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.

Previous Showing 21-28 of 28 results.

A357645 Triangle read by rows where T(n,k) is the number of integer compositions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 2, 2, 4, 0, 0, 3, 5, 3, 5, 0, 0, 4, 8, 10, 4, 6, 0, 0, 5, 11, 18, 18, 5, 7, 0, 0, 6, 14, 28, 36, 30, 6, 8, 0, 0, 7, 17, 41, 63, 65, 47, 7, 9, 0, 0, 8, 20, 58, 104, 126, 108, 70, 8, 10, 0, 0, 9, 23, 80, 164, 230, 230, 168, 100, 9, 11
Offset: 0

Views

Author

Gus Wiseman, Oct 12 2022

Keywords

Comments

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 - ...

Examples

			Triangle begins:
   1
   0   1
   0   0   2
   0   0   1   3
   0   0   2   2   4
   0   0   3   5   3   5
   0   0   4   8  10   4   6
   0   0   5  11  18  18   5   7
   0   0   6  14  28  36  30   6   8
   0   0   7  17  41  63  65  47   7   9
   0   0   8  20  58 104 126 108  70   8  10
Row n = 6 counts the following compositions:
  (114)   (123)    (132)     (141)  (6)
  (1113)  (213)    (222)     (231)  (15)
  (1122)  (1212)   (312)     (321)  (24)
  (1131)  (1221)   (1311)    (411)  (33)
          (2112)   (2211)           (42)
          (2121)   (3111)           (51)
          (11121)  (11112)
          (11211)  (12111)
                   (21111)
                   (111111)

Crossrefs

Row sums are A011782.
For original alternating sum we have A097805, unordered A344651.
Column k = n-4 appears to be A177787.
The case of partitions is A357637, skew A357638.
The central column k=0 is A357641 (aerated).
The skew-alternating version is A357646.
The reverse version for partitions is A357704, skew A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.
Cf. A029862, A035363, A035544, A357136, A357631, A357639.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],halfats[#]==k&]],{n,0,10},{k,-n,n,2}]

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.
Cf. A357136, A357182, A357626, A357631, A357639, A357642, A357706.
(End)

Links

Crossrefs

First differences of A227161. - Alois P. Heinz, Jul 18 2013
Cf. A035363, A088218.

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

A357646 Triangle read by rows where T(n,k) is the number of integer compositions of n with skew-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 1, 1, 0, 3, 3, 1, 1, 0, 4, 5, 5, 1, 1, 0, 5, 7, 10, 8, 1, 1, 0, 6, 9, 17, 18, 12, 1, 1, 0, 7, 11, 27, 35, 29, 17, 1, 1, 0, 8, 13, 41, 63, 63, 43, 23, 1, 1, 0, 9, 15, 60, 106, 126, 104, 60, 30, 1, 1, 0, 10, 17, 85, 168, 232, 230, 162, 80, 38, 1, 1
Offset: 0

Views

Author

Gus Wiseman, Oct 12 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 + ...

Examples

			Triangle begins:
   1
   0   1
   0   1   1
   0   2   1   1
   0   3   3   1   1
   0   4   5   5   1   1
   0   5   7  10   8   1   1
   0   6   9  17  18  12   1   1
   0   7  11  27  35  29  17   1   1
   0   8  13  41  63  63  43  23   1   1
   0   9  15  60 106 126 104  60  30   1   1
Row n = 6 counts the following compositions:
  (15)   (24)    (33)      (42)     (51)  (6)
  (114)  (213)   (312)     (411)
  (123)  (222)   (321)     (1113)
  (132)  (231)   (1122)    (2112)
  (141)  (1131)  (1212)    (3111)
         (1221)  (2121)    (11112)
         (1311)  (2211)    (11121)
                 (11211)   (21111)
                 (12111)
                 (111111)

Crossrefs

The central column k=0 is A001700 (aerated), half A357641.
Row sums are A011782.
For original alternating sum we have A097805, unordered A344651.
The skew-alternating sum of standard compositions is A357623, half A357621.
The case of partitions is A357638, half A357637.
The half-alternating version is A357645.
The reverse version for partitions is A357705, half A357704.
Cf. A029862, A035363, A035544, A177787, A357136, A357630, A357631, A357634, A357639, A357643, A357644.

Programs

  • Mathematica
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],skats[#]==k&]],{n,0,10},{k,-n,n,2}]

A357704 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 2, 0, 3, 0, 0, 2, 2, 0, 3, 0, 0, 3, 1, 3, 0, 4, 0, 0, 3, 2, 4, 2, 0, 4, 0, 0, 4, 2, 6, 2, 3, 0, 5, 0, 0, 4, 3, 5, 7, 3, 3, 0, 5, 0, 0, 5, 3, 8, 4, 10, 2, 4, 0, 6, 0, 0, 5, 4, 8, 6, 11, 9, 3, 4, 0, 6, 0, 0, 6, 4, 11, 5, 15, 8, 13, 3, 5, 0, 7
Offset: 0

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

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 - ...

Examples

			Triangle begins:
  1
  0  1
  0  0  2
  0  0  1  2
  0  0  2  0  3
  0  0  2  2  0  3
  0  0  3  1  3  0  4
  0  0  3  2  4  2  0  4
  0  0  4  2  6  2  3  0  5
  0  0  4  3  5  7  3  3  0  5
  0  0  5  3  8  4 10  2  4  0  6
  0  0  5  4  8  6 11  9  3  4  0  6
  0  0  6  4 11  5 15  8 13  3  5  0  7
  0  0  6  5 11  8 13 19 10 13  4  5  0  7
  0  0  7  5 14  8 19 13 25  9 17  4  6  0  8
  0  0  7  6 14 11 19 17 29 23 13 18  5  6  0  8
Row n = 7 counts the following reversed partitions:
  .  .  (115)   (124)   (133)      (11113)   .  (7)
        (1114)  (1222)  (223)      (111112)     (16)
        (1123)          (11122)                 (25)
                        (1111111)               (34)

Crossrefs

Row sums are A000041.
Last entry of row n is A008619(n).
The central column in the non-reverse case is A035363, skew A035544.
For original reverse-alternating sum we have A344612.
For original alternating sum we have A344651, ordered A097805.
The non-reverse version is A357637, skew A357638.
The central column is A357639, skew A357640.
The non-reverse ordered version (compositions) is A357645, skew A357646.
The skew-alternating version is A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.
Cf. A029862, A053251, A357136, A357189, A357487, A357488, A357631, A357632, A357641.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
Table[Length[Select[Reverse/@IntegerPartitions[n],halfats[#]==k&]],{n,0,15},{k,-n,n,2}]

A357705 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with skew-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 10 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 + ...

Examples

			Triangle begins:
  1
  0  1
  0  1  1
  0  2  0  1
  0  2  2  0  1
  0  3  1  2  0  1
  0  3  2  3  2  0  1
  0  4  2  4  1  3  0  1
  0  4  3  3  6  2  3  0  1
  0  5  3  5  3  7  2  4  0  1
  0  5  4  5  4  9  7  3  4  0  1
  0  6  4  7  3 12  5 10  3  5  0  1
  0  6  5  7  5 10 16  7 11  4  5  0  1
  0  7  5  9  5 14 11 18  7 14  4  6  0  1
Row n = 7 counts the following reversed partitions:
  .  (16)   (25)   (34)       (1123)  (1114)   .  (7)
     (115)  (223)  (1222)             (11113)
     (124)         (111112)           (11122)
     (133)         (1111111)

Crossrefs

Row sums are A000041.
First nonzero entry of each row is A004526.
The central column is A357640, half A357639.
For original alternating sum we have A344651, ordered A097805.
The half-alternating version is A357704.
The ordered non-reverse version (compositions) is A357646, half A357645.
The non-reverse version is A357638, half A357637.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.
Cf. A035363, A035594, A053251, A298311, A357136, A357189, A357487, A357488, A357624, A357631, A357632, A357636.

Programs

  • Mathematica
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Table[Length[Select[Reverse/@IntegerPartitions[n],skats[#]==k&]],{n,0,11},{k,-n,n,2}]

A228248 Number of 2n-step lattice paths from (0,0) to (0,0) using steps in {N, S, E, W} starting with East, then always moving straight ahead or turning left.

Original entry on oeis.org

1, 0, 1, 3, 9, 30, 103, 357, 1257, 4494, 16246, 59246, 217719, 805389, 2996113, 11200113, 42047593, 158452138, 599113966, 2272065638, 8639763574, 32933685102, 125817012366, 481631387438, 1847110931703, 7095928565405, 27302745922817, 105204285608025
Offset: 0

Views

Author

David Scambler and Alois P. Heinz, Aug 18 2013

Keywords

Comments

From Gus Wiseman, Oct 13 2022: (Start)
Also the number of integer compositions of 2n whose half-alternating and skew-alternating sums are both 0, 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 - ..., and the skew-alternating sum to be A - B - C + D + E - F - G + ... For example, the a(0) = 1 through a(4) = 9 compositions are:
() . (1111) (1212) (1313)
(2121) (2222)
(11211) (3131)
(11312)
(12221)
(21311)
(112211)
(1112111)
(11111111)
For skew-alternating only: A001700, ranked by A357627, reverse A357628.
For partitions: A035363, half only A357639, skew only A357640.
For half-alternating only: A357641, ranked by A357625, reverse A357626.
These compositions are ranked by A357706.
(End)

Examples

			a(0) = 1: [], the empty path.
a(1) = 0.
a(2) = 1: ENWS.
a(3) = 3: EENWWS, ENNWSS, ENWWSE.

Links

Crossrefs

Cf. A004006 (same rules, but self-avoiding).
Cf. A000583, A001511, A001700, A088218, A357136, A357182, A357642.

Programs

  • Maple
    b:= proc(x, y, n) option remember; `if`(abs(x)+abs(y)>n, 0,
      `if`(n=0, 1, b(x+1, y, n-1) +b(y+1, -x, n-1)))
    end:
a:= n-> ceil(b(0, 0, 2*n)/2):
seq(a(n), n=0..40);
# second Maple program:
a:= proc(n) option remember; `if`(n<5, [1, 0, 1, 3, 9][n+1],
     ((n-1)*(414288-1901580*n+186029*n^6-869551*n^5+2393807*n^4
     -3938624*n^3+3753546*n^2+1050*n^8-21605*n^7)*a(n-1)
     +(-17751540*n-12215020*n^5+3494038*n^6+3777840+27478070*n^4
     -39711374*n^3+35488098*n^2-2700*n^9+62370*n^8-621126*n^7)*a(n-2)
     +(-18193248+77490792*n-9138800*n^6+35323128*n^5-88122332*n^4
     +141370392*n^3-140075264*n^2+5400*n^9-135540*n^8+1476432*n^7)*a(n-3)
     +(-192473328*n+48577536+17091500*n^6-70036368*n^5+184890672*n^4
     -313388816*n^3+328043052*n^2-8400*n^9+224440*n^8-2600032*n^7)*a(n-4)
     +8*(n-5)*(150*n^6-2015*n^5+10852*n^4-29867*n^3+44208*n^2-33540*n
     +10416)*(-9+2*n)^2*a(n-5)) / (n^2*(396988*n-487261*n^2+150*n^7
     -3065*n^6+26092*n^5-119602*n^4+317746*n^3-131048)))
    end:
seq(a(n), n=0..40);
  • Mathematica
    b[x_, y_, n_] := b[x, y, n] = If[Abs[x] + Abs[y] > n, 0, If[n == 0, 1, b[x + 1, y, n - 1] + b[y + 1, -x, n - 1]]];
a[n_] := Ceiling[b[0, 0, 2n]/2];
a /@ Range[0, 40] (* Jean-François Alcover, May 14 2020, after Maple *)
halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[2n],halfats[#]==0&&skats[#]==0&]],{n,0,7}] (* Gus Wiseman, Oct 12 2022 *)

Formula

a(n) ~ 2^(2n-1)/(Pi*n). - Vaclav Kotesovec, Jul 16 2014

A357706 Numbers k such that the k-th composition in standard order has half-alternating sum and skew-alternating sum both 0.

Original entry on oeis.org

0, 15, 45, 54, 59, 153, 170, 179, 204, 213, 230, 235, 247, 255, 561, 594, 611, 660, 677, 710, 715, 727, 735, 750, 765, 792, 809, 842, 851, 871, 879, 894, 908, 917, 934, 939, 951, 959, 973, 982, 987, 1005, 1014, 1019
Offset: 1

Views

Author

Gus Wiseman, Oct 13 2022

Keywords

Comments

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 - ..., and the skew-alternating sum 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.

Crossrefs

For partitions and half only (or both): A000583, counted by A035363.
These compositions are counted by A228248.
For half-alternating only: A357625, reverse A357626, counted by A357641.
For skew-alternating only: A357627, reverse A357628, counted by A001700.
For reversed partitions and half only: A357631, counted by A357639.
For reversed partitions and skew only A357632, counted by A357640.
For partitions and skew only: A357636, counted by A035594.
Cf. A001511, A004006, A088218, A357136, A357182, A357642.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
Select[Range[0,1000],halfats[stc[#]]==0&&skats[stc[#]]==0&]

Formula

Intersection of A357625 and A357627.

A357851 Numbers k such that the half-alternating sum of the prime indices of k is 1.

Original entry on oeis.org

2, 8, 18, 32, 45, 50, 72, 98, 105, 128, 162, 180, 200, 231, 242, 275, 288, 338, 392, 420, 429, 450, 455, 512, 578, 648, 663, 720, 722, 800, 833, 882, 924, 935, 968, 969, 1050, 1058, 1100, 1125, 1152, 1235, 1250, 1311, 1352, 1458, 1463, 1568, 1680, 1682, 1716
Offset: 1

Views

Author

Gus Wiseman, Oct 28 2022

Keywords

Comments

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 - ...
A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

Examples

			The terms together with their prime indices begin:
     2: {1}
     8: {1,1,1}
    18: {1,2,2}
    32: {1,1,1,1,1}
    45: {2,2,3}
    50: {1,3,3}
    72: {1,1,1,2,2}
    98: {1,4,4}
   105: {2,3,4}
   128: {1,1,1,1,1,1,1}
   162: {1,2,2,2,2}
   180: {1,1,2,2,3}
   200: {1,1,1,3,3}

Crossrefs

The version for k = 0 is A357631, standard compositions A357625-A357626.
The version for original alternating sum is A001105.
Positions of ones in A357629, reverse A357633.
The skew version for k = 0 is A357632, reverse A357636.
Partitions with these Heinz numbers are counted by A035444, skew A035544.
The reverse version is A357635, k = 0 version A000583.
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, even-length A357642.
Cf. A003963, A053251, A055932, A345958, A357621-A357624, A357630, A357634, A357637, A357639, A357640.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
Select[Range[1000],halfats[primeMS[#]]==1&]
Previous Showing 21-28 of 28 results.

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