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 1-3 of 3 results.

A346702 The a(n)-th composition in standard order is the odd bisection of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5, 10, 5, 1, 3, 6, 3, 12, 6, 3, 7, 32, 16, 8, 17, 4, 9, 18, 9, 2, 5, 10, 5, 20, 10, 5, 11, 1, 3, 6, 3, 12, 6, 3, 7, 24, 12, 6, 13, 3, 7, 14, 7, 64, 32, 16, 33, 8, 17, 34, 17, 4, 9, 18, 9, 36, 18
Offset: 0

Views

Author

Gus Wiseman, Aug 12 2021

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.
a(n) is the row number in A066099 of the odd bisection of the n-th row of A066099.

Examples

			Composition number 741 in standard order is (2,1,1,3,2,1), with odd bisection (2,1,2), which is composition number 22 in standard order, hence a(741) = 22.

Crossrefs

Length of the a(n)-th standard composition is A000120(n)/2 rounded up.
Positions of 1's are A003945.
Positions of 2's (and zero) are A083575.
Sum of the a(n)-th standard composition is A209281(n+1).
Positions of first appearances are A290259.
The version for prime indices is A346703.
The version for even bisection is A346705, with sums A346633.
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A011782 counts compositions.
A029837 gives length of binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000009, A000302, A000346, A025047, A088218, A124754, A294175, A346697 (reverse: A346699).

Programs

  • Mathematica
    Table[Total[2^Accumulate[Reverse[First/@Partition[Append[ Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse,0],2]]]]/2,{n,0,100}]

Formula

A029837(a(n)) = A209281(n).

A290258 Triangle read by rows: row n (>=2) contains in increasing order the integers for which the binary representation has length n and all runs of 1's have even length.

Original entry on oeis.org

3, 6, 12, 15, 24, 27, 30, 48, 51, 54, 60, 63, 96, 99, 102, 108, 111, 120, 123, 126, 192, 195, 198, 204, 207, 216, 219, 222, 240, 243, 246, 252, 255, 384, 387, 390, 396, 399, 408, 411, 414, 432, 435, 438, 444, 447, 480, 483, 486, 492, 495, 504, 507, 510
Offset: 2

Views

Author

Emeric Deutsch, Sep 12 2017

Keywords

Comments

The viabin numbers of integer partitions having only even parts. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [6,4,4,2]. The southeast border of its Ferrers board yields 110110011 (length is 9), leading to the viabin number 435 (a term in row 9).
Number of entries in row n is the Fibonacci number F(n-1) = A000045(n-1).
T(n,k) = A290259(n-1,k) + 2^(n-1).
Last entry in row n = A141023(n).
Basically the same as A277335.

Examples

			399 is in the sequence because all the runs of 1's of its binary representation, namely 110001111, have even lengths.
Triangle begins:
  3;
  6;
  12,15;
  24,27,30;
  48,51,54,60,63;
  96,99,102,108,111,120,123,126;
  ...

Crossrefs

Cf. A000045, A141023, A277335, A290259.

Programs

  • Maple
    A[2] := {3}; A[3] := {6}; for n from 4 to 10 do A[n] := `union`(map(proc (x) 2*x end proc, A[n-1]), map(proc (x) 4*x+3 end proc, A[n-2])) end do; # yields sequence in triangular form
  • Mathematica
    A[2] = {3}; A[3] = {6};
For[n = 4, n <= 10, n++, A[n] = Union[2 A[n-1], 4 A[n-2] + 3]];
Table[A[n], {n, 2, 10}] // Flatten (* Jean-François Alcover, Aug 19 2024, after Maple program *)

Formula

The entries in row n (n>=4) are: (i) 2x, where x is in row n-1 and (ii) 4y + 3, where y is in row n-2. The Maple program is based on this.

A346705 The a(n)-th composition in standard order is the even bisection of the n-th composition in standard order.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 4, 2, 1, 3, 0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5, 10, 5, 1, 3, 6, 3, 12, 6, 3, 7, 0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5
Offset: 0

Views

Author

Gus Wiseman, Aug 19 2021

Keywords

Comments

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again.
a(n) is the row number in A066099 of the even bisection (even-indexed terms) of the n-th row of A066099.

Examples

			Composition number 741 in standard order is (2,1,1,3,2,1), with even bisection (1,3,1), which is composition number 25 in standard order, so a(741) = 25.

Crossrefs

Length of the a(n)-th standard composition is A000120(n)/2 rounded down.
Positions of first appearances appear to be A088698, sorted: A277335.
The version for reversed prime indices appears to be A329888, sums A346700.
Sum of the a(n)-th standard composition is A346633.
An unordered reverse version for odd bisection is A346701, sums A346699.
The version for odd bisection is A346702, sums A209281(n+1).
An unordered version for odd bisection is A346703, sums A346697.
An unordered version is A346704, sums A346698.
A011782 counts compositions.
A029837 gives length of binary expansion, or sometimes A070939.
A066099 lists compositions in standard order.
A097805 counts compositions by alternating sum.
Cf. A000302, A025047, A088218, A124754, A124767, A228351, A290258, A290259, A344618, A345197.

Programs

  • Mathematica
    Table[Total[2^Accumulate[Reverse[Last/@Partition[ Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse,2]]]]/2,{n,0,100}]

Formula

A029837(a(n)) = A346633(n).
Showing 1-3 of 3 results.

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