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A383479 Number of lattice paths from (0,0) to (n,n) using steps (1,0),(3,0),(0,1).

%I A383479 #18 May 29 2025 08:06:11
%S A383479 1,2,6,24,100,420,1792,7752,33858,148940,658944,2929056,13070876,
%T A383479 58521344,262754040,1182619280,5334172518,24104916504,109111142376,
%U A383479 494630028200,2245300152480,10204575481320,46429481139000,211460450151600,963971663881200,4398118872144192
%N A383479 Number of lattice paths from (0,0) to (n,n) using steps (1,0),(3,0),(0,1).
%H A383479 Robert Israel, <a href="/A383479/b383479.txt">Table of n, a(n) for n = 0..1492</a>
%F A383479 a(n) = [x^n] 1/(1 - x - x^3)^(n+1).
%F A383479 a(n) = (n+1) * A049140(n+1).
%F A383479 a(n) = Sum_{k=0..floor(n/3)} binomial(n+k,k) * binomial(2*n-2*k,n-3*k).
%p A383479 f:= proc(x,y) option remember;
%p A383479      local t;
%p A383479      t:= 0;
%p A383479      if x >= 1 then t:= t + procname(x-1,y) fi;
%p A383479      if x >= 3 then t:= t + procname(x-3,y) fi;
%p A383479      if y >= 1 then t:= t + procname(x,y-1) fi;
%p A383479      t
%p A383479 end proc:
%p A383479 f(0,0):= 1:
%p A383479 seq(f(n,n),n=0..25); # _Robert Israel_, May 28 2025
%o A383479 (PARI) a(n) = sum(k=0, n\3, binomial(n+k, k)*binomial(2*n-2*k, n-3*k));
%Y A383479 Cf. A038112, A383480.
%Y A383479 Cf. A049140, A144401, A370624.
%K A383479 nonn
%O A383479 0,2
%A A383479 _Seiichi Manyama_, Apr 28 2025