|
| |
|
|
A111577
|
|
Galton triangle T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k) read by rows.
|
|
6
| |
|
|
1, 1, 1, 1, 5, 1, 1, 21, 12, 1, 1, 85, 105, 22, 1, 1, 341, 820, 325, 35, 1, 1, 1365, 6081, 4070, 780, 51, 1, 1, 5461, 43932, 46781, 14210, 1596, 70, 1, 1, 21845, 312985, 511742, 231511, 39746, 2926, 92, 1, 1, 87381, 2212740, 5430405, 3521385, 867447, 95340, 4950, 117, 1
(list; table; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,5
|
|
|
COMMENTS
| In triangles of analogues to Stirling numbers of the second kind, the multipliers of T(n-1,k)
in the recurrence are terms in arithmetic sequences: in Pascal's triangle A007318,
the multiplier = 1. In triangle A008277, the Stirling numbers of the second kind, the multipliers
are in the set (1,2,3...). For this sequence here, the multipliers are from A016777.
Riordan array [exp(x), (exp(3x)-1)/3]. [From Paul Barry (pbarry(AT)wit.ie), Nov 26 2008]
|
|
|
LINKS
| R. Suter, Two Analogues of a Classical Sequence, Journal of Integer Sequences, Article 00.1.8 [From Paul Barry (pbarry(AT)wit.ie), Nov 26 2008]
|
|
|
FORMULA
| T(n, k) = T(n-1, k-1) + (3k-2)*T(n-1, k).
E.g.f.: exp(x)*exp((y/3)*(exp(3x)-1)); [From Paul Barry (pbarry(AT)wit.ie), Nov 26 2008]
|
|
|
EXAMPLE
| T(5,3) = T(4,2)+7*T(4,3) = 21 + 7*12 = 105.
The triangle starts in row n=1 as:
1;
1,1;
1,5,1;
1,21,12,1;
1,85,105,22,1;
|
|
|
MAPLE
| A111577 := proc(n, k) option remember; if k = 1 or k = n then 1; else procname(n-1, k-1)+(3*k-2)*procname(n-1, k) ; fi; end:
seq( seq(A111577(n, k), k=1..n), n=1..10) ; # R. J. Mathar, Aug 22 2009
|
|
|
CROSSREFS
| Cf. A008277, A039755.
Sequence in context: A144397 A047909 A171243 * A176242 A036969 A080249
Adjacent sequences: A111574 A111575 A111576 * A111578 A111579 A111580
|
|
|
KEYWORD
| nonn,easy,tabl
|
|
|
AUTHOR
| Gary W. Adamson (qntmpkt(AT)yahoo.com), Aug 07 2005
|
|
|
EXTENSIONS
| Edited and extended by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 22 2009
|
| |
|
|
