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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A248120 Triangle read by rows: Lagrange (compositional) inversion of a function in terms of the coefficients of the Taylor series expansion of its reciprocal, scaled version of A248927, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 2, 6, 3, 24, 36, 4, 120, 360, 60, 80, 5, 720, 3600, 1800, 1200, 300, 150, 6, 5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7, 40320, 423360, 705600, 235200, 176400, 352800, 58800, 35280, 23520, 35280, 7056, 1960, 1176, 392, 8
Offset: 1

Views

Author

Tom Copeland, Oct 28 2014

Keywords

Comments

Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series and a scaled version of A248927. A refinement of A055302, which enumerates the number of labeled rooted trees with n nodes and k leaves, with row sums A000169.
Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
g(t) = [ 1 (0') ] * t
+ [ 2 (0') (1') ] * t^2/2!
+ [ 6 (0') (1')^2 + 3 (0')^2 (2') ] * t^3/3!
+ [24 (0') (1')^3 + 36 (0')^2 (1') (2') + 4 (0')^3 (3')] * t^4/4!
+ [120 (0') (1')^4 + 360 (0')^2 (1')^2 (2') + (0')^3 [60 (2')^2
+ 80 (1') (3')] + 5 (0')^4 (4')] * t^5/5!
+ [720 (0')(1')^5 + 3600 (0')^2 (1')^3(2') + (0')^3 [1800 (1')(2')^2 + 1200 ( 1')^2(3')] + (0')^4 [300 (2')(3') + 150 (1')(4')] + 6 (0')^5 (5')] * t^6/6! + ... .
Operating with [1/(n*(n-1))] d/d(1') = [1/(n*(n-1))] d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
Each n-th partition polynomial here is n times the (n-1)-th partition polynomial of A248927.
From Tom Copeland, Nov 24 2014: (Start)
The n-th row is a mapping of the homogeneous symmetric monomials generated by [x(1) + x(2) + ... + x(n)]^(n-1) under the umbral mapping x(m)^j = h_j, for any m. E.g., [a + b + c]^2 = [a^2 + b^2 + c^2] + 2 * [a*b + a*c + b*c] is mapped to [3 * h_2] + 2 * [3 * h_1 * h_1] = 3 * h_2 + 6 * h_1^2 = A248120(3) with h_0 = 1. (Example corrected Jul 14 2015.)
For another example and relations to A134264 and A036038, see A134264. The general relation is n * A134264(n) = A248120(n) / A036038(n-1) where the arithmetic is performed on the coefficients of matching partitions in each row n.
The Abramowitz and Stegun reference in A036038 gives combinatorial interpretations of A036038 and relations to other number arrays.
This can also be related to repeated umbral composition of Appell sequences and topology with the Bernoulli numbers playing a special role. See the Todd class link. (End)
As presented above and in the Copeland link, this entry is related to exponentiation of e.g.f.s and, therefore, to discussions in the Scott and Sokal preprint (see eqn. 3.1 on p. 10 and eqn. 3.62 p. 24). - Tom Copeland, Jan 17 2017

Examples

			Triangle begins
     1;
     2;
     6,     3;
    24,    36,     4;
   120,   360,    60,    80,    5;
   720,  3600,  1800,  1200,  300,   150,    6;
  5040, 37800, 37800, 16800, 3150, 12600, 3150, 420, 630, 252, 7;
  ...
For f(t)= e^t-1, h(t)= t/f(t)= t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t)= t - t^2/2 + t^3/3 + ... = log(1+t).
		

Crossrefs

Cf. A134264 and A248927, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); (n+1)*n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 02 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is (j-1)! P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
For example, from g(t) above, T(6,3) = 5! * [6!/(3!*2!)]/(2!)^2 = 1800 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = 5! * [6!/4!]/(2!*3!) = 300.
If the initial factorial and final denominator of T(n,k) are removed and the expression divided by j and the partitions reversed in order, then A134264 is obtained, a refinement of the Narayana numbers.
For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums are A000169(n) = n^(n-1) = n* A000272(n).
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/[d{x/[h_0+h_1*x+ ...]/dx]. Then the partition polynomials above are given by (W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)=exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(y)d/dy] exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t)= n * PS(n-1,t) are R = t * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
Following the notes connected to the Lagrange reversion theorem in A248927, a generator for the n-th partition polynomial P_n of this entry is (d/dx)^(n-1) (h (x))^n, and -log(1-t*P.) = (t*Q.) / (1 - t*Q.), umbrally, where (Q.)^n = Q_n is the n-th partition polynomial of A248927. - Tom Copeland, Nov 25 2016
With h_0 = 1, the n-th partition polynomial is obtained as the n-th element (with initial index 0) of the first column of M^{n+1}, where M is the matrix with M_{i,j}= binomial(i,j) h_{i-j}, i.e., the lower triangular Pascal matrix with its n-th diagonal multiplied by h_n. This follows from the Lagrange inversion theorem and the relation between powers of matrices such as M and powers of formal Taylor series discussed in A133314. This is equivalent to repeated binomial convolution of the coefficients of the Taylor series with itself. - Tom Copeland, Nov 13 2019
T(n,k) = n*A248927(n,k). - Andrew Howroyd, Feb 02 2022

Extensions

Terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

A333483 Sort all positive integers, first by sum of prime indices (A056239), then by number of prime indices (A001222).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 14, 15, 18, 20, 24, 32, 13, 21, 22, 25, 27, 28, 30, 36, 40, 48, 64, 17, 26, 33, 35, 42, 44, 45, 50, 54, 56, 60, 72, 80, 96, 128, 19, 34, 39, 49, 55, 52, 63, 66, 70, 75, 81, 84, 88, 90, 100, 108, 112, 120, 144, 160, 192, 256, 23, 38, 51, 65, 77, 68, 78, 98, 99, 105, 110, 125, 104, 126, 132, 135, 140, 150, 162, 168, 176, 180, 200, 216, 224, 240, 288, 320, 384, 512
Offset: 0

Views

Author

Gus Wiseman, May 10 2020

Keywords

Comments

A refinement of A215366, from which it first differs at a(49) = 55, A215366(49) = 52.
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

			Triangle begins:
   1
   2
   3   4
   5   6   8
   7   9  10  12  16
  11  14  15  18  20  24  32
  13  21  22  25  27  28  30  36  40  48  64
  17  26  33  35  42  44  45  50  54  56  60  72  80  96 128
		

Crossrefs

Row lengths are A000041.
Ignoring length gives A215366 (graded Heinz numbers).
Sorting by decreasing length gives A333484.
Finally sorting lexicographically by prime indices gives A185974.
Finally sorting colexicographically by prime indices gives A334433.
Finally sorting reverse-lexicographically by prime indices gives A334435.
Finally sorting reverse-colexicographically by prime indices gives A334438.
Number of prime indices is A001222.
Reversed partitions in Abramowitz-Stegun (sum/length/lex) order are A036036.
Partitions in (sum/length/colex) order are A036037.
Sum of prime indices is A056239.
Sorting reversed partitions by Heinz number gives A112798.
Sorting partitions by Heinz number gives A296150.

Programs

  • Mathematica
    Join@@@Table[Sort[Times@@Prime/@#&/@IntegerPartitions[n,{k}]],{n,0,8},{k,0,n}]

A344089 Flattened tetrangle of reversed strict integer partitions, sorted first by length and then colexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 12 2021

Keywords

Comments

First differs from the revlex (instead of colex) version for partitions of 12.
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)(12)
  4: (4)(13)
  5: (5)(23)(14)
  6: (6)(24)(15)(123)
  7: (7)(34)(25)(16)(124)
  8: (8)(35)(26)(17)(134)(125)
  9: (9)(45)(36)(27)(18)(234)(135)(126)
		

Crossrefs

Positions of first appearances are A015724 plus one.
Taking lex instead of colex gives A026793 (non-reversed: A118457).
Triangle sums are A066189.
Reversing all partitions gives A344090.
The non-strict version is A344091.
A319247 sorts strict partitions by Heinz number.
A329631 sorts reversed strict partitions by Heinz number.

Programs

  • Mathematica
    Table[Reverse/@Sort[Select[IntegerPartitions[n],UnsameQ@@#&]],{n,0,30}]

A344085 Triangle of squarefree numbers first grouped by greatest prime factor, then sorted by omega, then in increasing order, read by rows.

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 15, 30, 7, 14, 21, 35, 42, 70, 105, 210, 11, 22, 33, 55, 77, 66, 110, 154, 165, 231, 385, 330, 462, 770, 1155, 2310, 13, 26, 39, 65, 91, 143, 78, 130, 182, 195, 273, 286, 429, 455, 715, 1001, 390, 546, 858, 910, 1365, 1430, 2002, 2145, 3003, 5005, 2730, 4290, 6006, 10010, 15015, 30030
Offset: 1

Views

Author

Gus Wiseman, May 11 2021

Keywords

Comments

Differs from A339195 in having 77 before 66.

Examples

			Triangle begins:
   1
   2
   3   6
   5  10  15  30
   7  14  21  35  42  70 105 210
		

Crossrefs

Programs

  • Mathematica
    nn=4;
    GatherBy[SortBy[Select[Range[Times@@Prime/@Range[nn]],SquareFreeQ[#]&&PrimePi[FactorInteger[#][[-1,1]]]<=nn&],PrimeOmega],FactorInteger[#][[-1,1]]&]

A248927 Triangle read by rows: T(n,k) are the coefficients of the Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its reciprocal, n >= 1, k = 1..A000041(n-1).

Original entry on oeis.org

1, 1, 2, 1, 6, 9, 1, 24, 72, 12, 16, 1, 120, 600, 300, 200, 50, 25, 1, 720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1, 5040, 52920, 88200, 29400, 22050, 44100, 7350, 4410, 2940, 4410, 882, 245, 147, 49, 1, 40320, 564480, 1411200, 376320, 705600, 940800
Offset: 1

Views

Author

Tom Copeland, Oct 16 2014

Keywords

Comments

Coefficients are listed in reverse graded colexicographic order (A228100). This is the reverse of Abramowitz and Stegun order (A036036).
Coefficients for Lagrange (compositional) inversion of a function in terms of the Taylor series expansion of its shifted reciprocal. Complementary to A134264 for formal power series. A refinement of A141618 with row sums A000272.
Given an invertible function f(t) analytic about t=0 with f(0)=0 and df(0)/dt not 0, form h(t) = t / f(t) and denote h_n = (n') as the coefficient of t^n/n! in h(t). Then the compositional inverse of f(t), g(t), as a formal Taylor series, or e.g.f., is given up to the first few orders by
g(t)/t = [ 1 (0') ]
+ [ 1 (0') (1') ] * t
+ [ 2 (0') (1')^2 + 1 (0')^2 (2') ] * t^2/2!
+ [ 6 (0') (1')^3 + 9 (0')^2 (1') (2') + 1 (0')^3 (3') ] * t^3/3!
+ [24 (0') (1')^4 + 72 (0')^2 (1')^2 (2') + (0')^3 [12 (2')^2
+ 16 (1') (3')] + (0')^4 (4')] * t^4/4!
+ [120 (0')(1')^5 + 600 (0')^2 (1')^3(2') + (0')^3 [300 (1')(2')^2 + 200 ( 1')^2(3')] + (0')^4 [50 (2')(3') + 25 (1')(4')] + (0')^5 (5')] * t^5/5! + [720 (0')(1')^6 + (0')^2 (1')^4(2')+(0')^3 [5400 (1')^2(2')^2 + 2400 (1')^3(3')] + (0')^4 [450 (2')^3+ 1800 (1')(2')(3') + 450( 1')^2(4')]+ (0')^5 [60 (3')^2 + 90 (2')(4') + 36 (1')(5')] + (0')^6 (6')] * t^6/6! + ...
..........
From Tom Copeland, Oct 28 2014: (Start)
Expressing g(t) as a Taylor series or formal e.g.f. in the indeterminates h_n generates a refinement of A055302, which enumerates the number of labeled root trees with n nodes and k leaves, with row sum A000169.
Operating with (1/n^2) d/d(1') = (1/n^2) d/d(h_1) on the n-th partition polynomial in square brackets above associated with t^n/n! generates the (n-1)-th partition polynomial.
Multiplying the n-th partition polynomial here by (n + 1) gives the (n + 1)-th partition polynomial of A248120. (End)
These are also the coefficients in the expansion of a series related to the Lagrange reversion theorem presented in Wikipedia of which the Lagrange inversion formula about the origin is a special case. Cf. Copeland link. - Tom Copeland, Nov 01 2016

Examples

			Triangle T(n,k) begins:
    1;
    1;
    2,    1;
    6,    9,    1;
   24,   72,   12,   16,   1;
  120,  600,  300,  200,  50,   25,   1;
  720, 5400, 5400, 2400, 450, 1800, 450, 60, 90, 36, 1;
  ...
For f(t) = e^t-1, h(t) = t/f(t) = t/(e^t-1), the e.g.f. for the Bernoulli numbers, and plugging the Bernoulli numbers into the Lagrange inversion formula gives g(t) = t - t^2/2 + t^3/3 + ... = log(1+t).
		

Crossrefs

Cf. A134264 and A248120, "scaled" versions of this Lagrange inversion.
Cf. A036038.

Programs

  • PARI
    C(v)={my(n=vecsum(v), S=Set(v)); n!^2/((n-#v+1)!*prod(i=1, #S, my(x=S[i], c=#select(y->y==x, v)); x!^c*c!))}
    row(n)=[C(Vec(p)) | p<-Vecrev(partitions(n-1))]
    { for(n=1, 7, print(row(n))) } \\ Andrew Howroyd, Feb 02 2022

Formula

For j>1, there are P(j,m;a...) = j! / [ (j-m)! (a_1)! (a_2)! ... (a_(j-1))! ] permutations of h_0 through h_(j-1) in which h_0 is repeated (j-m) times; h_1, repeated a_1 times; and so on with a_1 + a_2 + ... + a_(j-1) = m.
If, in addition, a_1 + 2 * a_2 + ... + (j-1) * a_(j-1) = j-1, then each distinct combination of these arrangements is correlated with a partition of j-1.
T(j,k) is [(j-1)!/j]* P(j,m;a...) / [(2!)^a_2 (3!)^a_3 ... ((j-1)!)^a_(j-1) ] for the k-th partition of j-1. The partitions are in reverse order--from bottom to top--from the order in Abramowitz and Stegun (page 831).
For example, from g(t) above, T(6,3) = [5!/6][6!/(3!*2!)]/(2!)^2 = 300 for the 3rd partition from the bottom under n=6-1=5 with m=3 parts, and T(6,5) = [5!/6][6!/4!]/(2!*3!) = 50.
If the initial factorial and final denominator are removed and the partitions reversed in order, A134264 is obtained, a refinement of the Narayana numbers.
For f(t) = t*e^(-t), g(t) = T(t), the Tree function, which is the e.g.f. of A000169, and h(t) = t/f(t) = e^t, so h_n = 1 for all n in this case; therefore, the row sums of A248927 are A000169(n)/n = n^(n-2) = A000272(n).
Let W(x) = 1/(df(x)/dx)= 1/{d[x/h(x)]/dx}=1/{d[x/[h_0+h_1*x+ ...]]/dx}. Then the partition polynomials above are given by (1/n)(W(x)*d/dx)^n x, evaluated at x=0, and the compositional inverse of f(t) is g(t)= exp(t*W(x)*d/dx) x, evaluated at x=0. Also, dg(t)/dt = W(g(t)). See A145271.
With exp[x* PS(.,t)] = exp[t*g(x)]=exp[x*W(y)d/dy] exp(t*y) eval. at y=0, the raising (creation) and lowering (annihilation) operators defined by R PS(n,t) = PS(n+1,t) and L PS(n,t)= n * PS(n-1,t) are R = t * W(d/dt) and L =(d/dt)/h(d/dt)=(d/dt) 1/[(h_0)+(h_1)*d/dt+(h_2)*(d/dt)^2/2!+...], which will give a lowering operator associated to the refined f-vectors of permutohedra (cf. A133314 and A049019).
Then [dPS(n,z)/dz]/n eval. at z=0 are the row partition polynomials of this entry. (Cf. A139605, A145271, and link therein to Mathemagical Forests for relation to planted trees on p. 13.)
As noted in A248120 and A134264, this entry is given by the Hadamard product by partition of A134264 and A036038. For example, (1,4,2,6,1)*(1,4,6,12,24) = (1,16,12,72,24). - Tom Copeland, Nov 25 2016
T(n,k) = ((n-1)!)^2/((n-j)!*Product_{i>=1} s_i!*(i!)^s_i), where (1*s_1 + 2*s_2 + ... = n-1) is the k-th partition of n-1 and j = s_1 + s_2 ... is the number of parts. - Andrew Howroyd, Feb 02 2022

Extensions

Name edited and terms a(31) and beyond from Andrew Howroyd, Feb 02 2022

A344087 Flattened tetrangle of strict integer partitions sorted first by sum, then colexicographically.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, May 11 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: (21)(3)
  4: (31)(4)
  5: (41)(32)(5)
  6: (321)(51)(42)(6)
  7: (421)(61)(52)(43)(7)
  8: (521)(431)(71)(62)(53)(8)
  9: (621)(531)(81)(432)(72)(63)(54)(9)
		

Crossrefs

Positions of first appearances are A015724.
Triangle sums are A066189.
Taking revlex instead of colex gives A118457.
The not necessarily strict version is A211992.
Taking lex instead of colex gives A344086.
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[Select[IntegerPartitions[n],UnsameQ@@#&],colex],{n,0,10}]

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}]

A182937 Triangle in which n-th row lists all integer partitions of n, in order of traversing the periphery of the Fenner-Loizou tree in the clockwise sense.

Original entry on oeis.org

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

Views

Author

Peter Luschny, Jan 21 2011

Keywords

Comments

If the Fenner-Loizou tree is traversed in the counterclockwise sense (preorder traversal) the integer partitions are in lexicographic order.

Examples

			First five rows are:
[[1]]
[[1, 1], [2]]
[[1, 1, 1], [2, 1], [3]]
[[1, 1, 1, 1], [2, 1, 1], [3, 1], [4], [2, 2]]
[[1, 1, 1, 1, 1], [2, 1, 1, 1], [3, 1, 1], [4, 1], [5], [3, 2], [2, 2,1]]
		

References

  • T. I. Fenner and G. Loizou, Comp. J. 23 (1980), 332-337.
  • D. E. Knuth, TAOCP 4 (2005), fasc. 3, 7.2.1.4, exercise 10.

Crossrefs

See A036036 for the Hindenburg (graded reflected colexicographic) ordering.
See A036037 for the graded colexicographic ordering.
See A080576 for the Maple (graded reflected lexicographic) ordering.
See A080577 for the Mathematica (graded reverse lexicographic) ordering.
See A193073 for the graded lexicographic ordering.
See A228100 for the Fenner-Loizou (binary tree) ordering.

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]

A228099 Triangle read by rows, T(n, k) = prime(1)^p(k,1)*...*prime(n)^p(k,n) where p(k,j) is the j-th part of the k-th partition of n, additionally T(0,0) = 1. The partitions of n are ordered such that partitions of n into r parts appear in lexicographic order previous to the partitions of n into s parts if s < r. (Fenner-Loizou tree).

Original entry on oeis.org

1, 2, 6, 4, 30, 12, 8, 210, 60, 36, 24, 16, 2310, 420, 180, 120, 72, 48, 32, 30030, 4620, 1260, 840, 900, 360, 240, 216, 144, 96, 64, 510510, 60060, 13860, 9240, 6300, 2520, 1680, 1800, 1080, 720, 480, 432, 288, 192, 128, 9699690, 1021020, 180180, 120120
Offset: 0

Views

Author

Peter Luschny, Aug 10 2013

Keywords

Comments

The partitions' representation (A228100) is a weakly decreasing list of parts.
The rows (read from left to right) are strongly decreasing. The T(n, 0) are the primorial numbers A002110(n). The right side of the triangle are the powers of 2,
T(n, A000041(n)) = A000079(n). The row sums are A074140.
The partition corresponding to a(n), n > 0, can be recovered as the exponents of the primes in the canonical prime factorization of a(n).

Examples

			The six-th row is:
[1, 1, 1, 1, 1, 1] -> 30030
[2, 1, 1, 1, 1] -> 4620
[2, 2, 1, 1] -> 1260
[3, 1, 1, 1] -> 840
[2, 2, 2] -> 900
[3, 2, 1] -> 360
[4, 1, 1] -> 240
[3, 3] -> 216
[4, 2] -> 144
[5, 1] -> 96
[6] -> 64
		

References

  • D. E. Knuth: The Art of Computer Programming. Generating all combinations and partitions, vol. 4, fasc. 3, 7.2.1.4, exercise 10.

Crossrefs

Cf. A228100.

Programs

  • Maple
    b:= proc(n, i) b(n, i):= `if`(n=0 or i=1, [[1$n]], [b(n, i-1)[],
          `if`(i>n, [], map(x-> [i, x[]], b(n-i, i)))[]])
        end:
    T:= n-> map(h-> mul(ithprime(j)^h[j], j=1..nops(h)), sort(b(n$2),
            proc(x, y) local i; if nops(x)<>nops(y) then return
            nops(x)>nops(y) else for i to nops(x) do if x[i]<>y[i]
            then return x[i]Alois P. Heinz, Aug 13 2013
  • Mathematica
    b[n_, i_] := If[n == 0 || i == 1, {Array[1&, n]}, Join[b[n, i-1], If[i>n, {}, Map[Function[x, Prepend[x, i]], b[n-i, i]]]]]; T[n_] := Map[Function[h, Times @@ ((Prime /@ Range[Length[h]])^h)], Sort[b[n, n], Which[Length[#1] > Length[#2], True, Length[#1] < Length[#2], False, True, OrderedQ[#1, #2]]&]]; Table[T[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Jan 27 2014, after Maple *)
  • Sage
    from collections import deque
    def Partitions_Fenner_Loizou(n):
        p = ([], 0, n)
        queue = deque()
        queue.append(p)
        yield p
        while len(queue) > 0 :
            (phead, pheadLen, pnum1s) = queue.popleft()
            if pnum1s != 1 :
                head = phead[:pheadLen] + [2]
                q = (head, pheadLen + 1, pnum1s - 2)
                if 1 <= q[2] : queue.append(q)
                yield q
            if pheadLen == 1 or (pheadLen > 1 and \
                        (phead[pheadLen - 1] != phead[pheadLen - 2])) :
                head = phead[:pheadLen]
                head[pheadLen - 1] += 1
                q = (head, pheadLen, pnum1s - 1)
                if 1 <= q[2] : queue.append(q)
                yield q
    def A228099_row(n):
        if n == 0: return [1]
        L = []
        P = primes_first_n(n)
        for p in Partitions_Fenner_Loizou(n):
            e = p[0] + [1 for i in range(p[2])]
            c = mul(P[i]^e[i] for i in range(len(e)))
            L.append(c)
        return L
    for n in (0..7): A228099_row(n)
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