sep 23 2018

nonn

a(n) in binary is either an alternating string of 11's and 00's, or exactly one of the 00's is replaced by 000 (see A319423). - Chai Wah Wu, Oct 04 2018

Chai Wah Wu, Table of n, a(n) for n = 1..10099
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018

Cf. A175046, A319423, A319424 (written in binary).

b[n_] := (Append[#, #[[1]]]& /@ Split[IntegerDigits[n, 2]]) // Flatten // FromDigits[#, 2]&;
Reap[For[rec = 0; k = 1, k < 3000, k++, bk = b[k]; If[bk > rec, rec = bk; Sow[rec]]]][[2, 1]] (* Jean-François Alcover, Nov 12 2018 *)

A319424 Records in A175046 (but written in base 2).

Original entry on oeis.org

11, 1100, 11000, 110011, 1100011, 11001100, 110001100, 110011000, 1100110011, 11000110011, 11001100011, 110011001100, 1100011001100, 1100110001100, 1100110011000, 11001100110011, 110001100110011, 110011000110011, 110011001100011, 1100110011001100, 11000110011001100, 11001100011001100, 11001100110001100, 11001100110011000, 110011001100110011, 1100011001100110011

Offset: 1

N. J. A. Sloane, Sep 23 2018

More terms than usual are given in order to display the pattern.
Conjecture. a(n) is either an alternating string of 11's and 00's, or exactly one of the 00's is replaced by 000.
Conjecture is true (see A319423). - Chai Wah Wu, Oct 04 2018

Chai Wah Wu, Table of n, a(n) for n = 1..10099
N. J. A. Sloane, Transforms
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018

Cf. A175406, A319422, A319423.

A320039 Write n in binary, then modify each run of 0's and each run of 1's by appending a 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

3, 13, 7, 25, 55, 29, 15, 49, 103, 221, 111, 57, 119, 61, 31, 97, 199, 413, 207, 441, 887, 445, 223, 113, 231, 477, 239, 121, 247, 125, 63, 193, 391, 797, 399, 825, 1655, 829, 415, 881, 1767, 3549, 1775, 889, 1783, 893, 447, 225, 455, 925, 463, 953, 1911, 957

Offset: 1

Chai Wah Wu, Oct 05 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 21 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) = 20.
Then f(k) = 21*6^(k-1) - 2^(k-1) for k >= 0.
Proof: the equation for f is true for k = 0. Looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n)-1, a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+1 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), 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} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 20 shows that f(k) = 21*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by appending a 1 to get 11101, which is 29 in decimal. So a(6) = 29.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038.

Array[FromDigits[Flatten@ Map[Append[#, 1] &, Split@ IntegerDigits[#, 2]], 2] &, 54] (* Michael De Vlieger, Nov 23 2018 *)

from re import split
def A320039(n):
    return int(''.join(d+'1' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2*a(2n)-1, a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+1 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 21 2018

A320261 Write n in binary, then modify each run of 0's and each run of 1's by prepending a 1. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

3, 14, 7, 28, 59, 30, 15, 56, 115, 238, 119, 60, 123, 62, 31, 112, 227, 462, 231, 476, 955, 478, 239, 120, 243, 494, 247, 124, 251, 126, 63, 224, 451, 910, 455, 924, 1851, 926, 463, 952, 1907, 3822, 1911, 956, 1915, 958, 479, 240, 483, 974, 487, 988, 1979, 990

Offset: 1

Chai Wah Wu, Oct 08 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 25 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) = 21, and f(2) = a(4)+a(5)+a(6)+a(7) = 132.
Then f(k) = 135*6^(k-2) - 3*2^(k-2) for k >= 0.
Proof: the formula is true for k = 0. Looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. By summing over the recurrence relations for a(n), 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} (6a(2i) + 6a(2i+1) + 6) = 6*f(k+1) + 3*2^(k+1). Solving this first-order recurrence relation with the initial condition f(1) = 21 shows that f(k) = 135*6^(k-2) - 3*2^(k-2) for k > 0.
(End)

			6 in binary is 110. Modify each run by prepending a 1 to get 11110, which is 30 in decimal. So a(6) = 30.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038, A320039.

Table[FromDigits[Flatten[Join[{1},#]&/@Split[IntegerDigits[n,2]]],2],{n,60}] (* Harvey P. Dale, Apr 26 2019 *)

from re import split
def A320261(n):
    return int(''.join('1'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+3, a(4n+2) = 4*a(2n+1)+2 and a(4n+3) = 2*a(2n+1)+1. - Chai Wah Wu, Nov 25 2018

A320262 Write n in binary, then modify each run of 0's and each run of 1's by appending a 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

2, 8, 6, 16, 34, 24, 14, 32, 66, 136, 70, 48, 98, 56, 30, 64, 130, 264, 134, 272, 546, 280, 142, 96, 194, 392, 198, 112, 226, 120, 62, 128, 258, 520, 262, 528, 1058, 536, 270, 544, 1090, 2184, 1094, 560, 1122, 568, 286, 192, 386, 776, 390, 784, 1570, 792, 398

Offset: 1

Chai Wah Wu, Oct 08 2018

A variation of A175046. Indices of record values are given by A319423.
From Chai Wah Wu, Nov 21 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) = 2 and f(1) = a(2) + a(3) = 14.
Then f(k) = 15*6^(k-1) - 2^(k-1) for k >= 0.
Proof: the equation for f is true for k = 0. Looking at the last 2 bits of n, it is easy to see that a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+2, a(4n+2) = 4*a(2n+1) and a(4n+3) = 2*a(2n+1)+2. By summing over the recurrence relations for a(n), 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} (6a(2i) + 6a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 14 shows that f(k) = 15*6^(k-1) - 2^(k-1) for k > 0.
(End)

			6 in binary is 110. Modify each run by appending a 0 to get 11000, which is 24 in decimal. So a(6) = 24.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038, A320039, A320261.

Array[FromDigits[Flatten@ Map[Append[#, 0] &, Split@ IntegerDigits[#, 2]], 2] &, 55] (* Michael De Vlieger, Nov 23 2018 *)

from re import split
def A320262(n):
    return int(''.join(d+'0' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(n) = 2*A320263(n).

a(4n) = 2*a(2n), a(4n+1) = 4*a(2n)+2, a(4n+2) = 4*a(2n+1) and a(4n+3) = 2*a(2n+1)+2. - Chai Wah Wu, Nov 21 2018

A320263 Write n in binary, then modify each run of 0's and each run of 1's by prepending a 0. a(n) is the decimal equivalent of the result.

Original entry on oeis.org

1, 4, 3, 8, 17, 12, 7, 16, 33, 68, 35, 24, 49, 28, 15, 32, 65, 132, 67, 136, 273, 140, 71, 48, 97, 196, 99, 56, 113, 60, 31, 64, 129, 260, 131, 264, 529, 268, 135, 272, 545, 1092, 547, 280, 561, 284, 143, 96, 193, 388, 195, 392, 785, 396, 199, 112, 225, 452, 227

Offset: 1

Chai Wah Wu, Oct 08 2018

A variation of A175046. Indices of record values are given by A319423.

			6 in binary is 110. Modify each run by prepending a 0 to get 01100, which is 12 in decimal. So a(6) = 12.

Chai Wah Wu, Table of n, a(n) for n = 1..10000
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175046, A319423, A320037, A320038, A320039, A320261, A320262.

from re import split
def A320263(n):
    return int(''.join('0'+d for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2)

a(n) = A320262(n)/2.

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.

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A320037 Write n in binary, then modify each run of 0's by appending one 1, and modify each run of 1's by appending one 0. a(n) is the decimal equivalent of the result.

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A320038 Write n in binary, then modify each run of 0's by prepending one 1, and modify each run of 1's by prepending one 0. a(n) is the decimal equivalent of the result.

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A319422 Records in A175046.

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A319424 Records in A175046 (but written in base 2).

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A320039 Write n in binary, then modify each run of 0's and each run of 1's by appending a 1. a(n) is the decimal equivalent of the result.

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A320261 Write n in binary, then modify each run of 0's and each run of 1's by prepending a 1. a(n) is the decimal equivalent of the result.

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