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A104856 Triangle read by rows: T(n,k)=binomial(n,k)binomial(k,floor(k/2))binomial(n-k,floor((n-k)/2)) (0<=k<=n). 0

%I

%S 1,1,1,2,2,2,3,6,6,3,6,12,24,12,6,10,30,60,60,30,10,20,60,180,180,180,

%T 60,20,35,140,420,630,630,420,140,35,70,280,1120,1680,2520,1680,1120,

%U 280,70,126,630,2520,5040,7560,7560,5040,2520,630,126,252,1260,6300

%N Triangle read by rows: T(n,k)=binomial(n,k)binomial(k,floor(k/2))binomial(n-k,floor((n-k)/2)) (0<=k<=n).

%C T(n,k) is the number of paths in the first quadrant, starting from the origin, with unit steps up, down, right, or left, having a total of n steps, exactly k of which are vertical (up or down). Example: T(3,2)=6 because we have NNE, NEN, ENN, NSE, ENS and NES. [From _Emeric Deutsch_, Nov 22 2008]

%D David M. Bloom, Problem 10921, Amer. Math. Monthly 110, (2003), 958-959.

%D E. Deutsch and D. Lovit, Math. Magazine, vol. 80, No. 1, 2007, p. 80, Problem 1739. [From _Emeric Deutsch_, Nov 22 2008]

%F T(n, k)=binomial(n, k)binomial(k, floor(k/2))binomial(n-k, floor((n-k)/2)) (0<=k<=n).

%p T:=(n,k)->binomial(n,k)*binomial(k,floor(k/2))*binomial(n-k,floor((n-k)/2)): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form

%Y Row sums yield A005566. T(n, 0)=T(n, n)=A001405(n).

%Y Cf. A005566, A001405.

%K nonn,tabl

%O 0,4

%A _Emeric Deutsch_, Apr 23 2005

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Last modified March 22 01:06 EDT 2019. Contains 321406 sequences. (Running on oeis4.)