A030304 -id:A030304 - cp's OEIS frontend

A296364 a(n) = A296349(n) - A030304(n).

0, 0, 0, 3, 0, 7, 11, 11, 0, 16, 0, 28, 29, 37, 38, 31, 0, 37, 54, 43, 7, 49, 3, 83, 73, 90, 75, 104, 106, 104, 105, 79, 0, 86, 124, 93, 0, 154, 144, 107, 32, 121, 164, 168, 39, 131, 212, 207, 177, 215, 233, 210, 181, 231, 183, 267, 258, 276, 218, 281

Offset: 0

N. J. A. Sloane, Dec 16 2017

Another measure of the binary "early-birdness" of n (cf. A296356, A116700).

Cf. A296349, A030304, A116700, A296356.

A031297 a(n) is the least k such that the base-10 representation of n begins at s(k), where s=A007376.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 1, 16, 18, 20, 22, 24, 26, 28, 30, 15, 34, 2, 38, 40, 42, 44, 46, 48, 50, 17, 37, 56, 3, 60, 62, 64, 66, 68, 70, 19, 39, 59, 78, 4, 82, 84, 86, 88, 90, 21, 41, 61, 81, 100, 5, 104, 106, 108, 110, 23, 43, 63, 83

Offset: 1

Clark Kimberling

A229186 is the same sequence including the a(0) term.

			a(1) = a(12) = a(123) = 1 since they each start at index 1 in 0123.
a(21) = 15 since it appears first at index 15 in 012345678910111213.

Nathaniel Johnston, Table of n, a(n) for n = 1..10000

Cf. A007376, A229186 (same sequence but including the a(0) term).
Cf. A165449.
Cf. A030304 (binary variant).

with(StringTools): s:="": for n from 1 to 70 do s:=cat(s,convert(n,string)): od: seq(Search(convert(n, string), s), n=1..70); # Nathaniel Johnston, May 26 2011

nmax = 100;
s = Table[IntegerDigits[n], {n, 0, nmax}] // Flatten;
a[n_] := SequencePosition[s, IntegerDigits[n], 1][[1, 1]] - 1;
Array[a, nmax] (* Jean-François Alcover, Feb 21 2021 *)

from itertools import count, islice
def agen():
    k, chap = 0, "0"
    for n in count(1):
        target = str(n)
        while chap.find(target) == -1: k += 1; chap += str(k)
        yield chap.find(target)
print(list(islice(agen(), 70))) # Michael S. Branicky, Oct 06 2022

A083655 Numbers which do not appear prematurely in the binary Champernowne word (A030190).

Original entry on oeis.org

0, 1, 2, 4, 8, 10, 16, 32, 36, 64, 128, 136, 256, 512, 528, 1024, 2048, 2080, 4096, 8192, 8256, 16384, 32768, 32896, 65536, 131072, 131328, 262144, 524288, 524800, 1048576, 2097152, 2098176, 4194304, 8388608, 8390656, 16777216, 33554432

Offset: 0

Reinhard Zumkeller, May 01 2003

In other words, numbers k whose binary expansion first appears in A030190 at its expected place, i.e., n appears first starting at position A296349(n). - N. J. A. Sloane, Dec 17 2017

a(n) are the Base 2 "Punctual Bird" numbers: write the nonnegative integers, base 2, in a string 011011100101110111.... Sequence gives numbers which do not occur in the string ahead of their natural place. - Graeme McRae, Aug 11 2007

Graeme McRae, Aug 11 2007, Table of n, a(n) for n = 0..113 - Corrected by _Rémy Sigrist_, Jun 14 2020
Index entries for linear recurrences with constant coefficients, signature (0,0,6,0,0,-8).

A000079 is a subsequence.
Cf. A030190, A030304, A007088, A131881, A132131.
For the complement, the "Early Bird" numbers, see A296365.

LinearRecurrence[{0,0,6,0,0,-8},{0,1,2,4,8,10,16,32,36},50] (* Harvey P. Dale, Aug 19 2020 *)

a(n)= if (n<=2, n, my (m=n\3); if (n%3==0, 2^(2*m), n%3==1, 2^(2*m+1), 2^m + 2^(2*m+1)))  \\ Rémy Sigrist, Jun 14 2020

concat(0, Vec(x*(1 + 2*x + 4*x^2 + 2*x^3 - 2*x^4 - 8*x^5 - 8*x^6 - 8*x^7) / ((1 - 2*x^3)*(1 - 4*x^3)) + O(x^40))) \\ Colin Barker, Jun 14 2020

a(n) = 2^((2*n+1)\3) + (n%3==2)<<(n\3) - (n<3) \\ Charles R Greathouse IV, Dec 16 2022

A083653(a(n))=a(n), A083654(a(n))=1.
a(0)=0, a(1)=1, a(2)=2; then for n>=1, a(3n)=2^(2n), a(3n+1)=2^(2n+1), a(3n+2)=2^(2n+1)+2^n. - Graeme McRae, Aug 11 2007
From Colin Barker, Jun 14 2020: (Start)
G.f.: x*(1 + 2*x + 4*x^2 + 2*x^3 - 2*x^4 - 8*x^5 - 8*x^6 - 8*x^7) / ((1 - 2*x^3)*(1 - 4*x^3)).
a(n) = 6*a(n-3) - 8*a(n-6) for n>8. (End)
a(n) = 2^floor(2*(n+2)/3-1) + (floor((n+1)/3)-floor(n/3))*2^(floor(n/3)) - floor(5/(n+3)). - Alan Michael Gómez Calderón, Dec 15 2022

More terms from Graeme McRae, Aug 11 2007

A083653 Consider the binary Champernowne sequence (A030190): smallest number m such that in binary representation n is contained in the concatenation of m and its successors.

Original entry on oeis.org

0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 10, 2, 3, 1, 3, 7, 16, 8, 4, 9, 18, 10, 21, 2, 7, 3, 9, 1, 3, 5, 7, 15, 32, 16, 8, 17, 36, 4, 9, 19, 34, 18, 10, 10, 37, 21, 2, 6, 15, 7, 3, 12, 19, 9, 21, 1, 7, 3, 19, 5, 7, 13, 15, 31, 64, 32, 16, 33, 8, 34, 17, 35, 68, 36, 18, 4, 73, 9, 19, 39, 66, 34, 20, 18

Offset: 0

Reinhard Zumkeller, May 01 2003

a(n)<=n; see A083655 for numbers m with a(m)=m;
a(A055143(n))=1;
A083654(n)-1 = number of successors of a(n) to cover n.

			n=24: '11000'=24 is a suffix of the concatenation of the first 8 numbers: '0'1'10'11'100'101'110'111'1000', therefore a(24)=7 and A083654(24)=2.

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Rémy Sigrist, PARI program for A083653

Cf. A030190, A030304, A007088, A345672 (decimal analog).

Cf. A055143, A083654, A083655.

PARI
```
See Links section.
```

Edited by Charles R Greathouse IV, Apr 26 2010

A083654 Consider the binary Champernowne sequence (A030190): number of successive numbers to be concatenated beginning with A083653(n) such that in binary representation n is contained.

Original entry on oeis.org

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

Offset: 0

Reinhard Zumkeller, May 01 2003

a(2^k)=1, see A083655 for all numbers m with a(m)=1;

			n=24: '11000'=24 is a suffix of the concatenation of the first 8 numbers: '0'1'10'11'100'101'110'111'1000', therefore a(24)=2 and A083653(24)=7.

Cf. A030304, A007088.

A341798 a(n) is the starting position of the first occurrence of the binary reversal of n in the binary Champernowne word (A030190).

Original entry on oeis.org

0, 1, 0, 1, 7, 2, 0, 4, 19, 6, 8, 1, 23, 2, 3, 15, 51, 18, 21, 5, 7, 26, 0, 11, 56, 22, 8, 1, 66, 2, 14, 46, 131, 50, 54, 17, 20, 63, 58, 4, 143, 6, 27, 25, 23, 77, 10, 41, 137, 55, 21, 34, 7, 28, 0, 11, 149, 65, 8, 1, 173, 13, 45, 125, 323, 130, 135, 49, 53

Offset: 0

Rémy Sigrist, Feb 20 2021

This sequence is a variant of A030304.

When considering the binary reversal of a positive number, the trailing zeros in that number turn into leading zeros in its binary reversal; we keep those leading zeros.

			For n = 4:
- the binary reversal of 4 is "001",
- the binary Champernowne word begins "0110111001011...",
- the first occurrence of "001" in this word starts at position 7,
- so a(4) = 7.

Rémy Sigrist, Table of n, a(n) for n = 0..8192
Rémy Sigrist, Perl program for A341798

Cf. A006995, A030101, A030190, A030304.

Perl
```
See Links section.
```

from itertools import count, islice
def agen():
    k, chap = 0, "0"
    for n in count(0):
        target = bin(n)[2:][::-1]
        while chap.find(target) == -1: k += 1; chap += bin(k)[2:]
        yield chap.find(target)
print(list(islice(agen(), 70))) # Michael S. Branicky, Oct 06 2022

a(n) = A030304(n) iff n is a binary palindrome (A006995).

A347196 Let c(k) be the infinite binary string 010111001... (A030308), the concatenation of reverse order integer binary words ( 0;1;01;11;001;101;... ). a(n) is the bit index k of the first occurrence of the reverse order binary word of n ( n = 2^0c(a(n)) + 2^1c(a(n)+1) + ... ).

Original entry on oeis.org

0, 1, 0, 3, 6, 1, 2, 3, 18, 5, 0, 8, 6, 1, 2, 13, 50, 17, 32, 4, 23, 9, 7, 29, 18, 5, 0, 37, 34, 1, 12, 13, 130, 49, 88, 16, 67, 31, 20, 3, 56, 22, 24, 8, 6, 38, 28, 84, 50, 17, 32, 4, 70, 9, 39, 36, 90, 33, 0, 40, 110, 11, 12, 43, 322, 129, 224, 48, 175, 87, 53, 15

Offset: 0

Thomas Scheuerle, Aug 22 2021

It is not surprising to see dyadic self-similarity in the graph of this sequence. For example the graph of a(0..2^9) looks like a rescaled version of a(0..2^8). Each of these intervals reminds a bit of particle traces in a cloud chamber.

			pos:0,1,2,3,4,5,6,7,8,9,...
c:  0|1|0,1|1,1|0,0,1|1,0,1...
    0                           a(0) = 0
    . 1                         a(1) = 1
    0 1                         a(2) = 0
    . . . 1 1                   a(3) = 3
    . . . . . . 0 0 1           a(4) = 6
    . 1 0 1                     a(5) = 1
    . . 0 1 1                   a(6) = 2

Thomas Scheuerle, Table of n, a(n) for n = 0..5000

Cf. A030304, A030308, A030311, A070939.

function a = A347196( max_n)
    c = 0; a = 0;
    for n = 1:max_n
        b = bitget(n,1:64);
        c = [c b(1:find(b == 1, 1, 'last' ))];
    end
    for n = 1:max_n
        b = bitget(n,1:64);
        word = b(1:find(b == 1, 1, 'last' ));
        pos = strfind(c, word);
        a(n+1) = pos(1)-1;
    end
end

a(n) <= Sum_{k=0..n} A070939(k).

cp's OEIS Frontend

A296364 a(n) = A296349(n) - A030304(n).

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A031297 a(n) is the least k such that the base-10 representation of n begins at s(k), where s=A007376.

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A083655 Numbers which do not appear prematurely in the binary Champernowne word (A030190).

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A083653 Consider the binary Champernowne sequence (A030190): smallest number m such that in binary representation n is contained in the concatenation of m and its successors.

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A083654 Consider the binary Champernowne sequence (A030190): number of successive numbers to be concatenated beginning with A083653(n) such that in binary representation n is contained.

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A341798 a(n) is the starting position of the first occurrence of the binary reversal of n in the binary Champernowne word (A030190).

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Perl

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A347196 Let c(k) be the infinite binary string 010111001... (A030308), the concatenation of reverse order integer binary words ( 0;1;01;11;001;101;... ). a(n) is the bit index k of the first occurrence of the reverse order binary word of n ( n = 2^0*c(a(n)) + 2^1*c(a(n)+1) + ... ).

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MATLAB

Formula

A347196 Let c(k) be the infinite binary string 010111001... (A030308), the concatenation of reverse order integer binary words ( 0;1;01;11;001;101;... ). a(n) is the bit index k of the first occurrence of the reverse order binary word of n ( n = 2^0c(a(n)) + 2^1c(a(n)+1) + ... ).