A144906
a(0) = 1; thereafter a(n) = A144422(n)/n.
Original entry on oeis.org
1, 3, 31, 1684, 271776, 97484904, 65617109160, 74248657560720, 130752443907524880, 338450307621257099520, 1232284889962378714855680, 6094200542431309662145478400, 39788645361978802435089535468800, 334957784448996146804912925763507200
Offset: 0
A189886
a(n) is the number of compositions of the set {1, 2, ..., n} into blocks, each of size 1, 2 or 3 (n >= 0).
Original entry on oeis.org
1, 1, 3, 13, 74, 530, 4550, 45570, 521640, 6717480, 96117000, 1512819000, 25975395600, 483169486800, 9678799930800, 207733600074000, 4755768505488000, 115681418156304000, 2979408725813520000, 80998627977002736000, 2317937034142810080000, 69649003197501567840000, 2192459412316607834400000, 72152830779716793506400000, 2477756318984329979756160000
Offset: 0
a(3) = 13 because all compositions of set {a,b,c} into blocks of size 1, 2, or 3 are:
1: ({a,b,c}),
2: ({a},{b,c}),
3: ({b,c},{a}),
4: ({b},{a,c}),
5: ({a,c},{b}),
6: ({c},{a,b}),
7: ({a,b},{c}),
8: ({a},{b},{c}),
9: ({a},{c},{b}),
10: ({b},{a},{c}),
11: ({b},{c},{a}),
12: ({c},{a},{b}),
13: ({c},{b},{a}).
- Alois P. Heinz, Table of n, a(n) for n = 0..424
- Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394, 2017.
- David Applegate and N. J. A. Sloane, The Gift Exchange Problem (arXiv:0907.0513, 2009)
- Adi Dani, Compositions and partitions of sets
-
A189886 := proc(n) local m, j; add(add(2^(2*m-n-j)*3^(j-m)*n!
*binomial(m,j)*binomial(j,2*j-(3*m-n)),j=0..3*m-n),m=0..n) end:
seq(A189886(n),n=0..24); # Peter Luschny, May 02 2011
# second Maple program:
a:= proc(n) option remember; `if`(n=0, 1, add(
a(n-i)*binomial(n, i), i=1..min(n, 3)))
end:
seq(a(n), n=0..25); # Alois P. Heinz, Sep 22 2016
# third Maple program:
a:= n-> n! * (<<0|1|0>, <0|0|1>, <1/6|1/2|1>>^n)[3, 3]:
seq(a(n), n=0..25); # Alois P. Heinz, Sep 22 2016
-
Table[Sum[n!/(2^(n+j-2m)3^(m-j))*Binomial[m,j]*Binomial[j,n+2j-3m], {m,0,n},{j,0,3m-n}],{n,0,15}]
-
a(n)=sum(m=0,n, sum(j=0,3*m-n, n!/(2^(n+j-2*m) *3^(m-j)) *binomial(m,j) *binomial(j,n+2*j-3*m))); /* Joerg Arndt, May 03 2011 */
A189804
Triangle read by rows: T(n,k) is the number of compositions of set {1, 2, ..., k} into exactly n blocks, each of size 1, 2 or 3 (n >= 0, 0 <= k <= 3*n).
Original entry on oeis.org
1, 0, 1, 1, 1, 0, 0, 2, 6, 14, 20, 20, 0, 0, 0, 6, 36, 150, 450, 1050, 1680, 1680, 0, 0, 0, 0, 24, 240, 1560, 7560, 29400, 90720, 218400, 369600, 369600, 0, 0, 0, 0, 0, 120, 1800, 16800, 117600, 667800, 3137400, 12243000, 38808000, 96096000, 168168000, 168168000
Offset: 0
Triangle begins:
[1]
[0, 1, 1, 1]
[0, 0, 2, 6, 14, 20, 20]
[0, 0, 0, 6, 36, 150, 450, 1050, 1680, 1680]
[0, 0, 0, 0, 24, 240, 1560, 7560, 29400, 90720, 218400, 369600, 369600]
[0, 0, 0, 0, 0, 120, 1800, 16800, 117600, 667880, 3137400, 12243000, 3880800, 96096000, 168168000, 168168000]
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
- Moa Apagodu, David Applegate, N. J. A. Sloane, and Doron Zeilberger, Analysis of the Gift Exchange Problem, arXiv:1701.08394 [math.CO], 2017.
- David Applegate and N. J. A. Sloane, The Gift Exchange Problem, arXiv:0907.0513 [math.CO], 2009.
-
T := proc(n, k)
option remember;
if n = k then 1;
elif k < n then 0;
elif n < 1 then 0;
else =k *T(n - 1, k - 1) + (1/2)*k*(k - 1)*T(n - 1, k - 2)+ (1/6)*k* (k - 1)*(k - 2)*T(n - 1, k - 3);
end if;
end proc; for n from 0 to 12 do lprint([seq(T(n, k), k=0..3*n)]); od:
-
Table[Sum[ n!/(2^(n + j - 2m)3^(m - j))Binomial[m, j]Binomial[j, n + 2j - 3m], {j, 0, 3m - n}], {m, 0, 5}, {n, 0, 3m}]//Flatten
-
for(m=0,7, for(n=0,3*m, print1(sum(j=0,3*m-n, (n!/(2^(n+j-2*m)*3^(m-j)))*binomial(m, j)*binomial(j, n+2*j-3*m)), ", "))) \\ G. C. Greubel, Jan 16 2018
Showing 1-3 of 3 results.
Comments