A177255
a(n) = Sum_{j=1..n} j*B(j-1), where B(k) = A000110(k) are the Bell numbers.
Original entry on oeis.org
0, 1, 3, 9, 29, 104, 416, 1837, 8853, 46113, 257583, 1533308, 9676148, 64452909, 451475027, 3314964857, 25442301577, 203604718076, 1695172374548, 14654631691569, 131309475792709, 1217516798735521, 11664652754184043, 115319114738472472, 1174967255260496776
Offset: 0
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[n eq 0 select 0 else (&+[j*Bell(j-1): j in [1..n]]): n in [0..30]]; // G. C. Greubel, May 11 2024
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with(combinat): a := proc (n) options operator, arrow: sum(j*bell(j-1), j = 1 .. n) end proc; seq(a(n), n = 0 .. 23);
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With[{nn=30},Join[{0},Accumulate[BellB[Range[0,nn-1]]Range[nn]]]] (* Harvey P. Dale, Nov 10 2014 *)
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[sum(j*bell_number(j-1) for j in range(1,1+n)) for n in range(31)] # G. C. Greubel, May 11 2024
A177256
Triangle read by rows: T(n,k) is the number of partitions of the set {1,2,...,n} having exactly k blocks that do not consist of consecutive integers (0<=k<=floor(n/2); a singleton is considered a block of consecutive integers).
Original entry on oeis.org
1, 1, 2, 0, 4, 1, 8, 6, 1, 16, 25, 11, 32, 89, 77, 5, 64, 290, 433, 90, 128, 893, 2132, 951, 36, 256, 2645, 9602, 7710, 934, 512, 7618, 40589, 53137, 13790, 329, 1024, 21489, 163739, 328119, 152600, 11599, 2048, 59665, 637587, 1872748, 1409791, 228103
Offset: 0
T(4,1)=6 because we have 134-2, 124-3, 14-23, 1-24-3, 14-2-3, and 13-2-4.
Triangle starts:
1;
1;
2,0;
4,1;
8,6,1;
16,25,11;
32,89,77,5;
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Q[0] := 1: for n to 12 do Q[n] := expand(u*subs(w = v, diff(Q[n-1], u))+u*subs(w = v, diff(Q[n-1], v))+w*(diff(Q[n-1], w))+w*subs(w = v, Q[n-1])) end do: for n from 0 to 12 do P[n] := sort(expand(subs({v = 1, w = 1}, Q[n]))) end do: for n from 0 to 12 do seq(coeff(P[n], u, j), j = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form
A177257
a(n) = Sum_{j=0..n-1} (binomial(n,j) - (j+1))*A000110(j).
Original entry on oeis.org
0, 0, 0, 1, 8, 47, 258, 1426, 8154, 48715, 305012, 2001719, 13754692, 98801976, 740584196, 5782218745, 46942426080, 395607218279, 3455493024350, 31236784338746, 291836182128670, 2814329123555051, 27980637362452980
Offset: 0
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A177257:= func< n | n eq 0 select 0 else (&+[(Binomial(n,j)-(j+1))*Bell(j): j in [0..n-1]]) >;
[A177257(n): n in [0..30]]; // G. C. Greubel, May 12 2024
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with(combinat): a:= proc(n) add((binomial(n, j)-j-1)*bell(j), j = 0 .. n-1) end proc: seq(a(n), n = 0 .. 22);
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Table[Sum[(Binomial[n,j]-j-1)BellB[j],{j,0,n-1}],{n,0,30}] (* Harvey P. Dale, Oct 15 2015 *)
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def A177257(n): return sum((binomial(n,j) -(j+1))*bell_number(j) for j in range(n))
[A177257(n) for n in range(31)] # G. C. Greubel, May 12 2024
A168444
Number of partitions of the set {1,2,...,n} such that no block is a sequence of consecutive integers (including 1-element blocks).
Original entry on oeis.org
1, 0, 0, 0, 1, 5, 21, 91, 422, 2103, 11226, 63879, 385691, 2461004, 16535820, 116628147, 861033654, 6637143698, 53297137552, 444940442553, 3854539901147, 34592812084693, 321125878230123, 3079144039478532, 30457076370822777, 310407099470429818, 3255972198123974137, 35114803641531204063
Offset: 0
For n=5 the a(5) = 5 partitions are 13-245, 14-235, 24-135, 25-135, 35-124.
- Richard Stanley, Enumerative Combinatorics, volume 1, second edition, Cambridge Univ Press, 2011, page 192, solution 111.
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b:= func< n | n eq 0 select 1 else (&+[(-1)^(n+j)*Binomial(j,n-j)*Bell(j): j in [Ceiling(n/2)..n]]) >;
A168444:= func< n | n eq 0 select 1 else b(n)-b(n-1) >;
[A168444(n): n in [0..30]]; // G. C. Greubel, May 12 2024
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with(combinat): y:=sum(bell(n)*x^n,n=0..20): z:=(1-x)*subs(x=x*(1-x),y): taylor(z,x=0,21);
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nn = 20; b := Sum[BellB[n] (x - x^2)^n, {n, 0, nn}]; CoefficientList[ Series[ (1-x) b, {x, 0, nn}], x] (* Geoffrey Critzer, Jun 01 2013 *)
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b(n):=if n=0 then 1 else sum(binomial(k,n-k)*(-1)^(n-k)*belln(k),k,ceiling(n/2),n); a(n):=if n=0 then 1 else b(n)-b(n-1); /* Vladimir Kruchinin, Sep 09 2010 */
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N=66; x = 'x+O('x^N);
B = serlaplace(exp(exp(x)-1));
gf = (1-x)*subst(B,'x, x*(1-x));
Vec(gf) \\ Joerg Arndt, Jun 01 2013
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@CachedFunction
def b(n): return 1 if (n==0) else sum((-1)^(n+j)*binomial(j,n-j)*bell_number(j) for j in range((n//2), n+1))
def A168444(n): return 1 if (n==0) else b(n) - b(n-1)
[A168444(n) for n in range(31)] # G. C. Greubel, May 12 2024
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