A075497 Stirling2 triangle with scaled diagonals (powers of 2).
1, 2, 1, 4, 6, 1, 8, 28, 12, 1, 16, 120, 100, 20, 1, 32, 496, 720, 260, 30, 1, 64, 2016, 4816, 2800, 560, 42, 1, 128, 8128, 30912, 27216, 8400, 1064, 56, 1, 256, 32640, 193600, 248640, 111216, 21168, 1848, 72, 1
Offset: 1
Examples
Triangle begins: [1]; [2,1]; [4,6,1]; p(3,x) = x*(4 + 6*x + x^2). ...; Triangle (0, 2, 0, 4, 0, 6, 0, 8, ...) DELTA (1, 0, 1, 0, 1, 0, 1, 0, ...) begins: 1 0, 1 0, 2, 1 0, 4, 6, 1 0, 8, 28, 12, 1 0, 16, 120, 100, 20, 1. - _Philippe Deléham_, Feb 13 2013 From _Peter Bala_, Feb 23 2025: (Start) The array factorizes as / 1 \ /1 \ /1 \ /1 \ | 2 1 | | 2 1 ||0 1 ||0 1 | | 4 6 1 | = | 4 4 1 ||0 2 1 ||0 0 1 | ... | 8 28 12 1 | | 8 12 6 1 ||0 4 4 1 ||0 0 2 1 | |16 120 100 20 1| |16 32 24 8 1||0 8 12 6 1 ||0 0 4 4 1 | |... | |... ||... ||... | where, in the infinite product on the right-hand side, the first array is the Riordan array (1/(1 - 2*x), x/(1 - 2*x)) = P^2, where P denotes Pascal's triangle. See A038207. Cf. A143494. (End)
Links
- Alois P. Heinz, Rows n = 1..141, flattened
- Peter Bala, The white diamond product of power series
- Peter Bala, Factorising (r,b)-Stirling arrays
- Paul Barry, Three Études on a sequence transformation pipeline, arXiv:1803.06408 [math.CO], 2018.
- John R. Britnell and Mark Wildon, Bell numbers, partition moves and the eigenvalues of the random-to-top shuffle in types A, B and D, arXiv 1507.04803 [math.CO], 2015.
- Roberto B. Corcino, The (r, β)-Stirling Numbers, The Mindanao Forum, Vol. XIV, No.2, pp. 91-99, 1999.
- Roberto B. Corcino and Maribeth B. Montero, The (r, β)-Stirling Numbers in the Context of 0-1 Tableau, Jour. Math. Soc. of the Philippines, ISSN 0115-6926, Vol. 32, No. 1 (2009), pp. 45-52
- 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.
- Wolfdieter Lang, First 10 rows.
- Toufik Mansour, Generalization of some identities involving the Fibonacci numbers, arXiv:math/0301157 [math.CO], 2003.
- Emanuele Munarini, Characteristic, admittance and matching polynomials of an antiregular graph, Appl. Anal. Discrete Math 3 (1) (2009) 157-176.
Crossrefs
Programs
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Maple
with(combinat): b:= proc(n, i) option remember; expand(`if`(n=0, 1, `if`(i<1, 0, add(x^j*multinomial(n, n-i*j, i$j)/j!*add( binomial(i, 2*k), k=0..i/2)^j*b(n-i*j, i-1), j=0..n/i)))) end: T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)): seq(T(n), n=1..12); # Alois P. Heinz, Aug 13 2015 # Alternatively, giving the triangle in the form displayed in the Example section: gf := exp(x*exp(z)*sinh(z)): X := n -> series(gf, z, n+2): Z := n -> n!*expand(simplify(coeff(X(n), z, n))): A075497_row := n -> op(PolynomialTools:-CoefficientList(Z(n), x)): seq(A075497_row(n), n=0..9); # Peter Luschny, Jan 14 2018
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Mathematica
Table[(2^(n - m)) StirlingS2[n, m], {n, 9}, {m, n}] // Flatten (* Michael De Vlieger, Dec 31 2015 *)
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PARI
for(n=1, 11, for(m=1, n, print1(2^(n - m) * stirling(n, m, 2),", ");); print();) \\ Indranil Ghosh, Mar 25 2017
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Sage
# uses[inverse_bell_transform from A265605] multifact_2_2 = lambda n: prod(2*k + 2 for k in (0..n-1)) inverse_bell_matrix(multifact_2_2, 9) # Peter Luschny, Dec 31 2015
Formula
a(n, m) = (2^(n-m)) * Stirling2(n, m).
a(n, m) = (Sum_{p=0..m-1} A075513(m, p)*((p+1)*2)^(n-m))/(m-1)! for n >= m >= 1, else 0.
a(n, m) = 2*m*a(n-1, m) + a(n-1, m-1), n >= m >= 1, else 0, with a(n, 0) := 0 and a(1, 1)=1.
G.f. for m-th column: (x^m)/Product_{k=1..m}(1-2*k*x), m >= 1.
E.g.f. for m-th column: (((exp(2*x)-1)/2)^m)/m!, m >= 1.
The row polynomials in t are given by D^n(exp(x*t)) evaluated at x = 0, where D is the operator (1+2*x)*d/dx. Cf. A008277. - Peter Bala, Nov 25 2011
From Peter Bala, Jan 13 2018: (Start)
n-th row polynomial R(n,x)= x o x o ... o x (n factors), where o is the deformed Hadamard product of power series defined in Bala, section 3.1.
R(n+1,x)/x = (x + 2) o (x + 2) o...o (x + 2) (n factors).
R(n+1,x) = x*Sum_{k = 0..n} binomial(n,k)*2^(n-k)*R(k,x).
Dobinski-type formulas: R(n,x) = exp(-x/2)*Sum_{i >= 0} (2*i)^n* (x/2)^i/i!; 1/x*R(n+1,x) = exp(-x/2)*Sum_{i >= 0} (2 + 2*i)^n* (x/2)^i/i!. (End)
Comments