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%I #50 Apr 26 2024 09:05:33
%S 1,1,1,2,2,1,3,6,3,1,6,12,12,4,1,10,30,30,20,5,1,20,60,90,60,30,6,1,
%T 35,140,210,210,105,42,7,1,70,280,560,560,420,168,56,8,1,126,630,1260,
%U 1680,1260,756,252,72,9,1,252,1260,3150,4200,4200,2520,1260,360,90,10,1
%N A number triangle of inverse Chebyshev transforms.
%C First column is A001405, second column is A100071, third column is A107231. Row sums are A005773(n+1), diagonal sums are A026003. The inverse Chebyshev transform concerned takes a g.f. g(x)->(1/sqrt(1-4x^2))g(xc(x^2)) where c(x) is the g.f. of A000108. It transforms a(n) to b(n) = Sum_{k=0..floor(n/2)} binomial(n,k)*a(n-2k). Then a(n) = Sum_{k=0..floor(n/2)} (n/(n-k))*(-1)^k*binomial(n-k,k) *b(n-2k).
%C Triangle read by rows: T(n,k) is the number of paths of length n with steps U=(1,1), D=(1,-1) and H=(1,0), starting at (0,0), staying weakly above the x-axis (i.e., left factors of Motzkin paths) and having k H steps. Example: T(3,1)=6 because we have HUD. HUU, UDH, UHD, UHU and UUH. Sum_{k=0..n} k*T(n,k) = A132894(n). - _Emeric Deutsch_, Oct 07 2007
%H Jinyuan Wang, <a href="/A107230/b107230.txt">Rows n=0..200 of triangle, flattened</a>
%H Paul Barry, <a href="https://www.emis.de/journals/JIS/VOL22/Barry1/barry411.html">The Central Coefficients of a Family of Pascal-like Triangles and Colored Lattice Paths</a>, J. Int. Seq., Vol. 22 (2019), Article 19.1.3.
%F T(n,k) = binomial(n,k)*binomial(n-k, floor((n-k)/2)).
%F G.f.: G=G(t,z) satisfies z*(1-2*z-t*z)*G^2+(1-2*z-t*z)*G-1=0. - _Emeric Deutsch_, Oct 07 2007
%F E.g.f.: exp(x*y)*(BesselI(0,2*x)+BesselI(1,2*x)). - _Vladeta Jovovic_, Dec 02 2008
%e Triangle begins
%e 1;
%e 1, 1;
%e 2, 2, 1;
%e 3, 6, 3, 1;
%e 6, 12, 12, 4, 1;
%e 10, 30, 30, 20, 5, 1;
%p T:=proc(n,k) options operator, arrow: binomial(n, k)*binomial(n-k, floor((1/2)*n-(1/2)*k)) end proc: for n from 0 to 11 do seq(T(n,k),k=0..n) end do; # yields sequence in triangular form - _Emeric Deutsch_, Oct 07 2007
%t Table[Binomial[n, k]*Binomial[n-k, Floor[(n-k)/2]], {n, 0, 10}, {k, 0, n}]//Flatten (* _G. C. Greubel_, Feb 11 2019 *)
%o (PARI) T(n, k) = binomial(n, k)*binomial(n-k, (n-k)\2); \\ _Michel Marcus_, Feb 10 2019
%o (Magma) [[Binomial(n, k)*Binomial(n-k, Floor((n-k)/2)): k in [0..n]]: n in [0..10]]; // _G. C. Greubel_, Feb 11 2019
%o (Sage) [[binomial(n, k)*binomial(n-k, floor((n-k)/2)) for k in (0..n)] for n in (0..10)] # _G. C. Greubel_, Feb 11 2019
%Y Cf. A132894.
%K easy,nonn,tabl,changed
%O 0,4
%A _Paul Barry_, May 13 2005
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