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.

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A357628 Numbers k such that the reversed k-th composition in standard order has skew-alternating sum 0.

Original entry on oeis.org

0, 3, 10, 14, 15, 36, 43, 44, 45, 52, 54, 58, 59, 61, 63, 136, 147, 149, 152, 153, 166, 168, 170, 175, 178, 179, 181, 183, 185, 190, 200, 204, 211, 212, 213, 217, 219, 221, 228, 230, 234, 235, 237, 239, 242, 246, 247, 250, 254, 255, 528, 547, 549, 553, 560
Offset: 1

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
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. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The sequence together with the corresponding compositions begins:
    0: ()
    3: (1,1)
   10: (2,2)
   14: (1,1,2)
   15: (1,1,1,1)
   36: (3,3)
   43: (2,2,1,1)
   44: (2,1,3)
   45: (2,1,2,1)
   52: (1,2,3)
   54: (1,2,1,2)
   58: (1,1,2,2)
   59: (1,1,2,1,1)
   61: (1,1,1,2,1)
   63: (1,1,1,1,1,1)
		

Crossrefs

See link for sequences related to standard compositions.
The alternating form is A344619.
Positions of zeros are A357624, non-reverse A357623.
The half-alternating form is A357626, non-reverse A357625.
The non-reverse version is A357627.
The version for prime indices is A357632.
The version for Heinz numbers of partitions is A357636.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Select[Range[0,100],skats[Reverse[stc[#]]]==0&]

A357624 Skew-alternating sum of the reversed n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 0, 3, -1, 1, -1, 4, -2, 0, -2, 2, -2, 0, 0, 5, -3, -1, -3, 1, -3, -1, 1, 3, -3, -1, -1, 1, -1, 1, 1, 6, -4, -2, -4, 0, -4, -2, 2, 2, -4, -2, 0, 0, 0, 2, 2, 4, -4, -2, -2, 0, -2, 0, 2, 2, -2, 0, 0, 2, 0, 2, 0, 7, -5, -3, -5, -1, -5, -3, 3, 1, -5, -3, 1
Offset: 0

Views

Author

Gus Wiseman, Oct 08 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ....
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. This gives a bijective correspondence between nonnegative integers and integer compositions.

Examples

			The 357-th composition is (2,1,3,2,1) so a(357) = 1 - 2 - 3 + 2 + 1 = -1.
The 358-th composition is (2,1,3,1,2) so a(358) = 2 - 1 - 3 + 1 + 2 = 1.
		

Crossrefs

See link for sequences related to standard compositions.
The half-alternating form is A357622, non-reverse A357621.
The reverse version is A357623.
Positions of zeros are A357628, non-reverse A357627.
The version for prime indices is A357630.
The version for Heinz numbers of partitions is A357634.
A124754 gives alternating sum of standard compositions, reverse A344618.
A357637 counts partitions by half-alternating sum, skew A357638.
A357641 counts comps w/ half-alt sum 0, partitions A357639, even A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[skats[Reverse[stc[n]]],{n,0,100}]

A177787 Number of paths from (0,0) to (n+2,n) using only up and right steps and avoiding two or more consecutive moves up or three or more consecutive moves right.

Original entry on oeis.org

2, 5, 10, 18, 30, 47, 70, 100, 138, 185, 242, 310, 390, 483, 590, 712, 850, 1005, 1178, 1370, 1582, 1815, 2070, 2348, 2650, 2977, 3330, 3710, 4118, 4555, 5022, 5520, 6050, 6613, 7210, 7842, 8510, 9215, 9958, 10740, 11562, 12425, 13330, 14278, 15270
Offset: 1

Views

Author

Shanzhen Gao, May 13 2010

Keywords

Comments

Strings of length 2n+2 over the alphabet {U, R} with n Rs and avoiding UU or RRR as substrings.
Also number of binary words with 3 1's and n 0's that do not contain the substring 101. a(2) = 5: 00111, 10011, 11001, 11100, 01110. - Alois P. Heinz, Jul 18 2013
Let (b(n)) be the p-INVERT of A010892 using p(S) = 1 - S^2; then b(n) = a(n+1) for n >= 0. See A292301. - Clark Kimberling, Sep 30 2017
From Gus Wiseman, Oct 13 2022: (Start)
Also the number of integer compositions of n+3 with half-alternating sum n-1, where we define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ... For example, the a(1) = 2 through a(4) = 10 compositions are:
(112) (122) (132)
(1111) (212) (222)
(1211) (312)
(2111) (1311)
(11111) (2211)
(3111)
(11112)
(12111)
(21111)
(111111)
A001700/A138364 = compositions with alternating sum 0, ranked by A344619.
A357621 = half-alternating sum of standard compositions, reverse A357622.
A357641 = compositions with half-alternating sum 0, ranked by A357625.
(End)

Crossrefs

First differences of A227161. - Alois P. Heinz, Jul 18 2013

Programs

  • Magma
    I:=[2, 5, 10, 18]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Jul 04 2012
  • Maple
    a:= n-> n/6*(11+n^2): seq(a(n), n=1..40);
  • Mathematica
    CoefficientList[Series[(2-3*x+2*x^2)/(x-1)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)

Formula

a(n) = 1/6 * n (11 + n^2).
From R. J. Mathar, May 22 2010: (Start)
a(n) = A140226(n)/2.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: x*(2-3*x+2*x^2)/(x-1)^4. (End)

Extensions

More terms from R. J. Mathar, May 22 2010
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