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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A113109 Number of 5-tournament sequences: a(n) gives the number of increasing sequences of n positive integers (t_1,t_2,...,t_n) such that t_1 = 2 and t_i = 2 (mod 4) and t_{i+1} <= 5*t_i for 1

Original entry on oeis.org

1, 2, 16, 440, 43600, 16698560, 26098464448, 172513149018752, 4938593053649344000, 622793203804403960906240, 350552003258337075784341271552, 890153650520295355798989668668129280
Offset: 0

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Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Equals column 0 of triangle A113108, which is the matrix square of triangle A113106, which satisfies the recurrence: A113106(n,k) = [A113106^5](n-1,k-1) + [A113106^5](n-1,k).

Examples

			The tree of 5-tournament sequences of descendents
of a node labeled (2) begins:
[2]; generation 1: 2->[6,10]; generation 2:
6->[10,14,18,22,26,30], 10->[14,18,22,26,30,34,38,42,46,50]; ...
Then a(n) gives the number of nodes in generation n.
Also, a(n+1) = sum of labels of nodes in generation n.
		

Crossrefs

Programs

  • PARI
    {a(n)=local(M=matrix(n+1,n+1));for(r=1,n+1, for(c=1,r, M[r,c]=if(r==c,1,if(c>1,(M^5)[r-1,c-1])+(M^5)[r-1,c]))); return((M^2)[n+1,1])}

A113111 Number of 5-tournament sequences: a(n) gives the number of increasing sequences of n positive integers (t_1,t_2,...,t_n) such that t_1 = 3 and t_i = 3 (mod 4) and t_{i+1} <= 5*t_i for 1

Original entry on oeis.org

1, 3, 33, 1251, 173505, 94216515, 210576669921, 2002383115518243, 82856383278525698433, 15166287556997012904054915, 12437232461209961704387810340769
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Equals column 0 of triangle A113110, which is the matrix cube of triangle A113106, which satisfies the recurrence: A113106(n,k) = [A113106^5](n-1,k-1) + [A113106^5](n-1,k).

Examples

			The tree of 5-tournament sequences of descendents
of a node labeled (3) begins:
[3]; generation 1: 3->[7,11,15];
generation 2: 7->[11,15,19,23,27,31,35],
11->[15,19,23,27,31,35,39,43,47,51,55],
15->[19,23,27,31,35,39,43,47,51,55,59,63,67,71,75]; ...
Then a(n) gives the number of nodes in generation n.
Also, a(n+1) = sum of labels of nodes in generation n.
		

Crossrefs

Programs

  • PARI
    {a(n)=local(M=matrix(n+1,n+1));for(r=1,n+1, for(c=1,r, M[r,c]=if(r==c,1,if(c>1,(M^5)[r-1,c-1])+(M^5)[r-1,c]))); return((M^3)[n+1,1])}

A093729 Square table T, read by antidiagonals, where T(n,k) gives the number of n-th generation descendents of a node labeled (k) in the tree of tournament sequences.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 7, 7, 3, 1, 0, 41, 41, 15, 4, 1, 0, 397, 397, 123, 26, 5, 1, 0, 6377, 6377, 1656, 274, 40, 6, 1, 0, 171886, 171886, 36987, 4721, 515, 57, 7, 1, 0, 7892642, 7892642, 1391106, 134899, 10810, 867, 77, 8, 1, 0, 627340987, 627340987, 89574978, 6501536, 376175, 21456, 1351, 100, 9, 1
Offset: 0

Views

Author

Paul D. Hanna, Apr 14 2004; revised Oct 14 2005

Keywords

Comments

Column 1, of array T and antidiagonals, equals A008934, which is the number of tournament sequences.
A tournament sequence is an increasing sequence of positive integers (t_1,t_2,...) such that t_1 = 1 and t_{i+1} <= 2*t_i, where integer k>1.

Examples

			Array begins:
  1,      1,       1,       1,       1,      1,      1,     1,     1, ...],
  0,      1,       2,       3,       4,      5,      6,     7,     8, ...],
  0,      2,       7,      15,      26,     40,     57,    77,   100, ...],
  0,      7,      41,     123,     274,    515,    867,  1351,  1988, ...],
  0,     41,     397,    1656,    4721,  10810,  21456, 38507, 64126, ...],
  0,    397,    6377,   36987,  134899, 376175, 880032, .................],
  0,   6377,  171886, 1391106, 6501536, ...],
  0, 171886, 7892642, .....................];
Antidiagonals begin as:
  1;
  0,      1;
  0,      1,      1;
  0,      2,      2,     1;
  0,      7,      7,     3,    1;
  0,     41,     41,    15,    4,   1;
  0,    397,    397,   123,   26,   5,   1;
  0,   6377,   6377,  1656,  274,  40,   6,   1;
  0, 171886, 171886, 36987, 4721, 515,  57,   7,   1;
		

Crossrefs

Cf. A008934 (column k=1 of array and antidiagonals), A093730 (antidiagonal row sums).

Programs

  • Mathematica
    t[n_?Negative, ] = 0; t[0, ] = 1; t[n_, k_] /; k <= n := t[n, k] = t[n, k - 1] - t[n-1, k] + t[n - 1, 2 k - 1] + t[n - 1, 2 k]; t[n_, k_] := t[n, k] = Sum[(-1)^(j - 1)*Binomial[n + 1, j]*t[n, k - j], {j, 1, n + 1}]; Flatten[Table[t[i - k, k - 1], {i, 10}, {k, i}]] (* Jean-François Alcover, May 31 2011, after PARI prog. *)
  • PARI
    {T(n,k)=if(n<0,0,if(n==0,1,if(k==0,0, if(k<=n,T(n,k-1)-T(n-1,k)+T(n-1,2*k-1)+T(n-1,2*k), sum(j=1,n+1, (-1)^(j-1)*binomial(n+1,j)*T(n,k-j))))))}
    
  • PARI
    {a(n, m) = my(A=1); for(k=1, n, A = (A - q^k * r * subst( subst(A, q, q^2), r, r^2)) / (1-q)); subst(subst(A, r, q^(m-1)), q, 1)}; /* Michael Somos, Jun 19 2017 */
    
  • SageMath
    @CachedFunction
    def T(n, k):
        if n<0: return 0
        elif n==0: return 1
        elif k==0: return 0
        elif kA093729(n,k): return T(n-k,k)
    flatten([[A093729(n,k) for k in range(n+1)] for n in range(16)]) # G. C. Greubel, Feb 22 2024

Formula

T(0, k)=1 for k>=0, T(n, 0)=0 for n>=1; else T(n, k) = T(n, k-1) - T(n-1, k) + T(n-1, 2*k-1) + T(n-1, 2*k) for k<=n; else T(n, k) = Sum_{j=1..n+1} (-1)^(j-1)*C(n+1, j)*T(n, k-j) for k>n (Cook-Kleber).
Column k of T equals column 0 of the matrix k-th power of triangle A097710, which satisfies the matrix recurrence: A097710(n, k) = [A097710^2](n-1, k-1) + [A097710^2](n-1, k) for n>k>=0.
Sum_{k=0..n} T(n-k, k) = A093730(n) (antidiagonal row sums).

A113085 Number of 3-tournament sequences: a(n) gives the number of increasing sequences of n positive integers (t_1,t_2,...,t_n) such that t_1 = 1 and t_i = 1 (mod 2) and t_{i+1} <= 3*t_i for 1

Original entry on oeis.org

1, 1, 3, 21, 331, 11973, 1030091, 218626341, 118038692523, 166013096151621, 619176055256353291, 6207997057962300681573, 169117528577725378851523691, 12626174170113987651028630856581, 2602022118010488151483064379958957003
Offset: 0

Views

Author

Paul D. Hanna, Oct 14 2005

Keywords

Comments

Equals column 0 of triangle A113084, which satisfies: A113084(n,k) = [A113084^3](n-1,k-1) + [A113084^3](n-1,k).

Examples

			The tree of 3-tournament sequences of odd integer
descendents of a node labeled (1) begins:
[1]; generation 1: 1->[3]; generation 2: 3->[5,7,9];
generation 3: 5->[7,9,11,13,15], 7->[9,11,13,15,17,19,21],
9->[11,13,15,17,19,21,23,25,27]; ...
Then a(n) gives the number of nodes in generation n.
Also, a(n+1) = sum of labels of nodes in generation n.
		

Crossrefs

Programs

  • PARI
    {a(n)=local(M=matrix(n+1,n+1));for(r=1,n+1, for(c=1,r, M[r,c]=if(r==c,1,if(c>1,(M^3)[r-1,c-1])+(M^3)[r-1,c]))); return(M[n+1,1])}

A093654 Lower triangular matrix, read by rows, defined as the convergent of the concatenation of matrices using the iteration: M(n+1) = [[M(n),0*M(n)],[M(n)^2,M(n)^2]], with M(0) = [1].

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 2, 1, 2, 1, 1, 0, 0, 0, 1, 2, 1, 0, 0, 2, 1, 2, 0, 1, 0, 2, 0, 1, 7, 2, 4, 1, 7, 2, 4, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 0, 2, 1, 2, 0, 1, 0, 0, 0, 0, 0, 2, 0, 1, 7, 2, 4, 1, 0, 0, 0, 0, 7, 2, 4, 1, 2, 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 1, 7, 2, 0, 0, 4, 1, 0, 0, 7, 2, 0, 0, 4, 1
Offset: 1

Views

Author

Paul D. Hanna, Apr 08 2004

Keywords

Comments

Related to the number of tournament sequences (A008934). First column forms A093655, where A093655(2^n) = A008934(n) for n>=0. Row sums form A093656, where A093656(2^(n-1)) = A093657(n) for n>=1.

Examples

			Let M(n) be the lower triangular matrix formed from the first 2^n rows.
To generate M(3) from M(2), take the matrix square of M(2):
[1,0,0,0]^2=[1,0,0,0]
[1,1,0,0]...[2,1,0,0]
[1,0,1,0]...[2,0,1,0]
[2,1,2,1]...[7,2,4,1]
and append M(2)^2 to the bottom left and bottom right of M(2):
[1],
[1,1],
[1,0,1],
[2,1,2,1],
.........
[1,0,0,0],[1],
[2,1,0,0],[2,1],
[2,0,1,0],[2,0,1],
[7,2,4,1],[7,2,4,1].
Repeating this process converges to triangle A093654.
		

Crossrefs

Formula

First column: T(2^n, 1) = A008934(n) for n>=0.

A058222 Tree of tournament sequences read across rows.

Original entry on oeis.org

1, 2, 3, 4, 4, 5, 6, 5, 6, 7, 8, 5, 6, 7, 8, 6, 7, 8, 9, 10, 7, 8, 9, 10, 11, 12, 6, 7, 8, 9, 10, 7, 8, 9, 10, 11, 12, 8, 9, 10, 11, 12, 13, 14, 9, 10, 11, 12, 13, 14, 15, 16, 6, 7, 8, 9, 10, 7, 8, 9, 10, 11, 12, 8, 9, 10, 11, 12, 13, 14, 9, 10, 11, 12, 13, 14, 15, 16, 7, 8, 9
Offset: 0

Views

Author

N. J. A. Sloane, Dec 02 2000

Keywords

Examples

			Irregular triangle begins:
  1;
  2;
  3,4;
  4,5,6,5,6,7,8;
  ...
		

Crossrefs

A008934 gives number of children at level n. Cf. A058223.
Cf. A002083.

Formula

Top node is 1; each node k has children labeled k+1, k+2, ..., 2k at next level.

A093655 First column of lower triangular matrix A093654.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 2, 7, 1, 2, 2, 7, 2, 7, 7, 41, 1, 2, 2, 7, 2, 7, 7, 41, 2, 7, 7, 41, 7, 41, 41, 397, 1, 2, 2, 7, 2, 7, 7, 41, 2, 7, 7, 41, 7, 41, 41, 397, 2, 7, 7, 41, 7, 41, 41, 397, 7, 41, 41, 397, 41, 397, 397, 6377
Offset: 1

Views

Author

Paul D. Hanna, Apr 08 2004

Keywords

Comments

Related to the number of tournament sequences (A008934).
a(n) equals the number of tournament sequences (A008934) of length A000120(n-1), which is the number of 1's in the binary expansion of n-1.

Crossrefs

Formula

a(2^n) = A008934(n) for n>=0.
a(n) = A008934(A000120(n-1)) for n>=1.

A058223 Tree of Meeussen sequences read across rows.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 5, 6, 7, 8, 8, 10, 11, 12, 8, 9, 11, 12, 13, 8, 9, 10, 12, 13, 14, 9, 10, 11, 12, 13, 9, 10, 11, 12, 13, 14, 9, 10, 11, 12, 13, 14, 15, 9, 10, 11, 12, 13, 14, 15, 16, 13, 16, 18, 19, 20, 13, 15, 18, 20, 21, 22, 13, 14, 16, 19, 21, 22, 23, 13, 14, 15, 17, 20, 22, 23, 24
Offset: 0

Views

Author

N. J. A. Sloane, Dec 02 2000

Keywords

Comments

Labels on paths down the tree are Meeussen Sequences: b(1)=1 < b(2) <...< b(n) such that b(n)-1 has a unique representation as a sum of distinct b(i) (i

Examples

			Irregular triangle begins:
  1;
  2;
  3,4;
  5,6,7,5,6,7,8;
  ...
		

Crossrefs

A008934 gives number of children at level n. Cf. A058222.

A058311 Number of nodes at n-th level in tree in which top node is 1; each node k has children labeled k, k+1, ..., (k+1)^2 at next level.

Original entry on oeis.org

1, 4, 48, 7918, 463339346, 7134188685100826388, 13246386641449904934758023373599438217628, 643152870463337226096320122089499144560533929707886143570111588898313745804013188842
Offset: 0

Author

N. J. A. Sloane, Dec 09 2000

Keywords

Comments

Triggered by a comment from Michael Kleber, Dec 08 2009, who said: The algorithm in my paper with Cook lets you compute the equivalent sequence where the children of a node labeled (k) are labeled with all the integers in the interval [p(k), q(k)] where p,q are any polynomials you like (in the paper, p(k)=k+1 and q(k)=2k). For a bunch of p,q the resulting sequence is well known, e.g., p(k)=1, q(k)=k+1 is the Catalan numbers.

Crossrefs

Programs

  • Maple
    M:=4;
    L[0]:=[1]; a[0]:=1;
    for n from 1 to M do
    L[n]:=[];
    t1:=L[n-1];
    tc:=nops(t1);
    for i from 1 to tc do
    t2:=t1[i];
    for j from t2 to (t2+1)^2 do
    L[n]:=[op(L[n]),j]; od:
    a[n]:=nops(L[n]);
    #lprint(n,L[n],a[n]);
    od:
    od:
    [seq(a[n],n=0..M)];
    # See the reference for a better way to compute this!
    p := proc(n,k) option remember; local j ; if n = 1 then k^2+k+2; # (k+1)^2-(k-1) else sum( procname(n-1,j),j=k..(k+1)^2) ; fi; expand(%) ; end proc:
    A058311 := proc(n) if n = 0 then 1 ; else subs(k=1, p(n,k)) ; fi; end proc:
    for n from 0 do printf("%d,\n", A058311(n)) ; od: # R. J. Mathar, May 04 2009
  • Mathematica
    p[n_, k_] := p[n, k] = If[n == 1, k^2+k+2, Sum[p[n-1, j], {j, k, (k+1)^2}]];
    a[n_] := If[n == 0, 1, p[n, 1]];
    Table[Print[n, " ", a[n]]; a[n], {n, 0, 7}] (* Jean-François Alcover, Jun 26 2023, after R. J. Mathar *)

Extensions

Corrected, with Maple program, by N. J. A. Sloane, May 03 2009. Thanks to Max Alekseyev for pointing out that something was wrong.
Replaced a(4), added three more terms - R. J. Mathar, May 04 2009

A093657 2^(n-1)-th term of the row sums of triangle A093654.

Original entry on oeis.org

1, 2, 6, 28, 206, 2418, 45970, 1440746, 75840096, 6828414424, 1069361760254, 295609883371824, 146078092162147126, 130419475982163166640, 212257994312591826735888, 634463537260289571176650942
Offset: 1

Author

Paul D. Hanna, Apr 08 2004

Keywords

Crossrefs

Related to the number of tournament sequences (A008934).

Programs

  • Mathematica
    T[n_, k_]:= T[n,k]= If[n<0 || k>n, 0, If[n==k, 1, If[k==0, Sum[T[n-1,j]*T[j,0], {j,0,n-1}], Sum[T[n-1,j]*(T[j,k-1]+T[j,k]), {j,0,n-1}] ]]]; (* T = A097710 *)
    A093657[n_]:= A093657[n]= Sum[T[n,k], {k,0,n}];
    Table[A093657[n], {n,0,30}] (* G. C. Greubel, Feb 21 2024 *)
  • SageMath
    @CachedFunction
    def T(n, k): # T = A097710
        if n< 0 or k<0 or k>n: return 0
        elif k==n: return 1
        elif k==0: return sum(T(n-1,j)*T(j,0) for j in range(n))
        else: return sum(T(n-1, j)*(T(j, k-1)+T(j,k)) for j in range(n))
    def A093657(n): return sum(T(n,k) for k in range(n+1))
    [A093657(n) for n in range(31)] # G. C. Greubel, Feb 21 2024

Formula

a(n) = A093656(2^(n-1)) for n>=1.
a(n) = Sum_{k=0..n} A097710(n,k), row sums of triangle A097710.
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