A270236 Triangle T(n,p) read by rows: the number of occurrences of p in the restricted growth functions of length n.
1, 3, 1, 9, 5, 1, 30, 21, 8, 1, 112, 88, 47, 12, 1, 463, 387, 253, 97, 17, 1, 2095, 1816, 1345, 675, 184, 23, 1, 10279, 9123, 7304, 4418, 1641, 324, 30, 1, 54267, 48971, 41193, 28396, 13276, 3645, 536, 38, 1, 306298, 279855, 243152, 183615, 102244, 36223, 7473, 842, 47, 1
Offset: 1
A105488 Number of partitions of {1...n} containing 2 detached pairs of consecutive integers, i.e., partitions in which only 1- or 2-strings of consecutive integers can appear in a block and there are exactly two 2-strings.
1, 6, 30, 150, 780, 4263, 24556, 149040, 951615, 6378625, 44785620, 328660566, 2515643767, 20044428810, 165955025400, 1425299331992, 12678325080012, 116635133853189, 1108221018960830, 10862073229428120, 109694927532209481, 1140199081827172719
Offset: 4
Comments
Number of partitions enumerated by A105479 in which the maximal length of consecutive integers in a block is 2.
With offset 2t, number of partitions of {1...N} containing 2 detached strings of t consecutive integers, where N=n+2j, t=2+j, j = 0,1,2,..., i.e., partitions of [n] in which only v-strings of consecutive integers can appear in a block, where v=1 or v=t and there are exactly two t-strings.
Equals the minimum of the sum of the Rand distances over all A000110(n) set partitions of n elements. E.g. a(3) = 6 because over the 5 set partitions of {1, 2, 3} the sum of Rand distances from {{1}, {2}, {3}} to the rest is 6. - Andrey Goder (andy.goder(AT)gmail.com), Dec 08 2006
a(n+3) = A000110(n) * A000217(n) = Sum_{k=1..n} A285362(n,k) is the sum of the entries in all set partitions of [n]. - Alois P. Heinz, Apr 16 2017
Examples
a(5)=6 because the partitions of {1,2,3,4,5} with 2 detached pairs of consecutive integers are 145/23,125/34,1245/3,12/34/5,12/3/45,1/23/45.
Links
- Alois P. Heinz, Table of n, a(n) for n = 4..577
- A. O. Munagi, Set Partitions with Successions and Separations, IJMMS 2005:3 (2005), 451-463.
- W. Rand, Objective criteria for the evaluation of clustering methods, J. Amer. Stat. Assoc., 66 (336): 846-850, 1971.
Programs
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Maple
seq(binomial(n-2,2)*combinat[bell](n-3),n=4..28);
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Mathematica
a[n_] := Binomial[n-2, 2]*BellB[n-3]; Table[a[n], {n, 4, 25}] (* Jean-François Alcover, May 11 2019 *)
Formula
a(n) = binomial(n-2, 2)*Bell(n-3), which is the case r = 2 in the general case of r pairs, d(n, r)=binomial(n-r, r)*Bell(n-r-1), which is the case t=2 of the general formula d(n, r, t)=binomial(n-r*(t-1), r)*B(n-r*(t-1)-1).
A285439 Sum T(n,k) of the entries in the k-th cycles of all permutations of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
1, 4, 2, 21, 12, 3, 132, 76, 28, 4, 960, 545, 235, 55, 5, 7920, 4422, 2064, 612, 96, 6, 73080, 40194, 19607, 6692, 1386, 154, 7, 745920, 405072, 202792, 75944, 18736, 2816, 232, 8, 8346240, 4484808, 2280834, 911637, 254061, 46422, 5256, 333, 9
Offset: 1
Comments
Each cycle is written with the smallest element first and cycles are arranged in increasing order of their first elements.
Examples
T(3,1) = 21 because the sum of the entries in the first cycles of all permutations of [3] ((123), (132), (12)(3), (13)(2), (1)(23), (1)(2)(3)) is 6+6+3+4+1+1 = 21. Triangle T(n,k) begins: 1; 4, 2; 21, 12, 3; 132, 76, 28, 4; 960, 545, 235, 55, 5; 7920, 4422, 2064, 612, 96, 6; 73080, 40194, 19607, 6692, 1386, 154, 7; 745920, 405072, 202792, 75944, 18736, 2816, 232, 8; ...
Links
- Alois P. Heinz, Rows n = 1..23, flattened
- Wikipedia, Permutation
Crossrefs
Programs
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Maple
T:= proc(h) option remember; local b; b:= proc(n, l) option remember; `if`(n=0, [mul((i-1)!, i=l), 0], (p-> p+[0, (h-n+1)*p[1]*x^(nops(l)+1)])(b(n-1, [l[], 1]))+ add((p-> p+[0, (h-n+1)*p[1]*x^j])( b(n-1, subsop(j=l[j]+1, l))), j=1..nops(l))) end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2]) end: seq(T(n), n=1..10);
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Mathematica
T[h_] := T[h] = Module[{b}, b[n_, l_] := b[n, l] = If[n == 0, {Product[(i - 1)!, {i, l}], 0}, # + {0, (h - n + 1)*#[[1]]*x^(Length[l] + 1)}&[b[n - 1, Append[l, 1]]] + Sum[# + {0, (h-n+1)*#[[1]]*x^j}&[b[n - 1, ReplacePart[ l, j -> l[[j]] + 1]]], {j, 1, Length[l]}]]; Table[Coefficient[#, x, i], {i, 1, n}]&[b[h, {}][[2]]]]; Table[T[n], {n, 1, 10}] // Flatten (* Jean-François Alcover, May 25 2018, translated from Maple *)
Formula
Sum_{k=1..n} k * T(n,k) = n^2 * n! = A002775(n).
A285595 Sum T(n,k) of the k-th entries in all blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
1, 4, 2, 17, 10, 3, 76, 52, 18, 4, 362, 274, 111, 28, 5, 1842, 1500, 675, 200, 40, 6, 9991, 8614, 4185, 1380, 325, 54, 7, 57568, 51992, 26832, 9568, 2510, 492, 70, 8, 351125, 329650, 178755, 67820, 19255, 4206, 707, 88, 9, 2259302, 2192434, 1239351, 494828, 149605, 35382, 6629, 976, 108, 10
Offset: 1
Comments
T(n,k) is also k times the number of blocks of size >k in all set partitions of [n+1]. T(3,2) = 10 = 2 * 5 because there are 5 blocks of size >2 in all set partitions of [4], namely in 1234, 123|4, 124|3, 134|2, 1|234.
Examples
T(3,2) = 10 because the sum of the second entries in all blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 2+2+3+3+0 = 10. Triangle T(n,k) begins: 1; 4, 2; 17, 10, 3; 76, 52, 18, 4; 362, 274, 111, 28, 5; 1842, 1500, 675, 200, 40, 6; 9991, 8614, 4185, 1380, 325, 54, 7; 57568, 51992, 26832, 9568, 2510, 492, 70, 8; ...
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- Wikipedia, Partition of a set
Crossrefs
Programs
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Maple
T:= proc(h) option remember; local b; b:= proc(n, l) option remember; `if`(n=0, [1, 0], (p-> p+[0, (h-n+1)*p[1]*x^1])(b(n-1, [l[], 1]))+ add((p-> p+[0, (h-n+1)*p[1]*x^(l[j]+1)])(b(n-1, sort(subsop(j=l[j]+1, l), `>`))), j=1..nops(l))) end: (p-> seq(coeff(p, x, i), i=1..n))(b(h, [])[2]) end: seq(T(n), n=1..12); # second Maple program: b:= proc(n) option remember; `if`(n=0, [1, 0], add((p-> p+[0, p[1]*add(x^k, k=1..j-1)])( b(n-j)*binomial(n-1, j-1)), j=1..n)) end: T:= n-> (p-> seq(coeff(p, x, i)*i, i=1..n))(b(n+1)[2]): seq(T(n), n=1..12);
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Mathematica
b[n_] := b[n] = If[n == 0, {1, 0}, Sum[# + {0, #[[1]]*Sum[x^k, {k, 1, j-1} ]}&[b[n - j]*Binomial[n - 1, j - 1]], {j, 1, n}]]; T[n_] := Table[Coefficient[#, x, i]*i, {i, 1, n}] &[b[n + 1][[2]]]; Table[T[n], {n, 1, 12}] // Flatten (* Jean-François Alcover, May 23 2018, translated from 2nd Maple program *)
A285363 Sum of the entries in the first blocks of all set partitions of [n].
1, 4, 15, 60, 262, 1243, 6358, 34835, 203307, 1257913, 8216945, 56463487, 406868167, 3065920770, 24099977863, 197179545722, 1675846476148, 14769104672839, 134745258569108, 1270767279092285, 12371426210292311, 124173909409948575, 1283498833928098171
Offset: 1
Keywords
Examples
a(3) = 15 because the sum of the entries in the first blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 6+3+4+1+1 = 15.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..575
- Wikipedia, Partition of a set
Programs
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Maple
a:= proc(h) option remember; local b; b:= proc(n, m) option remember; `if`(n=0, [1, 0], add((p-> `if`(j=1, p+ [0, (h-n+1)*p[1]], p))(b(n-1, max(m, j))), j=1..m+1)) end: b(h, 0)[2] end: seq(a(n), n=1..30);
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Mathematica
a[h_] := a[h] = Module[{b}, b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[If[j == 1, # + {0, (h - n + 1)*#[[1]]}, #]&[b[n - 1, Max[m, j]]], {j, 1, m + 1}]]; b[h, 0][[2]]]; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, May 20 2018, translated from Maple *)
Formula
a(n) = A285362(n,1).
A285424 Sum of the entries in the last blocks of all set partitions of [n].
1, 5, 19, 75, 323, 1512, 7630, 41245, 237573, 1451359, 9365361, 63604596, 453206838, 3378581609, 26285755211, 212953670251, 1792896572319, 15658150745252, 141619251656826, 1324477898999161, 12791059496663293, 127395689514237279, 1307010496324272157
Offset: 1
Keywords
Examples
a(3) = 19 because the sum of the entries in the last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 6+3+2+5+3 = 19.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..100
- Wikipedia, Partition of a set
Programs
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Maple
a:= proc(h) option remember; local b; b:= proc(n, m, s) option remember; `if`(n=0, s, add(b(n-1, max(m, j), `if`(j
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Mathematica
a[h_] := a[h] = Module[{b}, b[n_, m_, s_] := b[n, m, s] = If[n == 0, s, Sum[b[n-1, Max[m, j], If[j < m, s, h - n + 1 + If[j == m, s, 0]]], {j, 1, m + 1}]]; b[h, 0, 0]]; Array[a, 25] (* Jean-François Alcover, May 22 2018, translated from Maple *)
A286232 Sum T(n,k) of the entries in the k-th last blocks of all set partitions of [n]; triangle T(n,k), n>=1, 1<=k<=n, read by rows.
1, 5, 1, 19, 10, 1, 75, 57, 17, 1, 323, 285, 145, 26, 1, 1512, 1421, 975, 317, 37, 1, 7630, 7395, 5999, 2865, 616, 50, 1, 41245, 40726, 36183, 22411, 7315, 1094, 65, 1, 237573, 237759, 221689, 163488, 72581, 16630, 1812, 82, 1, 1451359, 1468162, 1405001, 1160764, 649723, 206249, 34425, 2840, 101, 1
Offset: 1
Examples
T(3,2) = 10 because the sum of the entries in the second last blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+4+1+2 = 10. Triangle T(n,k) begins: 1; 5, 1; 19, 10, 1; 75, 57, 17, 1; 323, 285, 145, 26, 1; 1512, 1421, 975, 317, 37, 1; 7630, 7395, 5999, 2865, 616, 50, 1; 41245, 40726, 36183, 22411, 7315, 1094, 65, 1; ...
Links
- Alois P. Heinz, Rows n = 1..100, flattened
- Wikipedia, Partition of a set
Crossrefs
Programs
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Mathematica
app[P_, n_] := Module[{P0}, Table[P0 = Append[P, {}]; AppendTo[P0[[i]], n]; If[Last[P0] == {}, Most[P0], P0], {i, 1, Length[P]+1}]]; setPartitions[n_] := setPartitions[n] = If[n == 1, {{{1}}}, Flatten[app[#, n]& /@ setPartitions[n-1], 1]]; T[n_, k_] := Select[setPartitions[n], Length[#] >= k&][[All, -k]] // Flatten // Total; Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Aug 21 2021 *)
A285410 Sum of the entries in the (n+1)-th blocks of all set partitions of [2n+1].
1, 12, 185, 3757, 96454, 3018824, 111964040, 4813480830, 235727269842, 12967143328027, 792113203502422, 53224214308284463, 3902445739220008603, 310108348556403600064, 26551900616231571763742, 2437107937223749442138164, 238735439946016510599661488
Offset: 0
Keywords
Examples
a(1) = 12 because the sum of the entries in the second blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+2+5+2 = 12.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..345
- Wikipedia, Partition of a set
Programs
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Maple
a:= proc(h) option remember; local b; b:= proc(n, m) option remember; `if`(n=0, [1, 0], add((p-> `if`(j=h+1, p+ [0, (2*h-n+2)*p[1]], p))(b(n-1, max(m, j))), j=1..m+1)) end: b(2*h+1, 0)[2] end: seq(a(n), n=0..20);
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Mathematica
a[h_] := a[h] = Module[{b}, b[0, ] = {1, 0}; b[n, m_] := b[n, m] = Sum[ If[j == h + 1, # + {0, (2*h - n + 2)*#[[1]]}, #]&[b[n - 1, Max[m, j]]], {j, 1, m + 1}]; b[2*h + 1, 0][[2]]]; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, May 23 2018, translated from Maple *)
Formula
a(n) = A285362(2n+1,n+1).
A285364 Sum of the entries in the second blocks of all set partitions of [n].
2, 12, 58, 273, 1329, 6839, 37423, 217606, 1340597, 8719806, 59680387, 428481322, 3218109788, 25220647760, 205790862332, 1744755841379, 15342274425585, 139692065365753, 1314995731359189, 12780466391685166, 128081591768679823, 1322011886920066940
Offset: 2
Keywords
Examples
a(3) = 12 because the sum of the entries in the second blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+3+2+5+2 = 12.
Links
- Alois P. Heinz, Table of n, a(n) for n = 2..575
- Wikipedia, Partition of a set
Crossrefs
Column k=2 of A285362.
Programs
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Maple
a:= proc(h) option remember; local b; b:= proc(n, m) option remember; `if`(n=0, [1, 0], add((p-> `if`(j=2, p+ [0, (h-n+1)*p[1]], p))(b(n-1, max(m, j))), j=1..m+1)) end: b(h, 0)[2] end: seq(a(n), n=2..30);
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Mathematica
a[h_] := a[h] = Module[{b}, b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[Function[p, If[j == 2, p + {0, (h - n + 1)*p[[1]]}, p]][b[n - 1, Max[m, j]]], {j, 1, m + 1}]]; b[h, 0][[2]]]; Table[a[n], {n, 2, 30}] (* Jean-François Alcover, May 27 2018, from Maple *)
Formula
a(n) = A285362(n,2).
A285365 Sum of the entries in the third blocks of all set partitions of [n].
3, 28, 185, 1094, 6293, 36619, 219931, 1376929, 9023266, 61944014, 445076570, 3341575188, 26164558199, 213243368898, 1805626838935, 15856747810014, 144189514375955, 1355629263039685, 13159535002316403, 131729480987412527, 1358188539892586220
Offset: 3
Keywords
Examples
a(3) = 3 because the sum of the entries in the third blocks of all set partitions of [3] (123, 12|3, 13|2, 1|23, 1|2|3) is 0+0+0+0+3 = 3.
Links
- Alois P. Heinz, Table of n, a(n) for n = 3..400
- Wikipedia, Partition of a set
Crossrefs
Column k=3 of A285362.
Programs
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Maple
a:= proc(h) option remember; local b; b:= proc(n, m) option remember; `if`(n=0, [1, 0], add((p-> `if`(j=3, p+ [0, (h-n+1)*p[1]], p))(b(n-1, max(m, j))), j=1..m+1)) end: b(h, 0)[2] end: seq(a(n), n=3..30);
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Mathematica
a[h_] := a[h] = Module[{b}, b[n_, m_] := b[n, m] = If[n == 0, {1, 0}, Sum[Function[p, If[j == 3, p + {0, (h - n + 1)*p[[1]]}, p]][b[n - 1, Max[m, j]]], {j, 1, m + 1}]]; b[h, 0][[2]]]; Table[a[n], {n, 3, 30}] (* Jean-François Alcover, May 27 2018, from Maple *)
Formula
a(n) = A285362(n,3).
Comments
Examples
Links
Crossrefs
Programs
Maple
Mathematica
Formula