A186021
a(n) = Bell(n)*(2 - 0^n).
Original entry on oeis.org
1, 2, 4, 10, 30, 104, 406, 1754, 8280, 42294, 231950, 1357140, 8427194, 55288874, 381798644, 2765917090, 20960284294, 165729739608, 1364153612318, 11665484410114, 103448316470744, 949739632313502, 9013431476894646, 88304011710168692, 891917738589610578, 9277180664459998706
Offset: 0
a(4) = A060719(3) + 1 = 29 + 1 = 30.
-
[Bell(n)*(2-0^n): n in [0..50]]; // Vincenzo Librandi, Apr 06 2011
-
A186021List := proc(m) local A, P, n; A := [1,2]; P := [2];
for n from 1 to m - 2 do P := ListTools:-PartialSums([P[-1], op(P)]);
A := [op(A), P[-1]] od; A end: A186021List(26); # Peter Luschny, Mar 24 2022
-
Prepend[Table[2 Sum[Binomial[n, j] BellB[j], {j, 0, n}], {n, 0, 25}], 1] (* Geoffrey Critzer, Aug 28 2014 *)
With[{nmax = 50}, CoefficientList[Series[2*Exp[Exp[x] - 1] - 1, {x, 0, nmax}], x]*Range[0, nmax]!] (* G. C. Greubel, Jul 24 2017 *)
-
x='x+O('x^50); Vec(serlaplace(2*exp(exp(x) - 1) -1)) \\ G. C. Greubel, Jul 24 2017
-
from itertools import accumulate
def A186021_list(size):
if size < 1: return []
L, accu = [1], [2]
for _ in range(size-1):
accu = list(accumulate([accu[-1]] + accu))
L.append(accu[0])
return L
print(A186021_list(26)) # Peter Luschny, Apr 25 2016
A102661
Triangle of partial sums of Stirling numbers of 2nd kind (A008277): T(n,k) = Sum_{i=1..k} Stirling2(n,i), 1<=k<=n.
Original entry on oeis.org
1, 1, 2, 1, 4, 5, 1, 8, 14, 15, 1, 16, 41, 51, 52, 1, 32, 122, 187, 202, 203, 1, 64, 365, 715, 855, 876, 877, 1, 128, 1094, 2795, 3845, 4111, 4139, 4140, 1, 256, 3281, 11051, 18002, 20648, 21110, 21146, 21147, 1, 512, 9842, 43947, 86472, 109299, 115179, 115929, 115974, 115975
Offset: 1
Triangle begins:
1;
1, 2;
1, 4, 5;
1, 8, 14, 15;
1, 16, 41, 51, 52;
...
- Richard Stanley, Enumerative Combinatorics, Cambridge Univ. Press, 1997 page 38. (#7 of the twelvefold ways)
- Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
- John R. Britnell and Mark Wildon, Bell numbers, partition moves and the eigenvalues of the random-to-top shuffle in Dynkin Types A, B and D, arXiv:1507.04803 [math.CO], 2015.
- T. S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176. [Annotated, scanned copy]
- OEIS Wiki, Sorting numbers
-
a102661 n k = a102661_tabl !! (n-1) !! (k-1)
a102661_row n = a102661_tabl !! (n-1)
a102661_tabl = map (scanl1 (+) . tail) $ tail a048993_tabl
-- Reinhard Zumkeller, Jun 19 2015
-
with(combinat): A102661_row := proc(n) local k,j; seq(add(stirling2(n,j),j=1..k),k=1..n) end:
seq(print(A102661_row(r)),r=1..6); # Peter Luschny, Sep 30 2011
-
Table[Table[Sum[StirlingS2[n, i], {i, 1, k}], {k, 1, n}], {n, 1,10}] // Grid (* Geoffrey Critzer, Mar 22 2011*)
Table[Accumulate[StirlingS2[n,Range[n]]],{n,10}]//Flatten (* Harvey P. Dale, Oct 28 2019 *)
-
tabl(nn) = {for (n=1, nn, for (k=1, n, print1(sum(i=1, k, stirling(n,i, 2)), ", ");); print(););} \\ Michel Marcus, Aug 10 2015
-
def T(n,k):
return sum([stirling_number2(n,j) for j in range(1,k+1)])
# Danny Rorabaugh, Oct 13 2015
A256894
Triangle read by rows, T(n,k) = Sum_{j=0..n-k+1} C(n-1,j-1)*T(n-j,k-1) if k != 0 else 1, n>=0, 0<=k<=n.
Original entry on oeis.org
1, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 8, 13, 7, 1, 1, 16, 40, 35, 11, 1, 1, 32, 121, 155, 80, 16, 1, 1, 64, 364, 651, 490, 161, 22, 1, 1, 128, 1093, 2667, 2751, 1316, 294, 29, 1, 1, 256, 3280, 10795, 14721, 9597, 3108, 498, 37, 1, 1, 512, 9841, 43435, 76630, 65352
Offset: 0
Triangle starts:
1;
1, 1;
1, 2, 1;
1, 4, 4, 1;
1, 8, 13, 7, 1;
1, 16, 40, 35, 11, 1;
1, 32, 121, 155, 80, 16, 1;
1, 64, 364, 651, 490, 161, 22, 1;
The signed version is the inverse of A326326:
1;
-1, 1;
1, -2, 1;
-1, 4, -4, 1;
1, -8, 13, -7, 1;
-1, 16, -40, 35, -11, 1;
1, -32, 121, -155, 80, -16, 1;
-1, 64, -364, 651, -490, 161, -22, 1. - _Peter Luschny_, Jul 02 2019
T(4,3)=7 is the number of disjoint [4]-covering collections of 4 subsets:
{{1},{2},{3},{4}}
{{1,2},{3},{4},{}}
{{1,3},{2},{4},{}}
{{1,4},{2},{3},{}}
{{2,3},{1},{4},{}}
{{2,4},{1},{3},{}}
{{3,4},{1},{2},{}}. - _Manfred Boergens_, Mar 04 2025
-
# Implemented as a sequence transformation acting on f: n -> 1,1,1,1,... .
F := proc(n, k, f) option remember; `if`(k=0, f(0)^n,
add(binomial(n-1,j-1)*f(j)*F(n-j,k-1,f),j=0..n-k+1)) end:
for n from 0 to 7 do seq(F(n,k,j->1), k=0..n) od;
-
Table[StirlingS2[n, m+1]+StirlingS2[n, m], {n, 0, 10}, {m, 0, n}]//Flatten (* Manfred Boergens, Mar 04 2025 *)
A369950
Triangle read by rows: T(n,k) = number of j-covers of [n] with j<=k, k=1..2^n-1.
Original entry on oeis.org
1, 1, 4, 5, 1, 13, 45, 80, 101, 108, 109, 1, 40, 361, 1586, 4505, 9482, 15913, 22348, 27353, 30356, 31721, 32176, 32281, 32296, 32297, 1, 121, 2681, 27671, 182777, 894103, 3491513, 11348063, 31483113, 75820263, 160485753, 301604003
Offset: 1
Triangle (with rows n >= 1 and columns k >= 1) begins as follows:
1;
1, 4, 5;
1, 13, 45, 80, 101, 108, 109;
1, 40, 361, 1586, 4505, 9482, 15913, 22348, 27353, 30356, 31721, 32176, 32281, 32296, 32297;
...
There are T(3,2) = 13 covers of [3] consisting of up to 2 subsets (brackets and commas omitted):
123
123 1
123 2
123 3
123 12
123 13
123 23
12 13
12 23
13 23
12 3
13 2
23 1
-
Flatten[Table[Sum[Sum[StirlingS1[i+1, j+1] (2^j-1)^n, {j, 0, i}]/i!, {i, k}], {n, 6}, {k, 2^n-1}]]
-
from math import comb
def A369950(n,k): return sum((-1)**j*comb(n, j)*comb(2**(n-j)-1, i) for j in range(n+1) for i in range(1,k+1)) # John Tyler Rascoe, Mar 06 2025
A381683
Triangle read by rows: T(n,k) = number of collections of up to k subsets of [n] covering [n], with [0]={}; n>=0, k=0..2^n.
Original entry on oeis.org
1, 2, 0, 1, 2, 0, 1, 5, 9, 10, 0, 1, 14, 58, 125, 181, 209, 217, 218, 0, 1, 41, 401, 1947, 6091, 13987, 25395, 38261, 49701, 57709, 62077, 63897, 64457, 64577, 64593, 64594, 0, 1, 122, 2802, 30352, 210448, 1076880, 4385616, 14839576, 42831176, 107303376, 236306016, 462089756, 809460556, 1280895556, 1846618196, 2447698581
Offset: 0
Triangle begins:
1 2
0 1 2
0 1 5 9 10
0 1 14 58 125 181 209 217 218
0 1 41 401 1947 6091 13987 25395 38261 49701 57709 62077 63897 64457 64577 64593 64594
...
T(3,2)=14 is the number of covering collections of 1 or 2 subsets of [3]:
{{1,2,3}}
{{},{1,2,3}}
{{1},{2,3}}
{{1},{1,2,3}}
{{2},{1,3}}
{{2},{1,2,3}}
{{3},{1,2}}
{{3},{1,2,3}}
{{1,2},{1,3}}
{{1,2},{2,3}}
{{1,3},{2,3}}
{{1,2},{1,2,3}}
{{1,3},{1,2,3}}
{{2,3},{1,2,3}}.
-
Table[Sum[Sum[(-1)^(n-i)*Binomial[n, i]*Binomial[2^i, j], {i, 0, n}], {j, 0, k}], {n, 0, 4}, {k, 0, 2^n}]//Flatten
-
T(n,k) = sum(j=0,k, sum(i=0,n, (-1)^(n-i)*binomial(n,i)*binomial(2^i,j)));
for(n=0,5,for(k=0,2^n,print1(T(n,k),", "))); \\ Joerg Arndt, Mar 04 2025
Showing 1-5 of 5 results.
Comments