cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 20 results. Next

A216956 Triangle read by rows: A216955/2.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 6, 6, 2, 1, 1, 10, 13, 5, 2, 1, 1, 20, 26, 10, 4, 2, 1, 1, 37, 55, 19, 9, 4, 2, 1, 1, 74, 107, 41, 18, 8, 4, 2, 1, 1, 143, 219, 82, 35, 17, 8, 4, 2, 1, 1, 286, 438, 164, 70, 34, 16, 8, 4, 2, 1, 1, 562, 881, 330, 143, 67, 33, 16, 8, 4, 2, 1, 1, 1124, 1762, 660, 286, 134, 66, 32, 16, 8, 4, 2, 1, 1
Offset: 1

Views

Author

N. J. A. Sloane, Sep 26 2012

Keywords

Comments

It appears that reversed rows converge to A011782. - Omar E. Pol, Nov 20 2012

Examples

			Triangle begins:
1,
1, 1,
2, 1, 1,
3, 3, 1, 1,
6, 6, 2, 1, 1,
10, 13, 5, 2, 1, 1,
20, 26, 10, 4, 2, 1, 1,
37, 55, 19, 9, 4, 2, 1, 1,
74, 107, 41, 18, 8, 4, 2, 1, 1,
...
		

Crossrefs

First column is A093371.
Cf. A216955.

A217941 a(n) = Sum_{k=1..n} k*C(n,k), where C(n,k) = number of binary sequences of length n and curling number k (A216955).

Original entry on oeis.org

2, 6, 14, 32, 66, 140, 282, 574, 1156, 2326, 4654, 9348, 18698, 37436, 74904, 149896, 299794, 599780, 1199562, 2399448, 4798996, 9598556, 19197114, 38395584, 76791200, 153584626, 307169622, 614343808, 1228687618, 2457384892, 4914769786, 9829557516, 19659116482, 39318268388
Offset: 1

Views

Author

N. J. A. Sloane, Oct 23 2012

Keywords

Comments

a(n)/2^n appears to be converging to 2.2886...

Crossrefs

Formula

For n>=1, a(2n+1) = 2*a(2n)+2, while a(2n) is a mystery.

A217943 Triangle read by rows: T(n,k) = 2*C(n-1,k)-C(n,k) for kA216955(n,k).

Original entry on oeis.org

2, -2, 0, 2, -2, 2, -2, 2, -2, 0, 0, 0, 2, -2, 4, -2, -2, 0, 2, -2, 0, 0, 0, 0, 0, 2, -2, 6, -6, 2, -2, 0, 0, 2, -2, 0, 6, -6, 0, 0, 0, 0, 2, -2, 10, -10, 0, 2, -2, 0, 0, 0, 2, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, -2, 20, -10, -4, -6, 2, -2, 0, 0, 0, 0, 2, -2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, -2
Offset: 2

Views

Author

N. J. A. Sloane, following a suggestion from Allan Wilks, Oct 25 2012

Keywords

Examples

			Triangle begins:
[2, -2]
[0, 2, -2]
[2, -2, 2, -2]
[0, 0, 0, 2, -2]
[4, -2, -2, 0, 2, -2]
[0, 0, 0, 0, 0, 2, -2]
[6, -6, 2, -2, 0, 0, 2, -2]
[0, 6, -6, 0, 0, 0, 0, 2, -2]
[10, -10, 0, 2, -2, 0, 0, 0, 2, -2]
[0, 0, 0, 0, 0, 0, 0, 0, 0, 2, -2]
[20, -10, -4, -6, 2, -2, 0, 0, 0, 0, 2, -2]
...
		

Crossrefs

Cf. A216955.
The nonzero entries in the first column form A216958.

A122536 Number of binary sequences of length n with no initial repeats (or, with no final repeats).

Original entry on oeis.org

2, 2, 4, 6, 12, 20, 40, 74, 148, 286, 572, 1124, 2248, 4460, 8920, 17768, 35536, 70930, 141860, 283440, 566880, 1133200, 2266400, 4531686, 9063372, 18124522, 36249044, 72493652, 144987304, 289965744
Offset: 1

Views

Author

Sarah Nibs, Sep 18 2006

Keywords

Comments

An initial repeat of a string S is a number k>=1 such that S(i)=S(i+k) for i=0..k-1. In other words, the first k symbols are the same as the next k symbols, e.g., ABCDABCDZQQ has an initial repeat of size 4.
Equivalently, this is the number of binary sequences of length n with curling number 1. See A216955. - N. J. A. Sloane, Sep 26 2012

Examples

			a(4)=6: 0100, 0110, 0111, 1000, 1001 and 1011. (But not 00**, 11**, 0101, 1010.)
		

Crossrefs

Twice A093371. Leading column of each of the triangles A216955, A217209, A218869, A218870. Different from, but easily confused with, A003000 and A216957. - N. J. A. Sloane, Sep 26 2012
See A121880 for difference from 2^n.

Formula

Conjecture: a_n ~ C * 2^n where C is 0.27004339525895354325... [Chaffin, Linderman, Sloane, Wilks, 2012]
a(2n+1)=2*a(2n) = A211965(n+1), a(2n)=2*a(2n-1)-A216958(n) = A211966(n). - N. J. A. Sloane, Sep 28 2012
a(1) = 2; a(2n) = 2*[a(2n-1) - A216959(n)], n >= 1. - Daniel Forgues, Feb 25 2015

Extensions

a(31)-a(71) computed from recurrence and the first 30 terms of A216958 by N. J. A. Sloane, Sep 28 2012, Oct 25 2012

A093371 Start with any initial string of n numbers s(1), ..., s(n), with s(1) = 2, other s(i)'s = 2 or 3 (so there are 2^(n-1) starting strings). The rule for extending the string is this as follows: To get s(n+1), write the string s(1)s(2)...s(n) as xy^k for words x and y (where y has positive length) and k is maximized, i.e., k = the maximal number of repeating blocks at the end of the sequence. Then a(n) = number of starting strings for which k = 1.

Original entry on oeis.org

1, 1, 2, 3, 6, 10, 20, 37, 74, 143, 286, 562, 1124, 2230, 4460, 8884, 17768, 35465, 70930, 141720, 283440, 566600, 1133200, 2265843, 4531686, 9062261, 18124522, 36246826, 72493652, 144982872
Offset: 1

Views

Author

N. J. A. Sloane, Apr 28 2004

Keywords

Comments

See A122536 for many more terms. - N. J. A. Sloane, Oct 25 2012

Crossrefs

Equals A122536/2. - N. J. A. Sloane, Sep 25 2012
Different from, but easily confused with, A007148 and A045690.

Formula

a(n) = 2^(n-1) - A093370(n).

Extensions

More terms from N. J. A. Sloane, Sep 26 2012

A211027 Triangle of binary numbers >= 1 with no initial repeats.

Original entry on oeis.org

1, 10, 100, 101, 1000, 1001, 1011, 10000, 10001, 10010, 10011, 10110, 10111, 100000, 100001, 100010, 100011, 100101, 100110, 100111, 101100, 101110, 101111, 1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1011000, 1011001, 1011100, 1011101, 1011110, 1011111
Offset: 1

Views

Author

Omar E. Pol, Nov 30 2012

Keywords

Comments

Triangle read by rows in which row n lists the binary numbers with n digits and with no initial repeats.
Also triangle read by rows in which row n lists the binary words of length n with no initial repeats and with initial digit 1. See also A211029.

Examples

			Triangle begins:
1;
10;
100, 101;
1000, 1001, 1011;
10000, 10001, 10010, 10011, 10110, 10111;
100000, 100001, 100010, 100011, 100101, 100110, 100111, 101100, 101110, 101111;
1000000, 1000001, 1000010, 1000011, 1000100, 1000101, 1000110, 1000111, 1001010, 1001011, 1001100, 1001101, 1001110, 1001111, 1011000, 1011001, 1011100, 1011101, 1011110, 1011111;
		

Crossrefs

Column 1 is A011557. Row n has length A093371(n).

Programs

  • Maple
    s:= proc(n) s(n):= `if`(n=1, [[1]], map(x->
          [[x[], 0], [x[], 1]][], s(n-1))) end:
    T:= proc(n) map(x-> parse(cat(x[])), select(proc(l) local i;
          for i to iquo(nops(l), 2) do if l[1..i]=l[i+1..2*i]
          then return false fi od; true end, s(n)))[] end:
    seq(T(n), n=1..7);  # Alois P. Heinz, Dec 02 2012
  • Mathematica
    T[n_] := If[n == 1, {1}, FromDigits /@ Select[Range[2^(n-1), 2^n-2] // IntegerDigits[#, 2]&, FindTransientRepeat[Reverse[#], 2][[2]] == {}&]];
    Array[T, 7] // Flatten (* Jean-François Alcover, Feb 27 2021 *)

A218869 Triangle read by rows: T(n,k) = number of aperiodic binary sequences of length n with curling number k (1 <= k <= n).

Original entry on oeis.org

2, 2, 0, 4, 2, 0, 6, 4, 2, 0, 12, 12, 4, 2, 0, 20, 20, 8, 4, 2, 0, 40, 52, 20, 8, 4, 2, 0, 74, 100, 36, 16, 8, 4, 2, 0, 148, 214, 76, 36, 16, 8, 4, 2, 0, 286, 414, 160, 68, 32, 16, 8, 4, 2, 0, 572, 876, 328, 140, 68, 32, 16, 8, 4, 2, 0, 1124, 1722, 640, 276, 132, 64, 32, 16, 8, 4, 2, 0
Offset: 1

Views

Author

N. J. A. Sloane, Nov 07 2012

Keywords

Comments

S is aperiodic if it is not of the form S = T^m with m > 1.
Row sums are A027375. First column is A122536.
It appears that reversed rows converge to A155559. - Omar E. Pol, Nov 20 2012

Examples

			Triangle begins:
2,
2, 0,
4, 2, 0,
6, 4, 2, 0,
12, 12, 4, 2, 0,
20, 20, 8, 4, 2, 0,
40, 52, 20, 8, 4, 2, 0,
74, 100, 36, 16, 8, 4, 2, 0,
148, 214, 76, 36, 16, 8, 4, 2, 0,
286, 414, 160, 68, 32, 16, 8, 4, 2, 0,
572, 876, 328, 140, 68, 32, 16, 8, 4, 2, 0,
...
		

Crossrefs

A211029 Triangle read by rows in which row n lists the binary words of length n over the alphabet {1,2} with no initial repeats.

Original entry on oeis.org

1, 2, 12, 21, 121, 122, 211, 212, 1211, 1221, 1222, 2111, 2112, 2122, 12111, 12112, 12211, 12212, 12221, 12222, 21111, 21112, 21121, 21122, 21221, 21222, 121111, 121112, 121122, 122111, 122112, 122121, 122211, 122212, 122221, 122222, 211111, 211112
Offset: 1

Views

Author

Omar E. Pol, Nov 29 2012

Keywords

Comments

As usual in the OEIS, binary alphabets are encoded with {1,2} over the alphabet {0,1} the entries contain nonzero "numbers" beginning with 0.

Examples

			The fourth row of triangle of binary sequences is
0100, 0110, 0111, 1000, 1001, 1011 (see section example of A122536) therefore the fourth row of this triangle is
1211, 1221, 1222, 2111, 2112, 2122.
The first six rows of triangle are:
1, 2;
12, 21;
121, 122, 211, 212;
1211, 1221, 1222, 2111, 2112, 2122;
12111, 12112, 12211, 12212, 12221, 12222, 21111, 21112, 21121, 21122, 21221, 21222;
121111, 121112, 121122, 122111, 122112, 122121, 122211, 122212, 122221, 122222, 211111, 211112, 211121, 211122, 211212, 211221, 211222, 212211, 212221, 212222;
		

Crossrefs

Row n has length A122536(n).

Programs

  • Maple
    s:= proc(n) s(n):= `if`(n=0, [[]], map(x->
          [[x[], 1], [x[], 2]][], s(n-1))) end:
    T:= proc(n) map(x-> parse(cat(x[])), select(proc(l) local i;
          for i to iquo(nops(l), 2) do if l[1..i]=l[i+1..2*i]
          then return false fi od; true end, s(n)))[] end:
    seq(T(n), n=1..7);  # Alois P. Heinz, Dec 02 2012

Extensions

More terms and name improved by R. J. Mathar, Nov 30 2012

A211968 Triangle of binary numbers with some initial repeats.

Original entry on oeis.org

11, 110, 111, 1010, 1100, 1101, 1110, 1111, 10100, 10101, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111, 100100, 101000, 101001, 101010, 101011, 101101, 110000, 110001, 110010, 110011, 110100, 110101, 110110, 110111, 111000, 111001, 111010, 111011
Offset: 2

Views

Author

Omar E. Pol, Dec 03 2012

Keywords

Comments

Triangle read by rows in which row n lists the binary numbers with n digits and with some initial repeats, n >= 2.
Also triangle read by rows in which row n lists the binary words of length n with some initial repeats and with initial digit 1, n >= 2.

Examples

			Triangle begins, starting at row 2:
  11;
  110, 111;
  1010, 1100, 1101, 1110, 1111;
  10100, 10101, 11000, 11001, 11010, 11011, 11100, 11101, 11110, 11111;
		

Crossrefs

Complement in base 2 of A211027.
Rows lengths give: A093370.

Programs

  • Maple
    s:= proc(n) s(n):= `if`(n=1, [[1]], map(x->
          [[x[], 0], [x[], 1]][], s(n-1))) end:
    T:= proc(n) map(x-> parse(cat(x[])), select(proc(l) local i;
          for i to iquo(nops(l), 2) do if l[1..i]=l[i+1..2*i]
          then return true fi od; false end, s(n)))[] end:
    seq(T(n), n=2..7);  # Alois P. Heinz, Dec 04 2012
  • Mathematica
    T[n_] := FromDigits /@ Select[Range[2^(n-1), 2^n-1] // IntegerDigits[#, 2]&, FindTransientRepeat[Reverse[#], 2][[2]] != {}&];
    Table[T[n], {n, 2, 7}] // Flatten (* Jean-François Alcover, Feb 12 2025 *)

A211969 Triangle of decimal equivalents of binary numbers with some initial repeats, A211968.

Original entry on oeis.org

3, 6, 7, 10, 12, 13, 14, 15, 20, 21, 24, 25, 26, 27, 28, 29, 30, 31, 36, 40, 41, 42, 43, 45, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 72, 73, 80, 81, 82, 83, 84, 85, 86, 87, 90, 91, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106
Offset: 2

Views

Author

Omar E. Pol, Dec 03 2012

Keywords

Examples

			Irregular triangle begins, starting at row 2:
3;
6, 7;
10, 12, 13, 14, 15;
20, 21, 24, 25, 26, 27, 28, 29, 30, 31;
36, 40, 41, 42, 43, 45, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63;
		

Crossrefs

Complement of A211967.
Row lengths give: A093370.
Column 1 gives: A005418(n+1).
Right border gives: A000225(n).

Programs

  • Maple
    s:= proc(n) s(n):= `if`(n=1, [[1]], map(x->
          [[x[], 0], [x[], 1]][], s(n-1))) end:
    T:= proc(n) map (x-> add(x[i]*2^(nops(x)-i), i=1..nops(x)), select
          (proc(l) local i; for i to iquo(nops(l), 2) do if l[1..i]=
          l[i+1..2*i] then return true fi od; false end, s(n)))[] end:
    seq (T(n), n=2..7);  # Alois P. Heinz, Dec 04 2012
Showing 1-10 of 20 results. Next