A026374 Triangular array T read by rows: T(n,0) = T(n,n) = 1 for all n >= 0, T(n,k) = T(n-1,k-1) + T(n-1,k) for odd n and 1< = k <= n-1, T(n,k) = T(n-1,k-1) + T(n-1,k) + T(n-2,k-1) for even n and 1 <= k <= n-1.
1, 1, 1, 1, 3, 1, 1, 4, 4, 1, 1, 6, 11, 6, 1, 1, 7, 17, 17, 7, 1, 1, 9, 30, 45, 30, 9, 1, 1, 10, 39, 75, 75, 39, 10, 1, 1, 12, 58, 144, 195, 144, 58, 12, 1, 1, 13, 70, 202, 339, 339, 202, 70, 13, 1, 1, 15, 95, 330, 685, 873, 685, 330, 95, 15, 1
Offset: 0
Examples
Triangle starts: 1; 1, 1; 1, 3, 1; 1, 4, 4, 1; 1, 6, 11, 6, 1; 1, 7, 17, 17, 7, 1; 1, 9, 30, 45, 30, 9, 1; 1, 10, 39, 75, 75, 39, 10, 1; 1, 12, 58, 144, 195, 144, 58, 12, 1; 1, 13, 70, 202, 339, 339, 202, 70, 13, 1; 1, 15, 95, 330, 685, 873, 685, 330, 95, 15, 1; 1, 16, 110, 425, 1015, 1558, 1558, 1015, 425, 110, 16, 1;
Links
- Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
- Paul Barry, Jacobsthal Decompositions of Pascal's Triangle, Ternary Trees, and Alternating Sign Matrices, Journal of Integer Sequences, 19, 2016, #16.3.5.
Programs
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Haskell
a026374 n k = a026374_tabl !! n !! k a026374_row n = a026374_tabl !! n a026374_tabl = [1] : map fst (map snd $ iterate f (1, ([1, 1], [1]))) where f (0, (us, vs)) = (1, (zipWith (+) ([0] ++ us) (us ++ [0]), us)) f (1, (us, vs)) = (0, (zipWith (+) ([0] ++ vs ++ [0]) $ zipWith (+) ([0] ++ us) (us ++ [0]), us)) -- Reinhard Zumkeller, Feb 22 2014
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Mathematica
p[x, 1] := 1; p[x_, n_] := p[x, n] = If[Mod[n, 2] == 0, (x + 1)*p[x, n - 1], (x^2 + 1)^Floor[n/2]]; a = Table[CoefficientList[p[x, n], x], {n, 1, 12}]; Flatten[a] (* Roger L. Bagula and Gary W. Adamson, Dec 04 2009 *)
Formula
T(n, k) = number of integer strings s(0), ..., s(n) such that s(0)=0, s(n) = n-2k, where, for 1 <= i <= n, s(i) is even if i is even and |s(i) - s(i-1)| <= 1.
From Emeric Deutsch, Jan 25 2004: (Start)
T(2n, k) = Sum_{j=ceiling(k/2)..k} 3^(2j-k)*binomial(n, j)*binomial(j, k-j);
T(2n+1, k) = T(2n, k-1) + T(2n, k).
G.f.: (1 + z + t*z)/(1 - (1+3*t+t^2)*z^2) = 1 + (1+t)*z + (1+3*t+t^2)*z^2+ ... .
Generating polynomial for row 2n is (1 + 3*t + t^2)^n;
Generating polynomial for row 2n+1 it is (1+t)*(1 + 3*t + t^2)^n. (End)
From Emeric Deutsch, Jan 30 2004: (Start)
T(2n, k) = Sum_{j=ceiling(k/2)..k} 3^(2j-k)*binomial(n, j)*binomial(j, k-j);
T(2n+1, k) = T(2n, k-1) + T(2n, k). (End)
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