A093127 Triangle read by rows: differences of Narayana numbers.
1, 1, 1, 2, 1, 5, 1, 9, 3, 1, 14, 14, 1, 20, 40, 4, 1, 27, 90, 30, 1, 35, 175, 125, 5, 1, 44, 308, 385, 55, 1, 54, 504, 980, 315, 6, 1, 65, 780, 2184, 1274, 91, 1, 77, 1155, 4410, 4116, 686, 7, 1, 90, 1650, 8250, 11340, 3528, 140, 1, 104, 2288, 14520, 27720, 14112, 1344, 8
Offset: 0
Examples
T(3,1) = 5 because there are 5 ways to insert a single diagonal into a pentagon. Triangle begins: 1; 1; 1, 2; 1, 5; 1, 9, 3; 1, 14, 14; 1, 20, 40, 4; 1, 27, 90, 30;
Links
- G. C. Greubel, Rows n = 0..100 of triangle, flattened
- M. Bona and A. Knopfmacher, On the probability that certain compositions have the same number of parts, Ann. Comb., 14 (2010), 291-306. - _Emeric Deutsch_, Sep 18 2014
Programs
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GAP
B:=Binomial;; T:= function(n,k) return B(n-k+2,k+1)*B(n-k+2,k)/(n-k+2) - B(n-k+1,k)*B(n-k+1,k-1)/(n-k+1); end; Flat(List([0..15], n-> List([0..Int(n/2)], k-> T(n,k) ))); # G. C. Greubel, Dec 28 2019
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Magma
B:=Binomial; T:= func< n, k | B(n-k+2,k+1)*B(n-k+2,k)/(n-k+2) - B(n-k+1,k)*B(n-k+1,k-1)/(n-k+1) >; [T(n,k): k in [0..Floor(n/2)], n in [0..15]]; // G. C. Greubel, Dec 28 2019
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Maple
eq := t*z^4*G^2-(1-z-2*t*z^2-t*z^3+t^2*z^4)*G+1 = 0: G := RootOf(eq, G): Gser := simplify(series(G, z = 0, 18)): for n from 0 to 15 do P[n] := sort(coeff(Gser, z, n)) end do: for n from 0 to 15 do seq(coeff(P[n], t, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form # Emeric Deutsch, Sep 18 2014 T := proc (n, k) if (1/2)*n+1/2 < k then 0 else binomial(n-k+2, k+1)*binomial(n-k+2, k)/(n-k+2)-binomial(n-k+1, k)*binomial(n-k+1, k-1)/(n-k+1) end if end proc: for n from 0 to 15 do seq(T(n, k), k = 0 .. floor((1/2)*n)) end do; # yields sequence in triangular form # Emeric Deutsch, Sep 18 2014
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Mathematica
Nara[n_, k_]:= Binomial[n, k]*Binomial[n, k-1]/n; T[n_, k_]:= Nara[n-k+2, k+1] - Nara[n-k+1, k]; Table[T[n, k], {n,0,15}, {k,0,Floor[n/2]}]//Flatten (* G. C. Greubel, Dec 28 2019 *)
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PARI
T(n,k) = my(b=binomial); b(n-k+2,k+1)*b(n-k+2,k)/(n-k+2) - b(n-k+1,k)* b(n-k+1,k-1)/(n-k+1); for(n=0,15, for(k=0, n\2, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 28 2019
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Sage
b=binomial; def T(n,k): return b(n-k+2,k+1)*b(n-k+2,k)/(n-k+2) - b(n-k+1,k)* b(n-k+1,k-1)/(n-k+1) [[T(n,k) for k in (0..floor(n/2))] for n in (0..15)] # G. C. Greubel, Dec 28 2019
Formula
T(n, k) = Narayana(n-k+2, k+1) - Narayana(n-k+1, k) where Narayana(n, k) = binomial(n, k)*binomial(n, k-1)/n is A001263.
G.f.: G=G(t,z) satisfies t*z^4*G^2 - (1 - z - 2*t*z^2 - t*z^3 + t^2*z^4)*G + 1 = 0. - Emeric Deutsch, Sep 18 2014
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