OFFSET
1,5
COMMENTS
In triangles of analogs 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]. - Paul Barry, Nov 26 2008
From Peter Bala, Jan 27 2015: (Start)
Working with an offset of 0, this is the triangle of connection constants between the polynomial basis sequences {x^n}, n>=0 and {n!*3^n*binomial((x - 1)/3,n)}, n>=0. An example is given below.
Call this array M and let P denote Pascal's triangle A007318, then P * M = A225468, P^2 * M = A075498. Also P^(-1) * M is a shifted version of A075498.
This triangle is the particular case a = 3, b = 0, c = 1 of the triangle of generalized Stirling numbers of the second kind S(a,b,c) defined in the Bala link. (End)
Named after the English scientist Francis Galton (1822-1911). - Amiram Eldar, Jun 13 2021
This is the array of (r, β)-Stirling numbers for r = 1 and β = 3. See Corcino. - Peter Bala, Feb 26 2025
LINKS
Peter Bala, A 3 parameter family of generalized Stirling numbers, 2015.
Peter Bala, Factorising (r,b)-Stirling arrays
Roberto B. Corcino, The (r, β)-Stirling Numbers, The Mindanao Forum, Vol. XIV, No.2, pp. 91-99, 1999.
Paweł Hitczenko, A class of polynomial recurrences resulting in (n/log n, n/log^2 n)-asymptotic normality, arXiv:2403.03422 [math.CO], 2024. See p. 8.
Ruedi Suter, Two Analogues of a Classical Sequence, Journal of Integer Sequences, Vol. 3 (2000), Article 00.1.8. [Paul Barry, 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)). - Paul Barry, Nov 26 2008
Let f(x) = exp(1/3*exp(3*x) + x). Then, with an offset of 0, the row polynomials R(n,x) are given by R(n,exp(3*x)) = 1/f(x)*(d/dx)^n(f(x)). Similar formulas hold for A008277, A039755, A105794, A143494 and A154537. - Peter Bala, Mar 01 2012
T(n, k) = 1/(3^k*k!)*Sum_{j=0..k}((-1)^(k-j)*binomial(k,j)*(3*j+1)^n). - Peter Luschny, May 20 2013
From Peter Bala, Jan 27 2015: (Start)
T(n,k) = Sum_{i = 0..n-1} 3^(i-k+1)*binomial(n-1,i)*Stirling2(i,k-1).
O.g.f. for n-th diagonal: exp(-x/3)*Sum_{k >= 0} (3*k + 1)^(k+n-1)*((x/3*exp(-x))^k)/k!.
O.g.f. column k (with offset 0): 1/( (1 - x)*(1 - 4*x)*...*(1 - (3*k + 1)*x) ). (End)
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;
Connection constants: Row 4: [1, 21, 12, 1] so
x^3 = 1 + 21*(x - 1) + 12*(x - 1)*(x - 4) + (x - 1)*(x - 4)*(x - 7). - Peter Bala, Jan 27 2015
From Peter Bala, Feb 26 2025: (Start)
The array factorizes as
/1 \ /1 \/1 \/1 \
|1 1 | |1 1 ||0 1 ||0 1 |
|1 5 1 | = |1 4 1 ||0 1 1 ||0 0 1 | ...
|1 21 12 1 | |1 13 7 1 ||0 1 4 1 ||0 0 1 1 |
|1 85 105 22 1| |1 44 34 10 1||0 1 13 7 1 ||0 0 1 4 1 |
|... | |... ||... ||... |
where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - x), x/(1 - 3*x)). Cf. A193843. (End)
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
MATHEMATICA
T[_, 1] = 1; T[n_, n_] = 1;
T[n_, k_] := T[n, k] = T[n-1, k-1] + (3k-2) T[n-1, k];
Table[T[n, k], {n, 1, 10}, {k, 1, n}] (* Jean-François Alcover, Jun 13 2019 *)
CROSSREFS
KEYWORD
AUTHOR
Gary W. Adamson, Aug 07 2005
EXTENSIONS
Edited and extended by R. J. Mathar, Aug 22 2009
STATUS
approved