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 14 results. Next

A099131 Quintisection and binomial transform of 1/(1-x^4-x^5).

Original entry on oeis.org

1, 1, 1, 1, 2, 7, 22, 57, 128, 264, 529, 1079, 2290, 5022, 11148, 24633, 53824, 116472, 250880, 540536, 1167937, 2531061, 5494247, 11928731, 25880583, 56101768, 121544393, 263289438, 570427339, 1236159756, 2679343966, 5807782301
Offset: 0

Views

Author

Paul Barry, Sep 29 2004

Keywords

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -4, 1}, {1, 1, 1, 1, 2}, 32] (* Jean-François Alcover, Sep 21 2017 *)

Formula

G.f.: (1-x)^4/((1-x)^5-x^4); a(n)=sum{k=0..floor(5n/4), binomial(k, 5n-4k)}; a(n)=A017827(5n).
a(n)=sum{k=0..floor((n+1)/2), binomial(n+k, 5k)}; - Paul Barry, May 09 2005

A306713 Square array A(n,k), n >= 0, k >= 1, read by antidiagonals, where column k is the expansion of g.f. 1/(1-x^k-x^(k+1)).

Original entry on oeis.org

1, 1, 1, 1, 0, 2, 1, 0, 1, 3, 1, 0, 0, 1, 5, 1, 0, 0, 1, 1, 8, 1, 0, 0, 0, 1, 2, 13, 1, 0, 0, 0, 1, 0, 2, 21, 1, 0, 0, 0, 0, 1, 1, 3, 34, 1, 0, 0, 0, 0, 1, 0, 2, 4, 55, 1, 0, 0, 0, 0, 0, 1, 0, 1, 5, 89, 1, 0, 0, 0, 0, 0, 1, 0, 1, 1, 7, 144, 1, 0, 0, 0, 0, 0, 0, 1, 0, 2, 3, 9, 233
Offset: 0

Views

Author

Seiichi Manyama, Mar 05 2019

Keywords

Comments

A(n,k) is the number of compositions of n into parts k and k+1.

Examples

			Square array begins:
    1, 1, 1, 1, 1, 1, 1, 1, 1, ...
    1, 0, 0, 0, 0, 0, 0, 0, 0, ...
    2, 1, 0, 0, 0, 0, 0, 0, 0, ...
    3, 1, 1, 0, 0, 0, 0, 0, 0, ...
    5, 1, 1, 1, 0, 0, 0, 0, 0, ...
    8, 2, 0, 1, 1, 0, 0, 0, 0, ...
   13, 2, 1, 0, 1, 1, 0, 0, 0, ...
   21, 3, 2, 0, 0, 1, 1, 0, 0, ...
   34, 4, 1, 1, 0, 0, 1, 1, 0, ...
   55, 5, 1, 2, 0, 0, 0, 1, 1, ...
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := Sum[Binomial[j, n-k*j], {j, 0, Floor[n/k]}]; Table[T[k, n - k + 1], {n, 0, 12}, {k, 0, n}] // Flatten (* Amiram Eldar, Jun 21 2021 *)

Formula

A(n,k) = Sum_{j=0..floor(n/k)} binomial(j,n-k*j).

A368475 Expansion of o.g.f. (1-x)^5/((1-x)^5 - x^4).

Original entry on oeis.org

1, 0, 0, 0, 1, 5, 15, 35, 71, 136, 265, 550, 1211, 2732, 6126, 13485, 29191, 62648, 134408, 289656, 627401, 1363124, 2963186, 6434484, 13951852, 30221185, 65442625, 141745045, 307137901, 665732417, 1443184210, 3128438335, 6780867186, 14696002913, 31848721632
Offset: 0

Views

Author

Enrique Navarrete, Dec 26 2023

Keywords

Comments

For n > 0, a(n) is the number of ways to split [n] into an unspecified number of intervals and then choose 4 blocks (i.e., subintervals) from each interval. For example, for n=12, a(12)=1211 since the number of ways to split [12] into intervals and then select 4 blocks from each interval is C(12,4) + C(8,4)*C(4,4) + C(7,4)*C(5,4) + C(6,4)*C(6,4) + C(5,4)*C(7,4) + C(4,4)*C(8,4) + C(4,4)*C(4,4)*C(4,4) for a total of 1211 ways.
For n > 0, a(n) is also the number of compositions of n using parts of size at least 4 where there are binomial(i,4) types of i, i >= 4 (see example).
Number of compositions of 5*n-4 into parts 4 and 5. - Seiichi Manyama, Feb 01 2024

Examples

			Since there are C(4,4) = 1 type of 4, C(5,4) = 5 types of 5, C(6,4) = 15 types of 6, C(7,4) = 35 types of 7, C(8,4) = 70 types of 8, and (12,4) = 495 types of 12, we can write 12 in the following ways:
  12: 495 ways;
  8+4: 70 ways;
  7+5: 175 ways;
  6+6: 225 ways;
  5+7: 175 ways;
  4+8: 70 ways;
  4+4+4: 1 way, for a total of 1211 ways.
		

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1 - x)^5/((1 - x)^5 - x^4), {x, 0, 50}], x] (* Wesley Ivan Hurt, Dec 26 2023 *)
  • PARI
    Vec((1-x)^5/((1-x)^5 - x^4) + O(x^40)) \\ Michel Marcus, Dec 27 2023

Formula

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5), n>=6; a(0)=1, a(1)=a(2)=a(3)=0, a(4)=1, a(5)=5.
G.f.: 1/(1-Sum_{k>=4} binomial(k,4)*x^k).
G.f.: 1/p(S), where p(S) = 1 - S^4 - S^5 and S = x/(1-x).
First differences of A099131. - R. J. Mathar, Jan 29 2024
a(n) = A017827(5*n-4) = Sum_{k=0..floor((5*n-4)/4)} binomial(k,5*n-4-4*k) for n > 0. - Seiichi Manyama, Feb 01 2024
a(n) = Sum_{k=0..floor(n/4)} binomial(n-1+k,n-4*k). - Seiichi Manyama, Feb 02 2024

A017867 Expansion of 1/(1 - x^8 - x^9).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 1, 3, 3, 1, 0, 0, 0, 0, 1, 4, 6, 4, 1, 0, 0, 0, 1, 5, 10, 10, 5, 1, 0, 0, 1, 6, 15, 20, 15, 6, 1, 0, 1, 7, 21, 35, 35, 21, 7, 1, 1, 8, 28, 56, 70, 56, 28, 8, 2, 9
Offset: 0

Views

Author

Keywords

Comments

Number of compositions of n into parts 8 and 9. - Joerg Arndt, Jun 29 2013

Crossrefs

Column k=8 of A306713.

Programs

  • Magma
    m:=80; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-x^8-x^9))); // Vincenzo Librandi, Jun 28 2013
    
  • Magma
    I:=[1,0,0,0,0,0,0,0,1]; [n le 9 select I[n] else Self(n-8)+Self(n-9): n in [1..80]]; // Vincenzo Librandi, Jun 28 2013
    
  • Mathematica
    CoefficientList[Series[1 / (1 - Total[x^Range[8, 9]]), {x, 0, 80}], x] (* Vincenzo Librandi, Jun 28 2013 *)
  • PARI
    x='x+O('x^66); Vec(1/(1-x^8-x^9)) \\ Altug Alkan, Oct 07 2018

Formula

a(n) = a(n-8) + a(n-9) for n>8. - Vincenzo Librandi, Jun 28 2013
a(n) = Sum_{k=0..floor(n/8)} binomial(k,n-8*k). - Seiichi Manyama, Oct 01 2024

A124789 Expansion of (1+x^2)/(1-x^4-x^5).

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 4, 4, 4, 5, 7, 8, 8, 9, 12, 15, 16, 17, 21, 27, 31, 33, 38, 48, 58, 64, 71, 86, 106, 122, 135, 157, 192, 228, 257, 292, 349, 420, 485, 549, 641, 769, 905, 1034, 1190, 1410, 1674, 1939, 2224, 2600, 3084, 3613
Offset: 0

Views

Author

Paul Barry, Nov 07 2006

Keywords

Comments

Diagonal sums of A124788.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1+x^2)/(1-x^4-x^5),{x,0,60}],x] (* or *) LinearRecurrence[ {0,0,0,1,1},{1,0,1,0,1},60] (* Harvey P. Dale, Aug 20 2013 *)

Formula

a(n) = Sum_{k=0..floor(n/2)} C(floor(k/2),n-2*k).
a(n) = A017827(n)+A017827(n-2). - R. J. Mathar, May 09 2013
a(n) = A103372(n-3) for n >= 4. - Georg Fischer, Nov 03 2018
a(n) = (-1)^n*A124746(n). - R. J. Mathar, Jun 30 2020

A369849 Number of compositions of 5*n-1 into parts 4 and 5.

Original entry on oeis.org

1, 2, 3, 4, 6, 13, 35, 92, 220, 484, 1013, 2092, 4382, 9404, 20552, 45185, 99009, 215481, 466361, 1006897, 2174834, 4705895, 10200142, 22128873, 48009456, 104111224, 225655617, 488945055, 1059372394, 2295532150, 4974876116, 10782658417, 23371307904, 50655960304
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -4, 1}, {1, 2, 3, 4, 6}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\4, binomial(n+k, n-1-4*k));

Formula

a(n) = A017827(5*n-1).
a(n) = Sum_{k=0..floor(n/4)} binomial(n+k,n-1-4*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5).
G.f.: x*(1-x)^3/((1-x)^5 - x^4).

A369850 Number of compositions of 5*n-2 into parts 4 and 5.

Original entry on oeis.org

0, 1, 3, 6, 10, 16, 29, 64, 156, 376, 860, 1873, 3965, 8347, 17751, 38303, 83488, 182497, 397978, 864339, 1871236, 4046070, 8751965, 18952107, 41080980, 89090436, 193201660, 418857277, 907802332, 1967174726, 4262706876, 9237582992, 20020241409, 43391549313
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -4, 1}, {0, 1, 3, 6, 10}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\4, binomial(n+k, n-2-4*k));

Formula

a(n) = A017827(5*n-2).
a(n) = Sum_{k=0..floor(n/4)} binomial(n+k,n-2-4*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5).
G.f.: x^2*(1-x)^2/((1-x)^5 - x^4).

A339087 Number of compositions (ordered partitions) of n into distinct parts congruent to 4 mod 5.

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 0, 2, 1, 0, 0, 0, 4, 1, 0, 0, 6, 4, 1, 0, 0, 6, 6, 1, 0, 0, 12, 6, 1, 0, 0, 18, 8, 1, 0, 24, 24, 8, 1, 0, 24, 30, 10, 1, 0, 48, 42, 10, 1, 0, 72, 48, 12, 1, 0, 120, 60, 12, 1, 120, 144, 72, 14, 1, 120, 216, 84, 14, 1, 240
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 23 2020

Keywords

Examples

			a(27) = 6 because we have [14, 9, 4], [14, 4, 9], [9, 14, 4], [9, 4, 14], [4, 14, 9] and [4, 9, 14].
		

Crossrefs

Programs

  • Mathematica
    nmax = 80; CoefficientList[Series[Sum[k! x^(k (5 k + 3)/2)/Product[1 - x^(5 j), {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x]

Formula

G.f.: Sum_{k>=0} k! * x^(k*(5*k + 3)/2) / Product_{j=1..k} (1 - x^(5*j)).

A369851 Number of compositions of 5*n-3 into parts 4 and 5.

Original entry on oeis.org

0, 0, 1, 4, 10, 20, 36, 65, 129, 285, 661, 1521, 3394, 7359, 15706, 33457, 71760, 155248, 337745, 735723, 1600062, 3471298, 7517368, 16269333, 35221440, 76302420, 165392856, 358594516, 777451793, 1685254125, 3652428851, 7915135727, 17152718719, 37172960128
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 10, -4, 1}, {0, 0, 1, 4, 10}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\4, binomial(n+k, n-3-4*k));

Formula

a(n) = A017827(5*n-3).
a(n) = Sum_{k=0..floor(n/4)} binomial(n+k,n-3-4*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 4*a(n-4) + a(n-5).
G.f.: x^3*(1-x)/((1-x)^5 - x^4).

A376546 G.f. A(x) satisfies A(x) = 1 + x^4*(1+x)*A(x)^4.

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 0, 4, 8, 4, 0, 22, 66, 66, 22, 140, 560, 840, 560, 1109, 4845, 9690, 9690, 11929, 43473, 106260, 141680, 160080, 419244, 1137304, 1883700, 2304432, 4496076, 12157236, 23614812, 32813500, 53821332, 132821856, 285795696, 451409380
Offset: 0

Views

Author

Seiichi Manyama, Sep 27 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(k, n-4*k)*binomial(4*k, k)/(3*k+1));

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(k,n-4*k) * binomial(4*k,k)/(3*k+1).
Showing 1-10 of 14 results. Next