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.
A291975 := proc(n) exp((cos(z) + cosh(z))/2 - 1): (4*n)!*coeff(series(%, z, 4*(n+1)), z, 4*n) end: seq(A291975(n), n=0..12);
P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
a[n_] := Module[{cl = CoefficientList[P[4, n], x]}, Sum[cl[[k + 1]]/k!, {k, 0, n}]];
Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Jul 23 2019, after Peter Luschny in A291452 *)
seq(n)={my(a=vector(n+1)); a[1]=1; for(n=1, n, a[1+n]=sum(k=1, n, binomial(4*n-1, 4*k-1) * a[1+n-k])); a} \\ Andrew Howroyd, Jan 21 2020
[ n ] [0 1 2 3 4 5 6]
[ m ] ------------------------------------------------------
[ 0 ] [1, 1, 2, 3, 5, 7, 11] A000041
[ 1 ] [1, 1, 2, 5, 15, 52, 203] A000110
[ 2 ] [1, 1, 4, 31, 379, 6556, 150349] A005046
[ 3 ] [1, 1, 11, 365, 25323, 3068521, 583027547] A291973
[ 4 ] [1, 1, 36, 6271, 3086331, 3309362716, 6626013560301] A291975
A260878,A309725, ...
For example the number of set partitions of {1,2,...,9} with sizes in [9], [6,3] and [3,3,3] is 1, 84 and 280 respectively. Thus A(3,3) = 365.
Formatted as a triangle:
[1]
[1, 1]
[1, 1, 2]
[1, 1, 2, 3]
[1, 1, 4, 5, 5]
[1, 1, 11, 31, 15, 7]
[1, 1, 36, 365, 379, 52, 11]
[1, 1, 127, 6271, 25323, 6556, 203, 15]
.
From _Peter Luschny_, Aug 14 2019: (Start)
For example consider the case n = 4. There are five integer partitions of 4:
P = [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]]. The shapes are m times the parts of the integer partitions: S(m) = [[4m], [3m, m], [2m, 2m], [2m, m, m], [m, m, m, m]].
* In the case m = 1 we look at set partitions of {1, 2, 3, 4} with sizes in [[4], [3, 1], [2, 2], [2, 1, 1], [1, 1, 1, 1]] which gives rise to [1, 4, 3, 6, 1] with sum 15.
* In the case m = 2 we look at set partitions of {1, 2, .., 8} with sizes in [[8], [6, 2], [4, 4], [4, 2, 2], [2, 2, 2, 2]] which gives rise to [1, 28, 35, 210, 105] with sum 379.
* In the case m = 0 we look at set partitions of {} with sizes in [[0], [0, 0], [0, 0], [0, 0, 0], [0, 0, 0, 0]] which gives rise to [1, 1, 1, 1, 1] with sum 5 (because the only partition of the empty set is the set that contains the empty set, thus from the definition T(0,4) = Sum_{S(0)} card({0}) = A000041(4) = 5).
If n runs through 0, 1, 2,... then the result is an irregular triangle in which the n-th row lists multinomials for partitions of [m*n] which have only parts which are multiples of m. These are the triangles A080575 (m = 1), A257490 (m = 2), A327003 (m = 3), A327004 (m = 4). In the case m = 0 the triangle is A000012 subdivided into rows of length A000041. See the cross references how this case integrates into the full picture.
(End)
A:= proc(m, n) option remember; `if`(m=0, combinat[numbpart](n),
`if`(n=0, 1, add(binomial(m*n-1, m*k-1)*A(m, n-k), k=1..n)))
end:
seq(seq(A(d-n, n), n=0..d), d=0..10); # Alois P. Heinz, Aug 14 2019
A[m_, n_] := A[m, n] = If[m==0, PartitionsP[n], If[n==0, 1, Sum[Binomial[m n - 1, m k - 1] A[m, n - k], {k, 1, n}]]];
Table[Table[A[d - n, n], {n, 0, d}], {d, 0, 10}] // Flatten (* Jean-François Alcover, Dec 06 2019, after Alois P. Heinz *)
def A260876(m, n): shapes = ([x*m for x in p] for p in Partitions(n)) return sum(SetPartitions(sum(s), s).cardinality() for s in shapes) for m in (0..4): print([A260876(m,n) for n in (0..6)])
Triangle starts: [1] [0, 1] [0, 1, 10] [0, 1, 84, 280] [0, 1, 682, 9240, 15400] [0, 1, 5460, 260260, 1401400, 1401400] [0, 1, 43690, 7128576, 99379280, 285885600, 190590400]
CL := (f, x) -> PolynomialTools:-CoefficientList(f, x): A291451_row := proc(n) exp(x*(exp(z)/3+2*exp(-z/2)*cos(z*sqrt(3)/2)/3-1)): series(%, z, 66): CL((3*n)!*coeff(series(%,z,3*(n+1)),z,3*n),x) end: for n from 0 to 7 do A291451_row(n) od; # Alternative: A291451row := proc(n) local P; P := proc(m, n) option remember; if n = 0 then 1 else add(binomial(m*n, m*k)*P(m, n-k)*x, k=1..n) fi end: CL(P(3, n), x); seq(%[k+1]/k!, k=0..n) end: # Peter Luschny, Sep 03 2018
P[m_, n_] := P[m, n] = If[n == 0, 1, Sum[Binomial[m*n, m*k]*P[m, n - k]*x, {k, 1, n}]];
row[n_] := Module[{cl = CoefficientList[P[3, n], x]}, Table[cl[[k + 1]]/k!, {k, 0, n}]];
Table[row[n], {n, 0, 7}] // Flatten (* Jean-François Alcover, Jul 23 2019, after Peter Luschny *)
A291976 := proc(n) exp(1 - (cos(z) + cosh(z))/2): (4*n)!*coeff(series(%, z, 4*(n+1)), z, 4*n) end: seq(A291976(n), n=0..12); # second Maple program: b:= proc(n, t) option remember; `if`(n=0, 1-2*t, add( b(n-4*j, 1-t)*binomial(n-1, 4*j-1), j=1..n/4)) end: a:= n-> b(4*n, 0): seq(a(n), n=0..20); # Alois P. Heinz, Aug 14 2019
b[n_, t_] := b[n, t] = If[n == 0, 1-2t, Sum[b[n-4j, 1-t] * Binomial[n-1, 4j-1], {j, 1, n/4}]];
a[n_] := b[4n, 0];
Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Jan 27 2023, after Alois P. Heinz *)
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]
Array A(n, k) starts: [0] 1, 1, 1, 1, 1, 1, ... [1] 1, 1, 1, 1, 1, 1, ... [2] 1, 1, 3, 15, 105, 945, ... A001147 [3] 1, 1, 10, 280, 15400, 1401400, ... A025035 [4] 1, 1, 35, 5775, 2627625, 2546168625, ... A025036 [5] 1, 1, 126, 126126, 488864376, 5194672859376, ... A025037 [6] 1, 1, 462, 2858856, 96197645544, 11423951396577720, ... A025038 . Triangle A(n-k, k) starts: [0] 1; [1] 1, 1; [2] 1, 1, 1; [3] 1, 1, 1, 1; [4] 1, 1, 3, 1, 1; [5] 1, 1, 10, 15, 1, 1; [6] 1, 1, 35, 280, 105, 1, 1;
A := (n, k) -> mul(binomial((j + 1)*n - 1, n - 1), j = 0..k-1): seq(seq(A(n-k, k), k = 0..n), n = 0..9); # Alternative, using recursion: A := proc(n, k) local P; P := proc(n, k) option remember; if n = 0 then return x^k*k! fi; if k = 0 then 1 else add(binomial(n*k, n*j)* P(n,k-j)*x, j=1..k) fi end: coeff(P(n, k), x, k) / k! end: seq(print(seq(A(n, k), k = 0..5)), n = 0..6); # Alternative, using exponential generating function: egf := n -> ifelse(n=0, 1, exp(x^n/n!)): ser := n -> series(egf(n), x, 8*n): row := n -> local k; seq((n*k)!*coeff(ser(n), x, n*k), k = 0..6): for n from 0 to 6 do [n], row(n) od; # Peter Luschny, Aug 15 2024
A[n_, k_] := Product[Binomial[n (j + 1) - 1, n - 1], {j, 0, k - 1}]; Table[A[n - k, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Michael De Vlieger, Apr 13 2023 *)
def Arow(n, size):
if n == 0: return [1] * size
return [prod(binomial((j + 1)*n - 1, n - 1) for j in range(k)) for k in range(size)]
for n in range(7): print(Arow(n, 7))
# Alternative, using exponential generating function:
def SetPolyLeadCoeff(m, n):
x, z = var("x, z")
if m == 0: return 1
w = exp(2 * pi * I / m)
o = sum(exp(z * w ** k) for k in range(m)) / m
t = exp(x * (o - 1)).taylor(z, 0, m*n)
p = factorial(m*n) * t.coefficient(z, m*n)
return p.leading_coefficient(x)
for m in range(7):
print([SetPolyLeadCoeff(m, k) for k in range(6)])
Triangle T(n, k) starts: [0] [1] [1] [0, 1] [2] [0, 1, 0] [3] [0, 1, 1, 0] [4] [0, 1, 1, 0, 0] [5] [0, 1, 2, 1, 0, 0] [6] [0, 1, 2, 1, 0, 0, 0] [7] [0, 1, 3, 2, 0, 0, 0, 0] [8] [0, 1, 3, 3, 1, 0, 0, 0, 0] [9] [0, 1, 4, 4, 2, 0, 0, 0, 0, 0]
T := (n, k) -> iquo(binomial(n - 1, k - 1), k!): seq(print(seq(T(n, k), k = 0..n)), n = 0..9);
R = PowerSeriesRing(ZZ, "x")
x = R.gen().O(33)
@cached_function
def p(n) -> Polynomial:
if n == 0: return R(1)
return sum(p(n - k) * x for k in range(1, n + 1))
def A362370_row(n) -> list[int]:
L = p(n).list()
return [L[k] // factorial(k) for k in range(n + 1)]
for n in range(10):
print(A362370_row(n))
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