A320037 -id:A320037 - cp's OEIS frontend

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.

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

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

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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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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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.

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Mathematica

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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.

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Python

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