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 1-7 of 7 results.

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

A369845 Number of compositions of 5*n into parts 3 and 5.

Original entry on oeis.org

1, 1, 1, 2, 7, 22, 58, 138, 319, 750, 1810, 4427, 10828, 26349, 63862, 154603, 374521, 908140, 2203162, 5344988, 12964858, 31443013, 76253683, 184929945, 448506736, 1087774536, 2638209313, 6398479259, 15518219326, 37636240539, 91279009255, 221378866489
Offset: 0

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 11, -5, 1}, {1, 1, 1, 2, 7}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+2*k, n-3*k));

Formula

a(n) = A052920(5*n).
a(n) = Sum_{k=0..floor(n/3)} binomial(n+2*k,n-3*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 11*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: (1-x)^4/((1-x)^5 - x^3).
a(n) = A369847(n+1)-A369847(n). - R. J. Mathar, Feb 14 2024

A369803 Expansion of 1/(1 - x^2/(1-x)^5).

Original entry on oeis.org

1, 0, 1, 5, 16, 45, 126, 361, 1046, 3032, 8771, 25348, 73252, 211724, 612009, 1769080, 5113647, 14781237, 42725841, 123501151, 356986401, 1031887518, 2982723523, 8621714049, 24921502864, 72036871920, 208226244217, 601888555723, 1739789499591, 5028950081882
Offset: 0

Views

Author

Seiichi Manyama, Feb 01 2024

Keywords

Comments

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

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/(1-x^2/(1-x)^5))
    
  • PARI
    a(n) = sum(k=0, n\2, binomial(n-1+3*k, n-2*k));

Formula

a(n) = A001687(5*n-1) for n > 0.
a(n) = 5*a(n-1) - 9*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
a(n) = Sum_{k=0..floor(n/2)} binomial(n-1+3*k,n-2*k).
a(n) = A369840(n)-A369840(n-1). - R. J. Mathar, Feb 14 2024

A369846 Number of compositions of 5*n-1 into parts 3 and 5.

Original entry on oeis.org

0, 1, 4, 10, 21, 44, 101, 250, 629, 1557, 3784, 9120, 21992, 53228, 129177, 313701, 761403, 1846804, 4478044, 10858285, 26332515, 63865592, 154900529, 375691009, 911166977, 2209835169, 5359470121, 12998281146, 31524747503, 76457088518, 185431544730
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 11, -5, 1}, {0, 1, 4, 10, 21}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+1+2*k, n-2-3*k));

Formula

a(n) = A052920(5*n-1).
a(n) = Sum_{k=0..floor(n/3)} binomial(n+1+2*k,n-2-3*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 11*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: x^2*(1-x)/((1-x)^5 - x^3).

A369847 Number of compositions of 5*n-2 into parts 3 and 5.

Original entry on oeis.org

1, 2, 3, 5, 12, 34, 92, 230, 549, 1299, 3109, 7536, 18364, 44713, 108575, 263178, 637699, 1545839, 3749001, 9093989, 22058847, 53501860, 129755543, 314685488, 763192224, 1850966760, 4489176073, 10887655332, 26405874658, 64042115197, 155321124452
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 11, -5, 1}, {1, 2, 3, 5, 12}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+2*k, n-1-3*k));

Formula

a(n) = A052920(5*n-2).
a(n) = Sum_{k=0..floor(n/3)} binomial(n+2*k,n-1-3*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 11*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: x*(1-x)^3/((1-x)^5 - x^3).

A369848 Number of compositions of 5*n-4 into parts 3 and 5.

Original entry on oeis.org

0, 1, 3, 6, 11, 23, 57, 149, 379, 928, 2227, 5336, 12872, 31236, 75949, 184524, 447702, 1085401, 2631240, 6380241, 15474230, 37533077, 91034937, 220790480, 535475968, 1298668192, 3149634952, 7638811025, 18526466357, 44932341015, 108974456212, 264295580664
Offset: 1

Views

Author

Seiichi Manyama, Feb 03 2024

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5, -10, 11, -5, 1}, {0, 1, 3, 6, 11}, 50] (* Paolo Xausa, Mar 15 2024 *)
  • PARI
    a(n) = sum(k=0, n\3, binomial(n+2*k, n-2-3*k));

Formula

a(n) = A052920(5*n-4).
a(n) = Sum_{k=0..floor(n/3)} binomial(n+2*k,n-2-3*k).
a(n) = 5*a(n-1) - 10*a(n-2) + 11*a(n-3) - 5*a(n-4) + a(n-5).
G.f.: x^2*(1-x)^2/((1-x)^5 - x^3).

A369686 LCM-transform of A359804 (see Comment and links).

Original entry on oeis.org

1, 2, 3, 5, 2, 1, 1, 7, 3, 2, 1, 1, 11, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 7, 13, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 17, 1, 1, 19, 1, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 29, 1, 1, 1, 1, 1, 1, 1, 11, 1, 5, 1, 1, 1, 1, 1, 1, 3, 31, 1
Offset: 1

Views

Author

David James Sycamore, Jan 28 2024

Keywords

Comments

Let b(k) be the Least Common Multiple (LCM) of the first k terms of A359804, then a(n) = b(n)/b(n-1), where sequence b(n) is A369685.
The property S (as defined in A368900) refers to what is observed in the positive integers (A000027), and also in the Doudna sequence (A005940), whereby each prime power appears prior to any of its multiples. The present sequence does not have this property since, for example, 26 = a(31) precedes 13 = a(42). Thus A369804 represents a significant disturbance of A000027 in that whereas it is conjectured to be a permutation of the positive integers, it does not preserve one of the basic properties of that sequence.

Crossrefs

Programs

  • Mathematica
    nn = 120; c[] = False; q[] = 1;
    Array[Set[{a[#], c[#]}, {#, True}] &, 2];
    Set[{i, j}, {1, 2}]; m = 2; u = 3;
    Do[
      (k = q[#]; While[c[k #], k++]; k *= #; While[c[# q[#]], q[#]++]) &[
      (p = 2; While[Divisible[i j, p], p = NextPrime[p]]; p)];
      Set[{a[n], c[k], i, j, m}, {#/m, True, j, k, #}] &[LCM[m, k]];
      If[k == u, While[c[u], u++]], {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Jan 29 2024 *)

Formula

a(n) = A369685(n)/A369685(n-1).

Extensions

More terms from Michael De Vlieger, Jan 29 2024
Showing 1-7 of 7 results.