A363591
a(n) = 3*(3^(n-1) - 2^n + 1)/2 - binomial(n,2), n >= 3.
Original entry on oeis.org
0, 12, 65, 255, 882, 2870, 9039, 27945, 85448, 259512, 784797, 2366819, 7125198, 21424938, 64373339, 193316877, 580344132, 1741819148, 5227030665, 15684238119, 47059006250, 141189602142, 423593972775, 1270832250545, 3812597415552, 11437993573920, 34314383375669
Offset: 3
2*a(5)=130 subtracting the 20 ordered set partitions of the type {1},{2},{3,4,5} from the 150 ordered set partitions of a 5-set into 3 parts.
-
LinearRecurrence[{8, -24, 34, -23, 6}, {0, 12, 65, 255, 882}, 30] (* or *)
A363591[n_] := (3^n - 3*2^n - n^2 + n + 3)/2;
Array[A363591, 30, 3] (* Paolo Xausa, Aug 30 2024 *)
A363603
Expansion of e.g.f. (1/4)*(exp(x)-x-1)*(exp(x)-1)^2.
Original entry on oeis.org
3, 20, 90, 343, 1197, 3966, 12720, 39941, 123651, 379132, 1154790, 3501219, 10581465, 31908218, 96068700, 288926977, 868288239, 2608010424, 7830584850, 23505386015, 70544469573, 211692128950, 635198021640, 1905845723133, 5718057263067
Offset: 4
4*a(5)=80 since the ordered set partitions are the following: 30 of type {1,2}{3,4},{5}; 30 of type {1,2},{3},{4,5}; 20 of type {1,2,3},{4},{5}.
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A363603[n_]:=(3^n-3(2^n-1))/4-(n/2)(2^(n-2)-1);Array[A363603,40,4] (* or *)
LinearRecurrence[{9,-31,51,-40,12},{3,20,90,343,1197},40] (* Paolo Xausa, Nov 18 2023 *)
A371568
Array read by ascending antidiagonals: A(n, k) is the number of paths of length k in Z^n from the origin to points such that x1+x2+...+xn = k with x1,...,xn > 0.
Original entry on oeis.org
1, 0, 1, 0, 2, 1, 0, 0, 6, 1, 0, 0, 6, 14, 1, 0, 0, 0, 36, 30, 1, 0, 0, 0, 24, 150, 62, 1, 0, 0, 0, 0, 240, 540, 126, 1, 0, 0, 0, 0, 120, 1560, 1806, 254, 1, 0, 0, 0, 0, 1800, 8400, 5796, 510, 1
Offset: 1
n\k 1 2 3 4 5 6 7 8 9 10
--------------------------------------------------
1| 1 1 1 1 1 1 1 1 1 1
2| 0 2 6 14 30 62 126 254 510 1022
3| 0 0 6 36 150 540 1806 5796 18150 55980
4| 0 0 0 24 240 1560 8400 40824 186480 818520
5| 0 0 0 0 120 1800 16800 126000 834120 5103000
6| 0 0 0 0 0 720 15120 191520 1905120 16435440
7| 0 0 0 0 0 0 5040 141120 2328480 29635200
8| 0 0 0 0 0 0 0 40320 1451520 30240000
9| 0 0 0 0 0 0 0 0 362880 16329600
10| 0 0 0 0 0 0 0 0 0 3628800
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A[n_,k_] := Sum[(-1)^(n-i) * i^k * Binomial[n,i], {i,1,n}]
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# The Akiyama-Tanigawa algorithm for the binomial generates the rows.
# Adds row(0) = 0^k and column(0) = 0^n.
from math import comb as binomial
def ATBinomial(n, len):
A = [0] * len
R = [0] * len
for k in range(len):
R[k] = binomial(k, n)
for j in range(k, 0, -1):
R[j - 1] = j * (R[j] - R[j - 1])
A[k] = R[0]
return A
for n in range(11): print([n], ATBinomial(n, 11)) # Peter Luschny, Apr 19 2024
A056268
Number of primitive (aperiodic) words of length n which contain exactly three different symbols.
Original entry on oeis.org
0, 0, 6, 36, 150, 534, 1806, 5760, 18144, 55830, 171006, 518580, 1569750, 4732014, 14250450, 42844320, 128746950, 386615376, 1160688606, 3483582660, 10454059938, 31368305694, 94118013006, 282378679920, 847187946000, 2541662931990, 7625194813656, 22875982414740
Offset: 1
- M. R. Nester (1999). Mathematical investigations of some plant interaction designs. PhD Thesis. University of Queensland, Brisbane, Australia. [See A056391 for pdf file of Chap. 2]
A179483
A(k,3) where A(k,n) = Sum_{m=1..k} (-1)^(m+1) *binomial(n,m)*m^k.
Original entry on oeis.org
3, -9, 6, 36, 150, 540, 1806, 5796, 18150, 55980, 171006, 519156, 1569750, 4733820, 14250606, 42850116, 128746950, 386634060, 1160688606, 3483638676, 10454061750, 31368476700, 94118013006, 282379204836, 847187946150, 2541664501740, 7625194831806
Offset: 1
-
A179483 := proc(n) add( (-1)^(m+1)*binomial(3,m)*m^n,m=1..n) ; end proc: # R. J. Mathar, Jan 31 2011
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Sum[(-1)^(m+1)Binomial[3,m]m^k,{m,1,k}]
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Vec(3*x*(1 - 9*x + 31*x^2 - 39*x^3 + 18*x^4) / ((1 - x)*(1 - 2*x)*(1 - 3*x)) + O(x^30)) \\ Colin Barker, May 21 2017
A380993
Irregular triangular array read by rows. T(n,k) is the number of ternary words of length n containing at least one copy of each letter and having exactly k inversions, n>=3, 0<=k<=floor(n^2/3).
Original entry on oeis.org
1, 2, 2, 1, 3, 6, 9, 9, 6, 3, 6, 12, 21, 27, 30, 24, 18, 9, 3, 10, 20, 38, 55, 74, 81, 80, 69, 53, 34, 17, 8, 1, 15, 30, 60, 93, 138, 174, 210, 216, 219, 195, 165, 120, 84, 48, 27, 9, 3, 21, 42, 87, 141, 222, 303, 405, 480, 546, 579, 588, 552, 498, 414, 324, 240, 162, 99, 54, 27, 9, 3
Offset: 3
Triangle T(n,k) begins:
1, 2, 2, 1;
3, 6, 9, 9, 6, 3;
6, 12, 21, 27, 30, 24, 18, 9, 3;
10, 20, 38, 55, 74, 81, 80, 69, 53, 34, 17, 8, 1;
...
T(4,2) = 9 because we have: {0, 1, 2, 0}, {0, 2, 0, 1}, {0, 2, 1, 1}, {0, 2, 2, 1}, {1, 0, 0, 2}, {1, 0, 2, 1}, {1, 1, 0, 2}, {1, 2, 0, 2}, {2, 0, 1, 2}.
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b:= proc(n, l) option remember; `if`(n=0, `if`(nops(subs(0=
[][], l))=3, 1, 0), add(expand(x^([0, l[1], l[1]+l[2]][j])*
b(n-1, subsop(j=`if`(j=3, 1, l[j]+1), l))), j=1..3))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, [0$3])):
seq(T(n), n=3..10); # Alois P. Heinz, Feb 12 2025
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nn = 8; B[n_] := FunctionExpand[QFactorial[n, u]];
e[z_] := Sum[z^n/B[n], {n, 0, nn}];
Drop[Map[CoefficientList[#, u] &,
Map[Normal[Series[#, {u, 0, Binomial[nn, 2]}]] &,
Table[B[n], {n, 0, nn}] CoefficientList[
Series[(e[z] - 1)^3, {z, 0, nn}], z]]], 3] // Grid
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