A036355 Fibonacci-Pascal triangle read by rows.
1, 1, 1, 2, 2, 2, 3, 5, 5, 3, 5, 10, 14, 10, 5, 8, 20, 32, 32, 20, 8, 13, 38, 71, 84, 71, 38, 13, 21, 71, 149, 207, 207, 149, 71, 21, 34, 130, 304, 478, 556, 478, 304, 130, 34, 55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55, 89, 420, 1177, 2272, 3310, 3736, 3310, 2272, 1177, 420, 89
Offset: 0
Examples
Triangle begins 1; 1, 1; 2, 2, 2; 3, 5, 5, 3; 5, 10, 14, 10, 5; 8, 20, 32, 32, 20, 8; 13, 38, 71, 84, 71, 38, 13; 21, 71, 149, 207, 207, 149, 71, 21; 34, 130, 304, 478, 556, 478, 304, 130, 34; 55, 235, 604, 1060, 1390, 1390, 1060, 604, 235, 55; with indices T(0,0); T(1,0), T(1,1); T(2,0), T(2,1), T(2,2); T(3,0), T(3,1), T(3,2), T(3,3); T(4,0), T(4,1), T(4,2), T(4,3), T(4,4); For example, T(4,2) = 14 and there are 14 lattice paths from (0,0) to (4-2,2) = (2,2) using steps (1,0),(2,0),(0,1),(0,2). - _Greg Dresden_, Aug 25 2020
Links
Crossrefs
Programs
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Haskell
a036355 n k = a036355_tabl !! n !! k a036355_row n = a036355_tabl !! n a036355_tabl = [1] : f [1] [1,1] where f us vs = vs : f vs (zipWith (+) (zipWith (+) ([0,0] ++ us) (us ++ [0,0])) (zipWith (+) ([0] ++ vs) (vs ++ [0]))) -- Reinhard Zumkeller, Apr 23 2013
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Mathematica
nmax = 11; t[n_, m_] := t[n, m] = tp[n-1, m-1] + tp[n-2, m-2] + tp[n-1, m] + tp[n-2, m]; tp[n_, m_] /; 0 <= m <= n && n >= 0 := t[n, m]; tp[n_, m_] = 0; t[0, 0] = 1; Flatten[ Table[t[n, m], {n, 0, nmax}, {m, 0, n}]] (* Jean-François Alcover, Nov 09 2011, after formula *)
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PARI
/* same as in A092566 but use */ steps=[[1,0], [2,0], [0,1], [0,2]]; /* Joerg Arndt, Jun 30 2011 */
Formula
T(n, m) = T'(n-1, m-1)+T'(n-2, m-2)+T'(n-1, m)+T'(n-2, m), where T'(n, m) = T(n, m) if 0<=m<=n and n >= 0 and T'(n, m)=0 otherwise. Initial term T(0, 0)=1.
G.f.: 1/(1-(1+y)*x-(1+y^2)*x^2). - Vladeta Jovovic, Oct 11 2003
Comments