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 41-50 of 72 results. Next

A135670 Triangular sequence of the coefficients of the denominator of the rational recursive sequence for tan(n*x).

Original entry on oeis.org

1, 1, -1, 0, 1, -1, 0, 3, 1, 0, -6, 0, 1, 1, 0, -10, 0, 5, -1, 0, 15, 0, -15, 0, 1, -1, 0, 21, 0, -35, 0, 7, 1, 0, -28, 0, 70, 0, -28, 0, 1, 1, 0, -36, 0, 126, 0, -84, 0, 9, -1, 0, 45, 0, -210, 0, 210, 0, -45, 0, 1, -1, 0, 55, 0, -330, 0, 462, 0, -165, 0, 11
Offset: 0

Views

Author

Roger L. Bagula, Feb 17 2008

Keywords

Comments

These are the denominators of the expansion of tan(n*x) as in A034839, but keeping the zeros with the terms in the denominator polynomials that vanish. Sign conventions differ slightly, maintaining either a positive coefficient [x^0], or a positive coefficient [x^n] or [x^(n-1)], resp.

Examples

			{1},
{1},
{-1, 0, 1},
{-1, 0, 3},
{1, 0, -6,0, 1},
{1, 0, -10, 0, 5},
{-1, 0, 15, 0, -15, 0, 1},
{-1, 0, 21, 0, -35, 0, 7},
{1, 0, -28, 0, 70, 0, -28, 0, 1},
{1, 0, -36,0, 126, 0, -84, 0, 9},
{-1, 0, 45, 0, -210, 0, 210, 0, -45, 0, 1},
{-1, 0, 55, 0, -330, 0, 462, 0, -165, 0, 11}
		

Programs

  • Mathematica
    Clear[p, x, a, b] p[x, 0] = 1; p[x, 1] = x; p[x, 2] = 2*x/(1 - x^2); p[x, 3] = (3*x - x^3)/(1 - 3*x^2); p[x_, n_] := p[x, n] = (p[x, n - 1] + x)/(1 - p[x, n - 1]*x); c = Table[CoefficientList[Denominator[FullSimplify[p[x, n]]], x], {n, 0, 11}]; Flatten[c]

Extensions

Edited by the Associate Editors of the OEIS, Aug 18 2009

A202023 Triangle T(n,k), read by rows, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 6, 1, 0, 0, 1, 10, 5, 0, 0, 0, 1, 15, 15, 1, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 1, 28, 70, 28, 1, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 1, 45, 210, 210, 45, 1, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 10 2011

Keywords

Comments

Riordan array (1/(1-x), x^2/(1-x)^2).
A skewed version of triangular array A085478.
Mirror image of triangle in A098158.
Sum_{k, 0<=k<=n} T(n,k)*x^k = A138229(n), A006495(n), A138230(n),A087455(n), A146559(n), A000012(n), A011782(n), A001333(n),A026150(n), A046717(n), A084057(n), A002533(n), A083098(n),A084058(n), A003665(n), A002535(n), A133294(n), A090042(n),A125816(n), A133343(n), A133345(n), A120612(n), A133356(n), A125818(n) for x = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A009116(n), A000007(n), A011782(n), A006012(n), A083881(n), A081335(n), A090139(n), A145301(n), A145302(n), A145303(n), A143079(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
From Gus Wiseman, Jul 08 2025: (Start)
After the first row this is also the number of subsets of {1..n-1} with k maximal runs (sequences of consecutive elements increasing by 1) for k = 0..n. For example, row n = 5 counts the following subsets:
{} {1} {1,3} . . .
{2} {1,4}
{3} {2,4}
{4} {1,2,4}
{1,2} {1,3,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
Requiring n-1 gives A202064.
For anti-runs instead of runs we have A384893.
(End)

Examples

			Triangle begins :
1
1, 0
1, 1, 0
1, 3, 0, 0
1, 6, 1, 0, 0
1, 10, 5, 0, 0, 0
1, 15, 15, 1, 0, 0, 0
1, 21, 35, 7, 0, 0, 0, 0
1, 28, 70, 28, 1, 0, 0, 0, 0
		

Crossrefs

Column k = 1 is A000217.
Column k = 2 is A000332.
Row sums are A011782 (or A000079 shifted right).
Removing all zeros gives A034839 (requiring n-1 A034867).
Last nonzero term in each row appears to be A093178, requiring n-1 A124625.
Reversing rows gives A098158, without zeros A109446.
Without the k = 0 column we get A210039.
Row maxima appear to be A214282.
A116674 counts strict partitions by number of maximal runs, for anti-runs A384905.
A268193 counts integer partitions by number of maximal runs, for anti-runs A384881.

Programs

  • Mathematica
    Table[Length[Select[Subsets[Range[n-1]],Length[Split[#,#2==#1+1&]]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jul 08 2025 *)

Formula

T(n,k) = binomial(n,2k).
G.f.: (1-x)/((1-x)^2-y*x^2).
T(n,k)= Sum_{j, j>=0} T(n-1-j,k-1)*j with T(n,0)=1 and T(n,k)= 0 if k<0 or if n
T(n,k) = 2*T(n-1,k) + T(n-2,k-1) - T(n-2,k) for n>1, T(0,0) = T(1,0) = 1, T(1,1) = 0, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, Nov 10 2013

A210039 Array of coefficients of polynomials u(n,x) jointly generated with A210040; see the Formula section.

Original entry on oeis.org

1, 3, 6, 1, 10, 5, 15, 15, 1, 21, 35, 7, 28, 70, 28, 1, 36, 126, 84, 9, 45, 210, 210, 45, 1, 55, 330, 462, 165, 11, 66, 495, 924, 495, 66, 1, 78, 715, 1716, 1287, 286, 13, 91, 1001, 3003, 3003, 1001, 91, 1, 105, 1365, 5005, 6435, 3003, 455, 15, 120, 1820
Offset: 1

Author

Clark Kimberling, Mar 17 2012

Keywords

Comments

Every term is a binomial coefficient.
Row sums: A000225
For a discussion and guide to related arrays, see A208510.

Examples

			First eight rows:
1
3
6....1
10...5
15...15....1
21...35....7
28...70....28...1
36...126...84...9
First five polynomials u(n,x):
1
3
6 + x
10 + 5x
21 + 35x + 7x^2.
		

Crossrefs

Programs

  • Mathematica
    u[1, x_] := 1; v[1, x_] := 1; z = 16;
    u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
    v[n_, x_] := x*u[n - 1, x] + v[n - 1, x] + 1;
    Table[Expand[u[n, x]], {n, 1, z/2}]
    Table[Expand[v[n, x]], {n, 1, z/2}]
    cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
    TableForm[cu]
    Flatten[%]    (* A210039 *)
    Table[Expand[v[n, x]], {n, 1, z}]
    cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
    TableForm[cv]
    Flatten[%]    (* A210040 *)
    Table[u[n, x] /. x -> 1, {n, 1, z}] (* A000225 *)
    Table[v[n, x] /. x -> 1, {n, 1, z}] (* A000225 *)

Formula

u(n,x)=u(n-1,x)+v(n-1,x)+1,
v(n,x)=x*u(n-1,x)+v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.
Also, writing the general term as T(n,m),
T(n,k)=C(n,2k) for 1<=k<=floor[(n+1)/2], for n>=1.

A177808 Triangle T(n,m) = binomial(4*n, 4*m), 0 <= m <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 70, 1, 1, 495, 495, 1, 1, 1820, 12870, 1820, 1, 1, 4845, 125970, 125970, 4845, 1, 1, 10626, 735471, 2704156, 735471, 10626, 1, 1, 20475, 3108105, 30421755, 30421755, 3108105, 20475, 1, 1, 35960, 10518300, 225792840, 601080390, 225792840, 10518300, 35960, 1, 1, 58905, 30260340, 1251677700, 7307872110, 7307872110, 1251677700, 30260340, 58905, 1
Offset: 0

Author

Roger L. Bagula, Dec 13 2010

Keywords

Comments

Row sums are A070775(n).

Examples

			1;
1, 1;
1, 70, 1;
1, 495, 495, 1;
1, 1820, 12870, 1820, 1;
1, 4845, 125970, 125970, 4845, 1;
1, 10626, 735471, 2704156, 735471, 10626, 1;
1, 20475, 3108105, 30421755, 30421755, 3108105, 20475, 1;
1, 35960, 10518300, 225792840, 601080390, 225792840, 10518300, 35960, 1;
1, 58905, 30260340, 1251677700, 7307872110, 7307872110, 1251677700,30260340, 58905, 1;
1, 91390, 76904685, 5586853480, 62852101650, 137846528820, 62852101650, 5586853480, 76904685, 91390, 1;
		

Crossrefs

Programs

  • Maple
    A177808 := proc(n,m) binomial(4*n,4*m) ; end proc: # R. J. Mathar, Dec 13 2010
  • Mathematica
    t[n_, m_] = Binomial[n, 4*m];
    Table[Table[t[n, m], {m, 0, Floor[n/4]}], {n, 0, 40, 4}];
    Flatten[%]

Formula

Right-left symmetric: T(n,m) = T(n,n-m).

A382300 a(n) = Sum_{k=0..floor(n/2)} (k+1) * binomial(2*k,2*n-4*k).

Original entry on oeis.org

1, 0, 2, 2, 3, 18, 7, 60, 65, 144, 356, 410, 1272, 1722, 3743, 7202, 11482, 25566, 40421, 81610, 147169, 259810, 507267, 867792, 1659112, 2961860, 5362592, 9940420, 17583485, 32564548, 58228386, 105606458, 191831767, 343313042, 625086891, 1119760040, 2023087045
Offset: 0

Author

Seiichi Manyama, Mar 29 2025

Keywords

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 37); Coefficients(R!( ((1-x^2-x^3)^2 + 4*x^5) / ((1-x^2-x^3)^2 - 4*x^5)^2)); // Vincenzo Librandi, May 11 2025
  • Mathematica
    Table[Sum[(k+1)*Binomial[2*k,2*n-4*k],{k,0,Floor[n/2]}],{n,0,30}] (* Vincenzo Librandi, May 11 2025 *)
  • PARI
    a(n) = sum(k=0, n\2, (k+1)*binomial(2*k, 2*n-4*k));
    
  • PARI
    my(N=1, M=40, x='x+O('x^M), X=1-x^2-x^3, Y=5); Vec(sum(k=0, (N+1)\2, 4^k*binomial(N+1, 2*k)*X^(N+1-2*k)*x^(Y*k))/(X^2-4*x^Y)^(N+1))
    

Formula

G.f.: ((1-x^2-x^3)^2 + 4*x^5) / ((1-x^2-x^3)^2 - 4*x^5)^2.
a(n) = 4*a(n-2) + 4*a(n-3) - 6*a(n-4) - 4*a(n-5) - 2*a(n-6) - 4*a(n-7) - 5*a(n-8) + 8*a(n-9) - 6*a(n-10) + 4*a(n-11) - a(n-12).

A384883 Number of maximal sparse subsets of the binary indices of n, where a set is sparse iff 1 is not a first difference.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 3, 4, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 4, 4, 2, 2, 2, 4, 3, 3, 4, 5, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 2
Offset: 0

Author

Gus Wiseman, Jul 02 2025

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The binary indices of 27 are {1,2,4,5}, with maximal sparse subsets {{1,4},{1,5},{2,4},{2,5}}, so a(27) = 4.
		

Crossrefs

For subsets of {1..n} we get A000931 (shifted), maximal case of A000045 (shifted).
This is the maximal case of A245564.
The greatest number whose binary indices are one of these subsets is A374356.
For prime instead of binary indices we have A385215, maximal case of A166469.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A202064 counts subsets containing n with k maximal runs.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
A384893 counts subsets by number of maximal anti-runs, for partitions A268193, A384905.

Programs

  • Mathematica
    spars[S_]:=Select[Subsets[S],FreeQ[Differences[#],1]&];
    bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
    maximize[sys_]:=Complement@@Prepend[Most[Subsets[#]]&/@sys,sys];
    Table[Length[maximize[spars[bpe[n]]]],{n,0,100}]

A384906 Number of maximal anti-runs of consecutive parts not increasing by 1 in the prime indices of n (with multiplicity).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1
Offset: 1

Author

Gus Wiseman, Jun 22 2025

Keywords

Comments

First differs from A300820 at a(462) = 3, A300820(462) = 2.
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 prime indices of 462 are {1,2,4,5}, with maximal anti-runs ((1),(2,4),(5)), so a(462) = 3.
		

Crossrefs

For the strict case we have A356228.
For binary instead of prime indices we have A384890 (for runs A069010).
For runs instead of anti-runs we have A385213.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Length[Split[prix[n],#2!=#1+1&]],{n,100}]

A385213 Number of maximal runs of consecutive parts increasing by 1 in the prime indices of n (with multiplicity).

Original entry on oeis.org

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

Author

Gus Wiseman, Jun 22 2025

Keywords

Comments

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 prime indices of 24 are {1,1,1,2}, with maximal runs ((1),(1),(1,2)), so a(24) = 3.
		

Crossrefs

Positions of first appearances are A000079.
For binary instead of prime indices we have A069010 (for anti-runs A384890).
For anti-runs instead of runs we have A384906.
A034839 counts subsets by number of maximal runs, for strict partitions A116674.
A055396 gives least prime index, greatest A061395.
A056239 adds up prime indices, row sums of A112798.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.

Programs

  • Mathematica
    prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Length[Split[prix[n],#2==#1+1&]],{n,100}]

A385572 Number of subsets of {1..n} with the same number of maximal runs (increasing by 1) as maximal anti-runs (increasing by more than 1).

Original entry on oeis.org

1, 2, 3, 4, 7, 12, 19, 34, 63, 112, 207, 394, 739, 1398, 2687, 5152, 9891, 19128, 37039, 71754, 139459, 271522, 528999, 1032308, 2017291, 3945186, 7723203, 15134440, 29679407, 58245068, 114389683, 224796210, 442021743, 869658304, 1711914351, 3371515306
Offset: 0

Author

Gus Wiseman, Jul 04 2025

Keywords

Comments

Also the number of subsets of {1..n} with the same number of adjacent elements increasing by 1 as adjacent elements increasing by more than 1.

Examples

			The set {2,3,5,6,8} has maximal runs ((2,3),(5,6),(8)) and maximal anti-runs ((2),(3,5),(6,8)) so is counted under a(8).
The a(0) = 1 through a(6) = 19 subsets:
  {}  {}   {}   {}   {}       {}       {}
      {1}  {1}  {1}  {1}      {1}      {1}
           {2}  {2}  {2}      {2}      {2}
                {3}  {3}      {3}      {3}
                     {4}      {4}      {4}
                     {1,2,4}  {5}      {5}
                     {1,3,4}  {1,2,4}  {6}
                              {1,2,5}  {1,2,4}
                              {1,3,4}  {1,2,5}
                              {1,4,5}  {1,2,6}
                              {2,3,5}  {1,3,4}
                              {2,4,5}  {1,4,5}
                                       {1,5,6}
                                       {2,3,5}
                                       {2,3,6}
                                       {2,4,5}
                                       {2,5,6}
                                       {3,4,6}
                                       {3,5,6}
		

Crossrefs

The LHS is counted by A034839 (for partitions A384881, strict A116674), rank statistic A069010.
The case containing n + 1 is A217615.
The RHS is counted by A384893 or A210034 (for partitions A268193, strict A384905), rank statistic A384890.
Subsets of this type are ranked by A385575.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<5, [1, 2, 3, 4, 7][n+1], ((3*n-4)*a(n-1)-
          (3*n-5)*a(n-2)+(5*n-12)*a(n-3)-2*(4*n-11)*a(n-4)+4*(n-3)*a(n-5))/(n-1))
        end:
    seq(a(n), n=0..35);  # Alois P. Heinz, Jul 06 2025
  • Mathematica
    Table[Length[Select[Subsets[Range[n]],Length[Split[#,#2==#1+1&]]==Length[Split[#,#2!=#1+1&]]&]],{n,0,10}]
  • PARI
    a(n)=polcoef([1,1,1]*[x,0,0;x,x^2,1;0,x,x]^n*[1,0,0]~,n) \\ Christian Sievers, Jul 06 2025

Formula

Let M be the matrix [1,0,0; 1,x,1/x; 0,1,1]. Then a(n) is the sum of the constant terms of the entries in the left column of M^n. - Christian Sievers, Jul 06 2025

Extensions

a(21) and beyond from Christian Sievers, Jul 06 2025

A062135 Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 2, 6, 3, 1, 0, 3, 10, 12, 4, 1, 0, 3, 19, 28, 20, 5, 1, 0, 4, 28, 66, 60, 30, 6, 1, 0, 4, 44, 126, 170, 110, 42, 7, 1, 0, 5, 60, 236, 396, 365, 182, 56, 8, 1, 0, 5, 85, 396, 868, 1001, 693, 280, 72, 9, 1
Offset: 0

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Because the sequence of column m=2*k, k >= 1, of A034851 is the partial sum sequence of the one of column m=2*k-1 the present triangle is essentially Losanitsch's triangle A034851.
Row sums give A051450 with A051450(0) := 1. Column sequences (without leading zeros) are for m=0..6: A000007, A008619, A005993, A005995, A018211, A018213, A062136.

Examples

			Triangle begins:
  {1};
  {0,1};
  {0,1,1};
  {0,2,2,1};
  ...
Pe(4,x^2)=1+6*x^2+x^4.
		

Crossrefs

Programs

  • Mathematica
    t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n - 1 + m, n - m], {n, 0, 12}, {m, 0, n}]] (* Michael De Vlieger, Sep 28 2024, after Jean-François Alcover at A034851  *)

Formula

T(n, m) = A034851(n-1+m, n-m), n >= m >= 0; A034851(n-1, n) := 0, n >= 1, A034851(-1, 0) := 1.
T(n, m) = 0 if n= 1; T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1) if n+m even and T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1)-binomial((n+m-3)/2, m-1) if n+m odd, n >= m >= 1.
G.f. for column m: x^m*Pe(m, x^2)/(((1-x)^(2*m))*(1+x)^m), m >= 0, with Pe(m, x^2)= sum(A034839(m, k)*x^(2*k), k=0..floor(n/2)), the row polynomial of array A034839 (even-indexed entries of the rows of Pascal's triangle).

Extensions

More terms from Michael De Vlieger, Sep 28 2024
Previous Showing 41-50 of 72 results. Next