A159853 Riordan array ((1-2*x+2*x^2)/(1-x), x/(1-x)).
1, -1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 3, 4, 4, 3, 1, 1, 4, 7, 8, 7, 4, 1, 1, 5, 11, 15, 15, 11, 5, 1, 1, 6, 16, 26, 30, 26, 16, 6, 1, 1, 7, 22, 42, 56, 56, 42, 22, 7, 1, 1, 8, 29, 64, 98, 112, 98, 64, 29, 8, 1, 1, 9, 37, 93, 162, 210, 210, 162, 93, 37, 9, 1, 1, 10, 46, 130, 255
Offset: 0
Examples
Triangle begins: 1; -1, 1; 1, 0, 1; 1, 1, 1, 1; 1, 2, 2, 2, 1; 1, 3, 4, 4, 3, 1; ...
Links
- Muniru A Asiru, Table of n, a(n) for n = 0..5151
- Peter Bala, A note on the diagonals of a proper Riordan Array
Crossrefs
Cf. A087698.
Programs
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GAP
Flat(List([0..12],n->List([0..n],k->Binomial(n,k)-2*Binomial(n-1,n-k-1)+2*Binomial(n-2,n-k-2)))); # Muniru A Asiru, Mar 22 2018
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Magma
/* As triangle */ [[Binomial(n, n-k)-2*Binomial(n-1, n-k-1)+2*Binomial(n-2, n-k-2): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Mar 22 2018
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Maple
C := proc (n, k) if 0 <= k and k <= n then factorial(n)/(factorial(k)*factorial(n-k)) else 0 end if; end proc: for n from 0 to 10 do seq(C(n, n-k) - 2*C(n-1, n-k-1) + 2*C(n-2, n-k-2), k = 0..n); end do; # Peter Bala, Mar 20 2018
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Mathematica
Join[{1, -1}, Rest[T[0, 0]=1; T[n_, k_]:=Binomial[n, n - k] - 2 Binomial[n - 1, n - k - 1] + 2 Binomial[n - 2, n - k - 2]; Table[T[n, k], {n, 1, 15}, {k, 0, n}]//Flatten]] (* Vincenzo Librandi, Mar 22 2018 *) -
Sage
# uses[riordan_array from A256893] riordan_array((1-2*x+2*x^2)/(1-x), x/(1-x), 8) # Peter Luschny, Mar 21 2018
Formula
From Peter Bala, Mar 20 2018: (Start)
T(n,k) = C(n,k) - 2*C(n-1,n-k-1) + 2*C(n-2,n-k-2), where C(n,k) = n!/(k!*(n-k)!) for 0 <= k <= n, otherwise 0.
Exp(x) * the e.g.f. for row n = the e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(1 + x + x^2/2! + x^3/3!) = 1 + 2*x + 2*x^2/2! + 4*x^3/3! + 8*x^4/4! + 15*x^5/5! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1-x) ). (End)
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