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A136011 Irregular triangle read by rows, Stirling numbers of the second kind: columns shifted to allow (1, 1, 2, 2, 3, 3, ...) terms per row. 3
1, 1, 1, 1, 1, 3, 1, 7, 1, 1, 15, 6, 1, 31, 25, 1, 1, 63, 90, 10, 1, 127, 301, 65, 1, 1, 255, 966, 350, 15, 1, 511, 3025, 1701, 140, 1, 1, 1023, 9330, 7770, 1050, 21, 1, 2047, 28501, 34105, 6951, 266, 1, 1, 4095, 86526, 145750, 42525, 2646, 28 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,6
COMMENTS
Row sums = A024427: (1, 1, 2, 4, 9, 22, 58, 164, 495, 1587, ...).
T(n,k) is the number of ways to partition {1,2,...,n+1} into exactly k blocks such that each block contains at least 2 elements and the smallest 2 elements in each block are consecutive integers. - Geoffrey Critzer, Dec 02 2013
LINKS
FORMULA
Given A008277, the Stirling number of the second kind triangle, left column = (1, 1, 1, ...); all other columns start at 3rd term of previous column.
O.g.f. for column k: Product_{i=1..k} x^2/(1 - i*x). - Geoffrey Critzer, Dec 02 2013
T(n,k) = Stirling2(n+1-k,k). - Alois P. Heinz, Dec 04 2013
EXAMPLE
First few rows of the triangle:
1;
1;
1, 1;
1, 3;
1, 7, 1;
1, 15, 6;
1, 31, 25, 1;
1, 63, 90, 10;
1, 127, 301, 65, 1;
1, 255, 966, 350, 15;
...
T(5,3) = 1 because we have {1,2},{3,4},{5,6}.
T(6,3) = 6 because we have {1,2,7},{3,4},{5,6}; {1,2},{3,4,7},{5,6}; {1,2},{3,4},{5,6,7}; {1,2},{3,4,5},{6,7}; {1,2,3},{4,5},{6,7}; {1,2,5},{3,4},{6,7}. - Geoffrey Critzer, Dec 02 2013
MAPLE
T:= (n, k)-> Stirling2(n+1-k, k):
seq(seq(T(n, k), k=1..(n+1)/2), n=1..20); # Alois P. Heinz, Dec 04 2013
MATHEMATICA
nn=15; Range[0, nn]!; Map[Select[#, #>0&]&, Drop[Transpose[Table[CoefficientList[Series[Product[x^2/(1-i x), {i, 1, k}], {x, 0, nn}], x], {k, 1, nn/2}]], 2]]//Grid (* Geoffrey Critzer, Dec 02 2013 *)
CROSSREFS
Sequence in context: A097229 A097862 A097612 * A227984 A021991 A112132
KEYWORD
nonn,tabf
AUTHOR
Gary W. Adamson, Dec 09 2007
STATUS
approved

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Last modified May 12 06:47 EDT 2024. Contains 372432 sequences. (Running on oeis4.)