A278074
Triangle read by rows, coefficients of the polynomials P(m, n) = Sum_{k=1..n} binomial(m*n, m*k)* P(m, n-k)*z with P(m, 0) = 1 and m = 4.
Original entry on oeis.org
1, 0, 1, 0, 1, 70, 0, 1, 990, 34650, 0, 1, 16510, 2702700, 63063000, 0, 1, 261630, 213519150, 17459442000, 305540235000, 0, 1, 4196350, 17651304000, 4350310965000, 231905038365000, 3246670537110000
Offset: 0
Triangle starts:
[1]
[0, 1]
[0, 1, 70]
[0, 1, 990, 34650]
[0, 1, 16510, 2702700, 63063000]
[0, 1, 261630, 213519150, 17459442000, 305540235000]
-
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:
for n from 0 to 6 do PolynomialTools:-CoefficientList(P(4,n), x) od;
# Alternatively:
A278074_row := proc(n) 1/(1-t*((cosh(x)+cos(x))/2-1)); expand(series(%,x,4*n+1));
(4*n)!*coeff(%,x,4*n); PolynomialTools:-CoefficientList(%,t) end:
for n from 0 to 5 do A278074_row(n) od;
-
With[{m = 4}, Table[Expand[j!*SeriesCoefficient[1/(1 - t*(MittagLefflerE[m, x^m] - 1)), {x, 0, j}]], {j, 0, 24, m}]];
Function[arg, CoefficientList[arg, t]] /@ % // Flatten
-
# uses [P from A278073]
def A278074_row(n): return list(P(4, n))
for n in (0..6): print(A278074_row(n)) # Peter Luschny, Mar 24 2020
A327022
Partition triangle read by rows. Number of ordered set partitions of the set {1, 2, ..., 2*n} with all block sizes divisible by 2.
Original entry on oeis.org
1, 1, 1, 6, 1, 30, 90, 1, 56, 70, 1260, 2520, 1, 90, 420, 3780, 9450, 75600, 113400, 1, 132, 990, 924, 8910, 83160, 34650, 332640, 1247400, 6237000, 7484400, 1, 182, 2002, 6006, 18018, 270270, 252252, 630630, 1081080, 15135120, 12612600, 37837800, 189189000, 681080400, 681080400
Offset: 0
Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1; 6]
[3] [1; 30; 90]
[4] [1; 56, 70; 1260; 2520]
[5] [1; 90, 420; 3780, 9450; 75600; 113400]
[6] [1; 132, 990, 924; 8910, 83160, 34650; 332640, 1247400; 6237000; 7484400]
.
T(4, 1) = 56 because [6, 2] is the integer partition 2*P(4, 1) in the canonical order and there are 28 set partitions which have the shape [6, 2] (an example is {{1, 3, 4, 5, 6, 8}, {2, 7}}). Finally, since the order of the sets is taken into account, one gets 2!*28 = 56.
-
def GenOrdSetPart(m, n):
shapes = ([x*m for x in p] for p in Partitions(n))
return [factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in shapes]
def A327022row(n): return GenOrdSetPart(2, n)
for n in (0..6): print(A327022row(n))
A327023
Ordered set partitions of the set {1, 2, ..., 3*n} with all block sizes divisible by 3, irregular triangle T(n, k) for n >= 0 and 0 <= k < A000041(n), read by rows.
Original entry on oeis.org
1, 1, 1, 20, 1, 168, 1680, 1, 440, 924, 55440, 369600, 1, 910, 10010, 300300, 1261260, 33633600, 168168000, 1, 1632, 37128, 48620, 1113840, 24504480, 17153136, 326726400, 2058376320, 34306272000, 137225088000
Offset: 0
Triangle starts (note the subdivisions by ';' (A072233)):
[0] [1]
[1] [1]
[2] [1; 20]
[3] [1; 168; 1680]
[4] [1; 440, 924; 55440; 369600]
[5] [1; 910, 10010; 300300, 1261260; 33633600; 168168000]
[6] [1; 1632, 37128, 48620; 1113840, 24504480, 17153136; 326726400, 2058376320;
34306272000; 137225088000]
.
T(4, 1) = 440 because [9, 3] is the integer partition 3*P(4, 1) in the canonical order and there are 220 set partitions which have the shape [9, 3]. Finally, since the order of the sets is taken into account, one gets 2!*220 = 440.
Showing 1-3 of 3 results.
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