A082137 Square array of transforms of binomial coefficients, read by antidiagonals.
1, 1, 1, 1, 2, 2, 1, 3, 6, 4, 1, 4, 12, 16, 8, 1, 5, 20, 40, 40, 16, 1, 6, 30, 80, 120, 96, 32, 1, 7, 42, 140, 280, 336, 224, 64, 1, 8, 56, 224, 560, 896, 896, 512, 128, 1, 9, 72, 336, 1008, 2016, 2688, 2304, 1152, 256, 1, 10, 90, 480, 1680, 4032, 6720, 7680, 5760, 2560, 512
Offset: 0
Examples
Rows begin 1 1 2 4 8 ... 1 2 6 16 40 ... 1 3 12 40 120 ... 1 4 20 80 280 ... 1 5 30 140 560 ... Read as a triangle, this begins: 1 1, 1 1, 2, 2 1, 3, 6, 4 1, 4, 12, 16, 8 1, 5, 20, 40, 40, 16 1, 6, 30, 80, 120, 96, 32 ... - _Philippe Deléham_, Nov 10 2013
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
- Laradji, A. and Umar, A. Combinatorial results for semigroups of order-preserving partial transformations, Journal of Algebra 278, (2004), 342-359.
- Laradji, A. and Umar, A. Combinatorial results for semigroups of order-decreasing partial transformations, J. Integer Seq. 7 (2004), 04.3.8.
Programs
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Maple
# As a triangular array: T := (n,k) -> 2^(k+0^k-1)*binomial(n,k): for n from 0 to 9 do seq(T(n,k), k=0..n) od; # Peter Luschny, Nov 10 2017
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Mathematica
rows = 11; t[n_, k_] := 2^(n-1)*(n+k)!/(n!*k!); t[0, ] = 1; tkn = Table[ t[n, k], {k, 0, rows}, {n, 0, rows}]; Flatten[ Table[ tkn[[ n-k+1, k ]], {n, 1, rows}, {k, 1, n}]] (* _Jean-François Alcover, Jan 20 2012 *)
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Sage
def A082137_row(n) : # as a triangular array var('z') s = (exp(z*x)/(1-tanh(x))).series(x,n+2) t = factorial(n)*s.coefficient(x,n) return [t.coefficient(z,n-k) for k in (0..n)] for n in (0..7) : print(A082137_row(n)) # Peter Luschny, Aug 01 2012
Formula
Square array defined by T(n, k)=(2^(n-1)+0^n/2)C(n + k, n)= Sum{k=0..n, C(n+k, k+j)C(k+j, k)(1+(-1)^j)/2 }.
O.g.f. for array read as a triangle: (1-x*(1+t))/((1-x)*(1-x*(1+2*t))) = 1 + x*(1+t) + x^2*(1+2*t+2*t^2) + x^3*(1+3*t+6*t^2+4*t^3) + .... - Peter Bala, Apr 26 2012
For array read as a triangle: T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k) -2*T(n-2,k-1), T(0,0) = T(1,0) = T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Nov 10 2013
Comments