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 11-19 of 19 results.

A072811 T(n,k) = multiplicity of the k-th partition of n in Mathematica order, defined to be the count of its permutations (compositions).

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 2, 2, 3, 3, 4, 1, 1, 2, 2, 3, 1, 6, 4, 1, 6, 5, 1, 1, 2, 2, 3, 2, 6, 4, 3, 3, 12, 5, 4, 10, 6, 1, 1, 2, 2, 3, 2, 6, 4, 1, 6, 3, 12, 5, 3, 6, 12, 20, 6, 1, 10, 15, 7, 1, 1, 2, 2, 3, 2, 6, 4, 2, 6, 3, 12, 5, 3, 6, 12, 12, 20, 6, 1, 12, 10, 4, 30, 30, 7, 5, 20, 21, 8, 1
Offset: 0

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Author

Wouter Meeussen, Aug 09 2002

Keywords

Comments

The sum of row n equals A011782(n). The first and last columns equal 1. The number of integers per row equals the partition number p(n). Row n is a vector of weights or multiplicities relating counts of ordered versus unordered objects classified according to the partitions of n.
a(n) is the multinomial coefficient of the signature of the n-th partition. - Franklin T. Adams-Watters, Apr 08 2008
Let f(x)=1/(1-sum(j>=1, c[j]*x^j))=sum(n>=0, w(n)*x^n), then the coefficients of wn=Pn(c[1],...,c[n]), listed in reverse lexicographic order, give row n of T(n,k). - Groux Roland, Mar 08 2011

Examples

			The partitions of 4 are {4}, {3,1}, {2,2}, {2,1,1}, {1,1,1,1}, so the fourth row equals 1,2,1,3,1 since these are the counts of the permutations of these lists.
Triangle begins:
1;
1;
1, 1;
1, 2, 1;
1, 2, 1, 3, 1;
1, 2, 2, 3, 3, 4, 1;
1, 2, 2, 3, 1, 6, 4, 1, 6, 5, 1;
		

Crossrefs

Programs

  • Mathematica
    mult[li:{__Integer}] := Apply[Multinomial, Length/@Split[ Sort[li] ] ]; Table[mult/@Partitions[n], {n, 12}]
  • PARI
    \\ here mulp(v) computes the multiplicity of the given partition.
    mulp(v) = {my(p=(#v)!, k=1); for(i=2, #v, k=if(v[i]==v[i-1], k+1, p/=k!; 1)); p/k!}
    Row(n)={apply(mulp, vecsort([Vecrev(p) | p<-partitions(n)], , 4))}
    { for(n=0, 9, print(Row(n))) } \\ Peter Dolland, Nov 11 2019

A276816 Irregular triangle read by rows: T(n,m) = coefficients in power/Fourier series expansion of an arbitrary anharmonic oscillator's exact period.

Original entry on oeis.org

-24, 480, -120, 6720, 3360, -241920, 1774080, -560, 40320, 40320, -1774080, 20160, -3548160, 61501440, -591360, 92252160, -1845043200, 8364195840, -2520, 221760, 221760, -11531520, 221760, -23063040, 461260800, 110880, -23063040, -11531520, 1383782400, -15682867200, -11531520, 691891200, 1383782400, -62731468800, 476759162880
Offset: 1

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Author

Bradley Klee, Sep 18 2016

Keywords

Comments

The phase space trajectory A276738 has phase space angular velocity A276814 and differential time dependence A276815. We calculate the period K = Int dt over the range [2*Pi, 0], trivial to compute from A276815 using A273496. Then K/(2*Pi) = 1 + sum b^(2n)*T(n,m)*f'(n,m); where the sum runs over n = 1, 2, 3 ... and m = 1, 2, 3, ... A000041(2n), and f'(n,m) = f(2n,m) of A276738 with Q=1/2. Choosing one point from the infinite dimensional coefficient space--v_i=0 for odd i, v_i=(-1)^(i/2-1)/2/(i!) otherwise--setting b^2 = 4*k, and summing over the entire table obtains the EllipticK expansion 2*A038534/A038533. For more details read "Plane Pendulum and Beyond by Phase Space Geometry" (Klee, 2016).

Examples

			n/m   1     2     3         4         5
------------------------------------------
1  | -24   480
2  | -120  6720  3360   -241920   1774080
------------------------------------------
For pendulum values, f'(1,*)={(-1/384), 0}, f'(2,*) = {1/46080, 0, 1/294912, 0, 0}. Then K/(2Pi) = 1+(-1/384)*(-24)*4*k+((1/46080)*(-120)+(1/294912)*3360)*16*k^2=1+(1/4)*k + (9/64)*k^2, the first few terms of EllipticK.
		

Crossrefs

Programs

  • Mathematica
    RExp[n_]:=Expand[b Plus[R[0], Total[b^# R[#] & /@ Range[n]]]]
    RCalc[n_]:=With[{basis =Subtract[Tally[Join[Range[n + 2], #]][[All, 2]],Table[1, {n + 2}]] & /@ IntegerPartitions[n + 2][[3 ;; -1]]},
    Total@ReplaceAll[Times[-2, Multinomial @@ #, v[Total[#]],Times @@ Power[RSet[# - 1] & /@ Range[n + 2], #]] & /@ basis, {Q^2 -> 1, v[2] -> 1/4}]]
    dt[n_] := With[{exp = Normal[Series[-1/(1 + x)/.x -> Total[(2 # v[#] RExp[n - 1]^(# - 2) &/@Range[3, n + 2])], {b, 0, n}]]},
    Expand@ReplaceAll[Coefficient[exp, b, #] & /@ Range[n], R -> RSet]]
    RingGens[n_] :=Times @@ (v /@ #) & /@ (IntegerPartitions[n]/. x_Integer :> x + 2)
    tri[m_] := MapThread[Function[{a, b},Times[-# /. v[n_] :> Q^n /. Q^n_ :>  Binomial[n, n/2],(1/2) Coefficient[a, #]] & /@ b], {dt[2 m][[2 #]] & /@ Range[m], RingGens[2 #] & /@ Range[m]}]
    RSet[0] = 1; Set[RSet[#], Expand@RCalc[#]] & /@ Range[2*7];
    tri7 = tri[7]; tri7 // TableForm
    PeriodExpansion[tri_, n_] := ReplaceAll[ 1 + Dot[MapThread[ Dot, {tri,
      2 RingGens[2 #] & /@ Range[n]}], (2 h)^(Range[n])], {v[m_] :> (v[m]*(1/2)^m)}]
    {#,SameQ[Normal@Series[(2/Pi)*EllipticK[k],{k,0,7}],#]}&@ReplaceAll[
    PeriodExpansion[tri7,7],{v[n_/;OddQ[n]]:>0,v[n_]:> (-1)^(n/2-1)/2/(n!),h->2 k}]

A124772 Number of set partitions associated with compositions in standard order.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 1, 2, 1, 1, 1, 4, 6, 6, 4, 8, 4, 4, 1, 3, 3, 3, 1, 2, 1, 1, 1, 5, 10, 10, 10, 20, 10, 10, 5, 15, 15, 15, 5, 10, 5, 5, 1, 4, 6, 6, 4, 8, 4, 4, 1, 3, 3, 3, 1, 2, 1, 1, 1, 6, 15, 15, 20, 40, 20, 20, 15, 45, 45, 45, 15, 30, 15, 15, 6, 24, 36, 36, 24, 48, 24, 24, 6, 18
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
Arrange the parts of the set partition by the smallest member of each part and read off the part sizes. E.g., for 1|24|3, the associated composition is 1,2,1. When the set partition is presented as the sequence of parts that each member is in, simply count the times each part number occurs. This representation for 1|24|3 is {1,2,3,2}.

Examples

			Composition number 11 is 2,1,1; the associated set partitions are 12|3|4, 13|2|4 and 14|2|3, so a(11) = 3.
The table starts:
1
1
1 1
1 2 1 1
		

Crossrefs

Cf. A066099, A124773, A011782 (row lengths), A000110 (row sums), A036040, A080575.

Formula

For composition b(1),...,b(k), a(n) = Product_{i=1}^k C((Sum_{j=i}^k b(j))-1, b(i)-1).

A198254 Number of maximum-diversity partitions of n.

Original entry on oeis.org

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

Views

Author

Olivier Gérard, Oct 22 2011

Keywords

Comments

A maximum-diversity partition of n is an integer partition whose part distribution maximizes the number of different compositions (=distinct partition orderings) that can be constructed from it.
An integer composition of n corresponds to a subgroup of the symmetric group on n element whose cycles are formed of contiguous integers.

Examples

			For n=17, there are 3 partitions reaching the maximum possible of 7!/2 =2520 distinct orderings : {4, 3, 2, 2, 2, 1, 1, 1, 1}, {4, 3, 2, 2, 1, 1, 1, 1, 1, 1} and {3, 3, 2, 2, 2, 1, 1, 1, 1, 1}.
		

Crossrefs

A102462 gives the number of compositions that can be constructed from a maximum-diversity partition of n.

A211350 Refined triangle A124323: T(n,k) is the number of partitions of an n-set that are of type k (k-th integer partition, defined by A194602).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 6, 4, 3, 1, 1, 10, 10, 15, 5, 10, 1, 1, 15, 20, 45, 15, 60, 6, 15, 15, 10, 1, 1, 21, 35, 105, 35, 210, 21, 105, 105, 70, 7, 105, 21, 35, 1, 1, 28, 56, 210, 70, 560, 56, 420, 420, 280, 28, 840, 168, 280, 8, 105, 210, 280, 28
Offset: 1

Views

Author

Tilman Piesk, Apr 09 2012

Keywords

Comments

Name could also be "Triangle of multinomial coefficients, read by rows (version 4)", compare A036040, A080575, A178867. The latter and this one differ only in the order of columns.
The rows are counted from 1, the columns from 0.
Row lengths: 1,2,3,5,7,11... (partition numbers A000041)
Row sums: 1,2,5,15,52,203... (Bell numbers A000110)
Row maxima: 1,1,3,6,15,60,210,840,3780,12600,69300,415800... (A102356)
Distinct entries per row: 1,1,2,4,4,7,7,13,17,23,26,40... (A102465)
Rightmost columns are those from Pascal's triangle A007318 without the second one (i.e. triangle A184049). The other columns - (always?) without a 1 at the top - are multiples of these columns from Pascal's triangle; so actually only the top elements of each column are needed to calculate the other entries; these top elements are in A211360. (The top elements of the related triangle A178867 are in A178866.)

Crossrefs

A327003 Irregular triangle read by rows in which the n-th row lists multinomials for partitions of 3n which have only parts which are multiples of 3, in Hindenburg order.

Original entry on oeis.org

1, 1, 1, 10, 1, 84, 280, 1, 220, 462, 9240, 15400, 1, 455, 5005, 50050, 210210, 1401400, 1401400, 1, 816, 18564, 185640, 24310, 4084080, 13613600, 2858856, 85765680, 285885600, 190590400, 1, 1330, 54264, 542640, 293930, 24690120, 82300400, 32332300, 135795660, 2715913200, 4526522000, 3802278480, 38022784800, 76045569600, 36212176000
Offset: 0

Views

Author

Peter Luschny, Aug 14 2019

Keywords

Comments

The Hindenburg order refers to the partition generating algorithm of C. F. Hindenburg (1779). [Knuth 7.2.1.4H]

Examples

			The irregular triangle starts:
[0] [1]
[1] [1]
[2] [1, 10]
[3] [1, 84, 280]
[4] [1, 220, 462, 9240, 15400]
[5] [1, 455, 5005, 50050, 210210, 1401400, 1401400]
[6] [1, 816, 18564, 185640, 24310, 4084080, 13613600, 2858856, 85765680, 285885600, 190590400]
		

Crossrefs

Cf. A000012 (m=0, subdivided into rows of length A000041), A080575 (m=1), A257490 (m=2), this sequence (m=3), A327004 (m=4).
Cf. A000041 (length of rows), A291973 (sum of rows), A291451 (coarser subdivision).
Cf. A260876.

Programs

  • SageMath
    def A327003row(n):
        shapes = ([3*x for x in p] for p in Partitions(n))
        return [SetPartitions(sum(s), s).cardinality() for s in shapes]
    for n in (0..7): print(A327003row(n))

Formula

Row of lengths are in A000041.

A327004 Irregular triangle read by rows in which the n-th row lists multinomials for partitions of 4n which have only parts which are multiples of 4, in Hindenburg order.

Original entry on oeis.org

1, 1, 1, 35, 1, 495, 5775, 1, 1820, 6435, 450450, 2627625, 1, 4845, 125970, 4408950, 31177575, 727476750, 2546168625, 1, 10626, 735471, 25741485, 1352078, 1338557220, 15616500900, 1577585295, 165646455975, 1932541986375, 4509264634875
Offset: 0

Views

Author

Peter Luschny, Aug 14 2019

Keywords

Comments

The Hindenburg order refers to the partition generating algorithm of C. F. Hindenburg (1779). [Knuth 7.2.1.4H]

Examples

			The irregular triangle starts:
[0] [1]
[1] [1]
[2] [1, 35]
[3] [1, 495, 5775]
[4] [1, 1820, 6435, 450450, 2627625]
[5] [1, 4845, 125970, 4408950, 31177575, 727476750, 2546168625]
[6] [1, 10626, 735471, 25741485, 1352078, 1338557220, 15616500900, 1577585295, 165646455975, 1932541986375, 4509264634875]
		

Crossrefs

Cf. A000012 (m=0, subdivided into rows of length A000041), A080575 (m=1), A257490 (m=2), A327003 (m=3), this sequence (m=4).
Cf. A000041 (length of rows), A291975 (sum of rows), A291452 (coarser subdivision).
Cf. A260876.

Programs

  • SageMath
    def A327004row(n):
        shapes = ([4*x for x in p] for p in Partitions(n))
        return [SetPartitions(sum(s), s).cardinality() for s in shapes]
    for n in (0..6): print((A327004row(n)))

Formula

Row of lengths are in A000041.

A356656 Partition triangle read by rows. The coefficients of the incomplete Bell polynomials.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 1, 3, 1, 0, 1, 4, 3, 6, 1, 0, 1, 5, 10, 10, 15, 10, 1, 0, 1, 6, 15, 10, 15, 60, 15, 20, 45, 15, 1, 0, 1, 7, 21, 35, 21, 105, 70, 105, 35, 210, 105, 35, 105, 21, 1, 0, 1, 8, 28, 56, 35, 28, 168, 280, 210, 280, 56, 420, 280, 840, 105, 70, 560, 420, 56, 210, 28, 1
Offset: 0

Views

Author

Peter Luschny, Aug 28 2022

Keywords

Comments

We call a triangle a 'partition triangle' if the rows have length A000041 or A000041 + 1.

Examples

			The triangle starts:
[0] 1;
[1] 0, 1;
[2] 0, 1, 1;
[3] 0, 1, 3,  1;
[4] 0, 1, [4,  3],  6,  1;
[5] 0, 1, [5, 10], [10, 15],  10,  1;
[6] 0, 1, [6, 15, 10], [15,  60, 15], [20, 45],  15,   1;
[7] 0, 1, [7, 21, 35], [21, 105, 70, 105], [35, 210, 105], [35, 105], 21, 1;
Summing the bracketed terms reduces the triangle to A048993.
The first few polynomials are:
[0] 1;
[1] 0, z[0];
[2] 0, z[1], z[0]^2;
[3] 0, z[2], 3*z[0]*z[1], z[0]^3;
[4] 0, z[3], 4*z[0]*z[2]+3*z[1]^2, 6*z[0]^2*z[1], z[0]^4;
[5] 0, z[4], 5*z[0]*z[3]+10*z[1]*z[2], 10*z[0]^2*z[2]+15*z[0]*z[1]^2, 10*z[0]^3* z[1], z[0]^5;
It is noteworthy that the substitution z[n] -> n! for n >= 0 yields A132393. More examples are given in the authors blog post (see links).
		

Crossrefs

Variants: A036040, A080575, A178867. Row sums: A000110.
A048993 (reduced triangle), A052810 (length of rows), A132393 (factorial substituion).

Programs

  • Maple
    aRow := n -> seq(coeffs(IncompleteBellB(n, k, seq(z[i], i = 0..n))), k = 0..n):
    seq(aRow(n), n = 0..8);
  • SageMath
    from functools import cache
    @cache
    def incomplete_bell_polynomial(n, k):
        Z = var(["z_" + str(i) for i in range(n - k + 1)])
        R = PolynomialRing(ZZ, Z, n - k + 1, order='lex')
        if k == 0: return R(k^n)
        return R(sum(binomial(n-1,j-1) * incomplete_bell_polynomial(n-j,k-1) * Z[j-1]
                for j in range(n - k + 2)).expand())
    def poly_row(n): return [incomplete_bell_polynomial(n, k) for k in range(n + 1)]
    def coeff_row(n): return flatten([[0] if (c := p.coefficients()) == [] else c for p in poly_row(n)])
    for n in range(8): print(coeff_row(n))

Formula

In row n the coefficients of IBell(n, k, Z_n) for k = 0..n are lined up. Z_n denotes the set of variables z[0], z[1], ... z[n] of the incomplete Bell polynomial IBell(n, k) of degree k.

A358165 Irregular triangular array read by rows. T(n,k) is the number of direct sum decompositions V_1 + V_2 + ... + V_m = GF(2)^n with the dimensions of the V_i corresponding to the k-th partition of n in canonical ordering, n >= 0, 1 <= k <= A000041(n).

Original entry on oeis.org

1, 1, 1, 3, 1, 28, 28, 1, 120, 280, 1680, 840, 1, 496, 9920, 29760, 138880, 277760, 83328, 1, 2016, 166656, 499968, 357120, 19998720, 19998720, 15554560, 139991040, 139991040, 27998208, 1, 8128, 2731008, 8193024, 48377856, 1354579968, 1354579968, 2902671360, 13545799680, 81274798080, 40637399040, 126427463680, 379282391040, 227569434624, 32509919232
Offset: 0

Views

Author

Geoffrey Critzer, Nov 01 2022

Keywords

Examples

			Triangle begins:
  1;
  1;
  1,   3;
  1,  28,   28;
  1, 120,  280,  1680,    840;
  1, 496, 9920, 29760, 138880, 277760, 83328;
  ...
T(4,3) = 280.  For n=4 the five partitions in canonical ordering are {4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}.  The third partition in this order is {2,2}.  So T(4,3) = A002884(4)/(A002884(2)^2*2!) = 280.
		

Crossrefs

Cf. A270880, A270881 (row sums), A279038, A080575, A000041, A002884, A053601 (main diagonal).

Programs

  • Mathematica
    dsd2[n_, signature_] := Product[2^n - 2^i, {i, 0, n - 1}]/ Product[Product[2^k - 2^i, {i, 0, k - 1}]^signature[[k]]*signature[[k]]!, {k, 1, n}];Table[Map[dsd2[n, #] &,Map[Table[Count[#, i], {i, 1, n}] &, IntegerPartitions[n]]], {n, 0,6}] // Grid

Formula

For i = 1,...,n let a_i be the number of parts of size i in the k-th partition of n in canonical ordering. T(n,k) = A002884(n)/Product_{j=1..n} A002884(j)^a_j*a_j!.
Previous Showing 11-19 of 19 results.