A130167
Another version of triangle in A127743.
Original entry on oeis.org
1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 6, 5, 3, 1, 0, 22, 16, 9, 4, 1, 0, 92, 60, 31, 14, 5, 1, 0, 426, 252, 120, 52, 20, 6, 1, 0, 2146, 1160, 510, 209, 80, 27, 7, 1, 0, 11624, 5776, 2348, 904, 335, 116, 35, 8, 1, 0, 67146, 30832, 11610, 4184, 1481, 507, 161, 44, 9, 1
Offset: 0
Triangle begins:
1;
0, 1;
0, 1, 1;
0, 2, 2, 1;
0, 6, 5, 3, 1;
0, 22, 16, 9, 4, 1;
0, 92, 60, 31, 14, 5, 1; ...
A179313
Triangle T(n,k) read by rows: product of the compositorial weight of the k-th partition of n times A074664(.) applied to each part.
Original entry on oeis.org
1, 1, 1, 2, 2, 1, 6, 4, 1, 3, 1, 22, 12, 4, 6, 3, 4, 1, 92, 44, 12, 4, 18, 12, 1, 8, 6, 5, 1, 426, 184, 44, 24, 66, 36, 12, 6, 24, 24, 4, 10, 10, 6, 1, 2146, 852, 184, 88, 36, 276, 132, 72, 18, 12, 88, 72, 24, 24, 1, 30, 40, 10, 12, 15, 7, 1, 11624, 4292, 852, 368, 264, 1278, 552
Offset: 1
T(6,3) represents the 3rd partition of 6, namely 2+4. A074664(2)*A074664(4) = 1*6 is multiplied
by the weight A048996([2,4]) = 2!/1!/1! =2, and T(6,3) =1*6*2=12.
T(6,5) represents the 5th partition of 6, namely 1+1+4. A074664(1)*A074664(1)*A074664(4) = 1*1*6 is multiplied
by the weight A048996([1,1,4]) = 3!/2!/1! =3, and T(6,5) =1*1*6*3.
T(7,6) represents the 6th partition of 7, namely 1+2+4. A074664(1)*A074664(2)*A074664(4) = 1*1*6 is multiplied
the weight A048996([1,2,4]) = 3!/1!/1!/1! =6, and T(7,6) =1*1*6*6.
The triangle starts
1;
1,1;
2,2,1;
6,4,1,3,1;
22,12,4,6,3,4,1;
92,44,12,4,18,12,1,8,6,5,1;
426,184,44,24,66,36,12,6,24,24,4,10,10,6,1;
- M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, p. 831
A127745
Counts Bell numbers (except for Catalans) associated with the partition number [n].
Original entry on oeis.org
0, 0, 0, 1, 8, 50, 294, 1717, 10194, 62284, 394346, 2597266, 17827166, 127575414, 951411752, 7386583917, 59623674472, 499648882838, 4340548090590, 39033489125836, 362871600781796, 3482858492844510, 34471940635650958, 351444263328831458
Offset: 1
There are 15 Bell objects when n = 4, 14 are also Catalans so a(4) = 1.
There are 52 Bell objects when n = 5, 42 are also Catalans; we know that 5 = 4+1 = 1+4 which accounts for two of the non-Catalan Bells so, a(5) = 52 - 42 - 2 = 8.
Showing 1-3 of 3 results.
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