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%I #25 Jul 31 2022 07:50:19
%S 1,0,1,0,0,1,1,0,0,1,0,2,0,0,1,0,0,3,0,0,1,3,0,0,4,0,0,1,0,7,0,0,5,0,
%T 0,1,0,0,12,0,0,6,0,0,1,12,0,0,18,0,0,7,0,0,1,0,30,0,0,25,0,0,8,0,0,1,
%U 0,0,55,0,0,33,0,0,9,0,0,1,55,0,0,88,0,0,42,0,0,10,0,0,1,0,143,0,0,130,0,0,52,0,0,11,0,0,1
%N A generalized Pascal triangle.
%C First diagonal is A000012, the all 1's sequence. Second nonzero diagonal is A000027 = n. Third nonzero diagonal is A027379 = n*(n+5)/2 for n>=1, or essentially A000217(n) - 3. Fourth nonzero diagonal is A111396. - _Jonathan Vos Post_, Nov 10 2005
%C Row sums are A126042. - _Paul Barry_, Dec 16 2006
%H G. C. Greubel, <a href="/A111373/b111373.txt">Rows n = 0..50 of the triangle, flattened</a>
%H I. Bajunaid et al., <a href="http://www.jstor.org/stable/30037599">Function series, Catalan numbers and random walks on trees</a>, Amer. Math. Monthly 112 (2005), 765-785.
%H Emeric Deutsch, L. Ferrari and S. Rinaldi, <a href="http://dx.doi.org/10.1016/j.aam.2004.05.002">Production Matrices</a>, Advances in Mathematics, 34 (2005) pp. 101-122.
%F Each term is the sum of the two terms above it to the left and two steps to the right.
%F From _Paul Barry_, Dec 16 2006: (Start)
%F Riordan array (g(x^3),x*g(x^3)) where g(x)=(2/sqrt(3x))*sin(asin(sqrt(27x/4))/3), the g.f. of A001764;
%F Number triangle T(n,k) = C(3*floor((n+2k)/3)-2k,floor((n+2k)/3)-k)*(k+1)/(2*floor((n+2k)/3)-k+ 1)(2*cos(2*pi*(n-k)/3)+1)/3. (End)
%F Inverse of Riordan array (1/(1+x^3), x/(1+x^3)), A126030. - _Paul Barry_, Dec 16 2006
%F G.f. (x*A(x))^k=sum{n>=k, T(n,k)*x^n}, where A(x)=1+x^3*A(x)^3. - Vladimir Kruchinin, Feb 18 2011
%F From _G. C. Greubel_, Jul 30 2022: (Start)
%F T(n, k) = (3/(n-k))*binomial(n, (n-k)/3 -1)*(k+1) for ( (n-k) mod 3 ) = 0, otherwise 0, with T(n, n) = 1.
%F T(n, n-3) = A000027(n-2), n >= 3.
%F T(n, n-6) = A027379(n-5), n >= 6.
%F T(n, n-9) = A111396(n-8), n >= 9.
%F T(n, n-12) = A167543(n+5), n >= 12.
%F Sum_{k=0..n} T(n, k) = A126042(n). (End)
%e Triangle begins:
%e 1;
%e 0, 1;
%e 0, 0, 1;
%e 1, 0, 0, 1;
%e 0, 2, 0, 0, 1;
%e 0, 0, 3, 0, 0, 1;
%e 3, 0, 0, 4, 0, 0, 1;
%e 0, 7, 0, 0, 5, 0, 0, 1;
%e 0, 0, 12, 0, 0, 6, 0, 0, 1;
%e 12, 0, 0, 18, 0, 0, 7, 0, 0, 1;
%e 0, 30, 0, 0, 25, 0, 0, 8, 0, 0, 1;
%e 0, 0, 55, 0, 0, 33, 0, 0, 9, 0, 0, 1;
%e 55, 0, 0, 88, 0, 0, 42, 0, 0, 10, 0, 0, 1;
%e Production matrix is
%e 0, 1;
%e 0, 0, 1;
%e 1, 0, 0, 1;
%e 0, 1, 0, 0, 1;
%e 0, 0, 1, 0, 0, 1;
%e 0, 0, 0, 1, 0, 0, 1;
%e 0, 0, 0, 0, 1, 0, 0, 1;
%e 0, 0, 0, 0, 0, 1, 0, 0, 1;
%e 0, 0, 0, 0, 0, 0, 1, 0, 0, 1;
%t T[n_, k_]= If[k==n, 1, If[Mod[n-k, 3]==0, (3/(n-k))*Binomial[n, (n-k)/3 -1]*(k + 1), 0]];
%t Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jul 30 2022 *)
%o (Magma)
%o function A111373(n,k)
%o if k eq n then return 1;
%o elif ((n-k) mod 3) eq 0 then return (3/(n-k))*Binomial(n, Floor((n-k-3)/3))*(k+1);
%o else return 0;
%o end if; return A111373;
%o end function;
%o [A111373(n,k): k in [0..n], n in [0..15]]; // _G. C. Greubel_, Jul 30 2022
%o (SageMath)
%o def A111373(n,k):
%o if(k==n): return 1
%o elif ((n-k)%3==0): return (3/(n-k))*binomial(n, (n-k)/3 -1)*(k+1)
%o else: return 0
%o flatten([[A111373(n,k) for k in (0..n)] for n in (0..15)]) # _G. C. Greubel_, Jul 30 2022
%Y First column is A001764. Bears same relation to A001764 as A053121 does to A000108.
%Y Cf. A000012, A000027, A027379, A111396, A126042 (row sums), A167543.
%K nonn,easy,tabl
%O 0,12
%A _N. J. A. Sloane_, Nov 09 2005
%E More terms from Kerri Sullivan (ksulliva(AT)ashland.edu), Jan 23 2006
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