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 41-47 of 47 results.

A344088 Flattened tetrangle of reversed strict integer partitions sorted first by sum, then colexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 12 2021

Keywords

Comments

The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
  0: ()
  1: (1)
  2: (2)
  3: (12)(3)
  4: (13)(4)
  5: (23)(14)(5)
  6: (123)(24)(15)(6)
  7: (124)(34)(25)(16)(7)
  8: (134)(125)(35)(26)(17)(8)
  9: (234)(135)(45)(126)(36)(27)(18)(9)
		

Crossrefs

Positions of first appearances are A015724.
Triangle sums are A066189.
The non-strict version is A080576.
Taking lex instead of colex gives A246688 (non-reversed: A344086).
The non-reversed version is A344087.
Taking revlex instead of colex gives A344089 (non-reversed: A118457).
A026793 gives reversed strict partitions in A-S order (sum/length/lex).
A319247 sorts strict partitions by Heinz number.
A329631 sorts reversed strict partitions by Heinz number.
A344090 gives strict partitions in A-S order (sum/length/lex).

Programs

  • Mathematica
    colex[f_,c_]:=OrderedQ[PadRight[{Reverse[f],Reverse[c]}]];
    Table[Sort[Reverse/@Select[IntegerPartitions[n],UnsameQ@@#&],colex],{n,0,10}]

A127743 Triangular array where T(n,k) is the number of set partitions of n with k atomic parts.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 6, 5, 3, 1, 22, 16, 9, 4, 1, 92, 60, 31, 14, 5, 1, 426, 252, 120, 52, 20, 6, 1, 2146, 1160, 510, 209, 80, 27, 7, 1, 11624, 5776, 2348, 904, 335, 116, 35, 8, 1, 67146, 30832, 11610, 4184, 1481, 507, 161, 44, 9, 1
Offset: 1

Views

Author

Alford Arnold, Feb 24 2007

Keywords

Comments

Triangular array distributing the Bell numbers (A000110). The value associated with each partition is the product of A074664(k) for each part of size k, times the number of compositions associated with the partition (A048996 & A072881). The value for T(n,k) is the total of these values for each partition of n into k parts.
Calculating the appropriate weights can be done by "working backward". Suppose for example we know the weights for 1 through 6 and desire the weight for the partitions of seven: Substitute the weights for each partition value and multiply. For example, 7 = 4+3 so f([4,3]) = 6*2 = 12; adjusting for the number of permutations of [4,3] we now have 2*12 = 24. Continuing in this manner for each partition of seven and summing to 451 we now know all of the values except that associated with the partition [7] which must be 877 - 451 = 426.
From Mike Zabrocki: (Start)
Every set partition can be uniquely split into "atomic" set partitions or is itself already atomic.
{{1},{2},{3}} = {{1}}|{{1}}|{{1}}
{{1},{23}} = {{1}}|{{12}}
{{12},{3}} = {{12}}|{{1}}
{{13},{2}} is already atomic
{{123}} is already atomic
where this operation | is defined as {A1,...,Ar}|{B1,...,Bs} = {A1,...,Ar,B1+n,...,Bs+n}
where Bi+n = {bi1+n,bi2+n,...,bik+n} if Bi = {bi1,bi2,...,bik} and n = |A1|+|A2|+...+|Ar|. (End)
Subtriangle (n >= 1 and 1 <= k <= n) of triangle given by [0,1,1,2,1,3,1,4,1,5,1,6,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Aug 03 2007
From Peter Bala, Aug 05 2014: (Start)
Let B(x) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + ... denote the o.g.f. for the Bell numbers A000110. Let f(x) = (B(x) - 1)/(x*B(x)) = 1 + x + x^2 + 2*x^3 + 6*x^4 + 22*x^5 + ..., the o.g.f. for the first column of this array. Then this array appears to be the Riordan array (f(x), x*f(x)).
If true, this gives the o.g.f. of the array as (B(x) - 1)/( x*(t + (1 - t)*B(x)) ) = 1 + (1 + t)*x + (2 + 2*t + t^2)*x^2 + ... and also the hockey-stick recurrence: T(n+1,k+1) = T(n,k) + T(n-1,k) + 2*T(n-2,k) + 6*T(n-3,k) + 22*T(n-4,k) + ..., n,k >= 1. (End)

Examples

			The partitions of 4 are
  4 31 22 211 1111
and the products are
  1*6 2*2 1*1 3*1 1*1
therefore row 4 of the table is
  6 5 3 1.
From _Philippe Deléham_, Aug 03 2007: (Start)
Triangle begins:
     1;
     1,    1;
     2,    2,   1;
     6,    5,   3,   1;
    22,   16,   9,   4,  1;
    92,   60,  31,  14,  5,  1;
   426,  252, 120,  52, 20,  6, 1;
  2146, 1160, 510, 209, 80, 27, 7, 1; ...
Triangle [0,1,1,2,1,3,1,4,1,...] DELTA [1,0,0,0,0,0,...] begins:
  1;
  0,    1;
  0,    1,    1;
  0,    2,    2,   1;
  0,    6,    5,   3,   1;
  0,   22,   16,   9,   4,  1;
  0,   92,   60,  31,  14,  5,  1;
  0,  426,  252, 120,  52, 20,  6, 1;
  0, 2146, 1160, 510, 209, 80, 27, 7, 1; ...
(End)
		

Crossrefs

Cf. A000041, A000110 (row sums), A074664 (1st column), A048996, A072881, A036043, A036042, A084938.

Programs

  • Mathematica
    T[n_, m_] := T[n, m] = Sum[Sum[T[k+i, k]*Binomial[n-m-k-1, n-m-k-i], {i, 1, n-m-k}]*Binomial[k+m-1, k], {k, 1, n-m}] + Binomial[n-1, n-m]; Table[T[n, m], {n, 1, 10}, {m, 1, n}] // Flatten (* Jean-François Alcover, Mar 23 2015, after Vladimir Kruchinin *)
  • Maxima
    T(n,m):=sum((sum(T(k+i,k)*binomial(n-m-k-1,n-m-k-i),i,1,n-m-k))*binomial(k+m-1,k),k,1,n-m)+binomial(n-1,n-m); /* Vladimir Kruchinin, Mar 21 2015 */
    
  • PARI
    {T(n,m) = sum(k=1,n-m, (sum(i=1, n-m-k, (T(k+i, k)*binomial(n-m-k-1, n-m-k-i))*binomial(k+m-1, k)))) + binomial(n-1, n-m)};
    for(n=1, 10, for(m=1, n, print1(T(n,m), ", "))) \\ G. C. Greubel, Dec 06 2018

Formula

T(n, m) = Sum_{k=1..n-m}( Sum_{i=1..n-m-k}(T(k+i, k)*C(n-m-k-1, n-m-k-i))*C(k+m-1, k) ) + C(n-1, n-m). - Vladimir Kruchinin, Mar 21 2015

Extensions

Edited by Franklin T. Adams-Watters, Jan 25 2010

A179972 Irregular table T(n,k) = A178886(n,k)/A048996(n,k) read by rows.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 6, 2, 2, 1, 1, 24, 6, 6, 2, 2, 1, 1, 120, 24, 24, 24, 6, 6, 6, 2, 2, 1, 1, 720, 120, 120, 120, 24, 24, 24, 24, 6, 6, 6, 2, 2, 1, 1, 5040, 720, 720, 720, 720, 120, 120, 120, 120, 120, 24, 24, 24, 24, 24, 6, 6, 6, 2, 2, 1, 1, 40320
Offset: 1

Views

Author

Alford Arnold, Aug 04 2010

Keywords

Comments

Row n has A000041(n) terms.
Consider the five partitions of the number 4:
4 3+1 2+2 2+1+1 and 1+1+1+1
rewriting as 4000 3100 2200 2110 and 1111
then a(n) counts the ways that the zeros can be permuted:
6,2,2,1,1
agreeing with the factorial of the difference between A036042 and A036043.

Examples

			Row four of A178886 begins: 6 4 2 3 1
Row four of A048996 begins: 1 2 1 3 1
so,
Row four of A179972 begins: 6 2 2 1 1
Triangle T(n,k) begins:
    1;
    1,  1;
    2,  1,  1;
    6,  2,  2,  1, 1;
   24,  6,  6,  2, 2, 1, 1;
  120, 24, 24, 24, 6, 6, 6, 2, 2, 1, 1;
  ...
		

Crossrefs

Cf. A178886, A048996, A036042, A036043, A179973 (row sums).

Formula

T(n,k) = ( A036042(n,k) - A036043(n,k))!.
T(n,k) = n!/A178888(n,k). - R. J. Mathar, Mar 03 2011

A179974 Triangle read by rows: T(n,k) = (n-A049085(n,k))! in columns 1<=k<=A000041(n), rows n>=1.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 6, 1, 1, 2, 2, 6, 6, 24, 1, 1, 2, 6, 2, 6, 24, 6, 24, 24, 120, 1, 1, 2, 6, 2, 6, 24, 24, 6, 24, 120, 24, 120, 120, 720, 1, 1, 2, 6, 24, 2, 6, 24, 24, 120, 6, 24, 120, 120, 720, 24, 120, 720, 120, 720, 720, 5040, 1, 1, 2, 6, 24, 2, 6, 24, 120, 24, 120, 720, 6, 24, 120, 120, 720, 720, 24, 120, 720, 720, 5040, 120, 720, 5040
Offset: 1

Views

Author

Alford Arnold, Aug 05 2010

Keywords

Comments

Since A049085 is a resortment of A036043 both A179972 and A179974 have row sums equal to A179973.

Examples

			Triangle begins
1;
1,1;
1,1,2;
1,1,2,2,6;
1,1,2,2,6,6,24;
1,1,2,6,2,6,24,6,24,24,120;
1,1,2,6,2,6,24,24,6,24,120,24,120,120,720;
1,1,2,6,24,2,6,24,24,120,6,24,120,120,720,24,120,720,120,720,720,5040;
1,1,2,6,24,2,6,24,120,24,120,720,6,24,120,120,720,720,24,120,720,720,5040,120,720,5040,720,5040,5040,40320,
		

Crossrefs

Cf. A000041 (row lengths), A179973 (row sums), A036042, A049085 (max part).

A305309 Array read by rows: a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 6, 4, 6, 1, 5, 8, 12, 9, 12, 8, 1, 6, 10, 16, 9, 12, 36, 8, 12, 24, 10, 1, 7, 12, 20, 24, 15, 48, 27, 36, 16, 72, 32, 15, 40, 12, 1, 8, 14, 24, 30, 16, 18, 60, 72, 48, 54, 20, 96, 54, 144, 16, 20, 120, 80, 18, 60, 14, 1, 9, 16, 28, 36, 40, 21, 72, 90, 48, 60, 144, 27, 24, 120, 144, 192, 216, 96, 25, 160, 90, 360, 80, 24, 180, 160, 21, 84, 16, 1, 10, 18, 32, 42, 48, 25, 24, 84, 108, 120, 72, 180, 96, 108, 28, 144, 180, 96, 240, 576, 108, 128, 216, 30, 200, 240, 480, 540, 480, 32, 30, 240, 135, 720, 240, 28, 252, 280, 24, 112, 18, 1
Offset: 1

Views

Author

Wolfdieter Lang, May 31 2018

Keywords

Comments

The Data section here is longer than usual. Do not shorten it! - N. J. A. Sloane, Jan 10 2019
The length of row n is A000041(n), the number of partitions of n.
Partitions follow the Abramowitz-Stegun (A-St) order (see the link).
The row sums give A001906(n) = Fibonacci(2*n).
The triangle T(n, m) obtained by summing in row n the entries of the columns k with identical part number m is A078812(n, m) = binomial(n+m-1, 2*m-1) (with offsets n >= 1, m = 1..n). The array of the number of parts m = m(n,k) = A036043(n, k) in A-St order.
This array is the elementwise product of the array A048996, the composition numbers, and A118851, the products of the parts of partitions, both arrays are in A-St order.
Therefore a(n, k) is the sum of the number of products of the block lengths of all the A048996(n, k) set partition of [n] := {1,2, ..., n} with m = m(n, k) blocks consisting of consecutive numbers corresponding to the k-th partition of n in A-St order. Because the block structure depends only on the exponents (signature) of the underlying partition this leads to the product of the two array entries. Equivalently, one can consider compositions. Then a(n, k) gives the sum of the products of the parts of all A048996(n, k) compositions originating from the k-th partition of n.
This array is the result of an attempt to understand the comment of Kevin Long, May 11 2018, on A001906.
This array is similar to A085643 but some pairs of numbers like (27, 36), (72,48), (54,144), ... are there swapped.

Examples

			For the rows n = 1..10, and comments on compositions and set partitions with blocks of consecutive numbers, see the link.
Example: n = 5, k = 4: the partition is (1^2, 3^1) = [1,1,3] with m = m(n,k) = 3. The A048996(5, 4) = 3 compositions are 1 + 1 + 3, 1 + 3 + 1 and 3 + 1 + 1. The corresponding three consecutive 3-block partitions of [5] := {1, 2, ..., 5} are {1}, {2}, {3,4,5} and {1}, {2,3,4}, {5} and {1,2,3}, {4}, {5}, Therefore, a(5, 4) = 1*1*3 + 1*3*1 + 3*1*1 = 3*3 = 9. For the compositions one has the same sum from the products of the parts.
		

Crossrefs

Formula

a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

A344084 Concatenated list of all finite nonempty sets of positive integers sorted first by maximum, then by length, and finally lexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 11 2021

Keywords

Examples

			The sets are the columns below:
  1 2 1 3 1 2 1 4 1 2 3 1 1 2 1 5 1 2 3 4 1 1 1 2 2 3 1
      2   3 3 2   4 4 4 2 3 3 2   5 5 5 5 2 3 4 3 4 4 2
              3         4 4 4 3           5 5 5 5 5 5 3
                              4                       5
As a tetrangle, the first four triangles are:
  {1}
  {2},{1,2}
  {3},{1,3},{2,3},{1,2,3}
  {4},{1,4},{2,4},{3,4},{1,2,4},{1,3,4},{2,3,4},{1,2,3,4}
		

Crossrefs

Triangle lengths are A000079.
Triangle sums are A001793.
Positions of first appearances are A005183.
Set maxima are A070939.
Set lengths are A124736.

Programs

  • Mathematica
    SortBy[Rest[Subsets[Range[5]]],Last]

A344092 Flattened tetrangle of strict integer partitions, sorted first by sum, then by length, and finally reverse-lexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 14 2021

Keywords

Comments

First differs from A118457 at a(53) = 4, A118457(53) = 2.
The zeroth row contains only the empty partition.
A tetrangle is a sequence of finite triangles.

Examples

			Tetrangle begins:
   0: ()
   1: (1)
   2: (2)
   3: (3)(21)
   4: (4)(31)
   5: (5)(41)(32)
   6: (6)(51)(42)(321)
   7: (7)(61)(52)(43)(421)
   8: (8)(71)(62)(53)(521)(431)
   9: (9)(81)(72)(63)(54)(621)(531)(432)
		

Crossrefs

Same as A026793 with rows reversed.
Ignoring length gives A118457.
The non-strict version is A334439 (reversed: A036036/A334302).
The version for lex instead of revlex is A344090.
A026791 reads off lexicographically ordered reversed partitions.
A080577 reads off reverse-lexicographically ordered partitions.
A112798 reads off reversed partitions by Heinz number.
A193073 reads off lexicographically ordered partitions.
A296150 reads off partitions by Heinz number.

Programs

  • Mathematica
    revlensort[f_,c_]:=If[Length[f]!=Length[c],Length[f]
    				
Previous Showing 41-47 of 47 results.