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Even row Pascal-square read by antidiagonals.
8

%I #74 Oct 02 2023 21:26:00

%S 1,1,2,1,4,3,1,6,10,4,1,8,21,20,5,1,10,36,56,35,6,1,12,55,120,126,56,

%T 7,1,14,78,220,330,252,84,8,1,16,105,364,715,792,462,120,9,1,18,136,

%U 560,1365,2002,1716,792,165,10

%N Even row Pascal-square read by antidiagonals.

%C Apart from signs identical to A053123. Mirror of A078812.

%C As a triangle, row n consists of the coefficients of Morgan-Voyce polynomial B(n,x); e.g., B(3,x)=x^3+6x^2+10x+4. As a triangle, rows 0 to 4 are as follows: 1 1...2 1...4...3 1...6...10...4 1...8...21...20...5 See A054142 for coefficients of Morgan-Voyce polynomial b(n,x).

%C Scaled version of A119900. - _Philippe Deléham_, Feb 24 2012

%C A172431 is jointly generated with A054142 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section. - _Clark Kimberling_, Mar 09 2012

%C Subtriangle of the triangle given by (1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - _Philippe Deléham_, Mar 22 2012

%H G. C. Greubel, <a href="/A172431/b172431.txt">Rows n = 1..100 of triangle, flattened</a>

%F As a decimal sequence: a(n)= 12*a(n-1)- a(n-2) with a(1)=1. [I interpret this remark as: 1, 12=1,2, 143=1,4,3, 1704=1,6,10,4,... taken from A004191 are decimals on the diagonal. - _R. J. Mathar_, Sep 08 2013]

%F As triangle T(n,k): T(n,k) = T(n-1,k) + 2*T(n-1,k-1) - T(n-2,k-2). - _Philippe Deléham_, Feb 24 2012

%F As DELTA-triangle T(n,k) with 0<=k<=n: G.f.: (1-y*x)^2/((1-y*x)^2-x). - _Philippe Deléham_, Mar 22 2012

%F T(n, k) = GegenbauerC(k, n-k, 1). - _Peter Luschny_, May 10 2016

%F As triangle T(n,k): Product_{k=1..n} T(n,k) = Product_{k=0..n-1} binomial(2*k,k) = A007685(n-1) for n >= 1. - _Werner Schulte_, Apr 26 2017

%F As triangle T(n,k) with 1 <= k <= n: T(n,k) = binomial(2*n-k, k-1). - _Paul Weisenhorn_, Nov 25 2019

%e Array begins:

%e 1, 2, 3, 4, 5, 6, ...

%e 1, 4, 10, 20, 35, ...

%e 1, 6, 21, 56, ...

%e 1, 8, 36, ...

%e 1, 10, ...

%e 1, ...

%e ...

%e Example:

%e Starting with 1, every entry is twice the one to the left minus the second one to the left, plus the one above.

%e For n = 9 the a(9) = 10 solution is 2*4 - 1 + 3.

%e From _Philippe Deléham_, Feb 24 2012: (Start)

%e Triangle T(n,k) begins:

%e 1;

%e 1, 2;

%e 1, 4, 3;

%e 1, 6, 10, 4;

%e 1, 8, 21, 20, 5;

%e 1, 10, 36, 56, 35, 6;

%e 1, 12, 55, 120, 126, 56, 7; (End)

%e From _Philippe Deléham_, Mar 22 2012: (Start)

%e (1, 0, 0, 0, 0, 0, ...) DELTA (0, 2, -1/2, 1/2, 0, 0, ...) begins:

%e 1;

%e 1, 0;

%e 1, 2, 0;

%e 1, 4, 3, 0;

%e 1, 6, 10, 4, 0;

%e 1, 8, 21, 20, 5, 0;

%e 1, 10, 36, 56, 35, 6, 0;

%e 1, 12, 55, 120, 126, 56, 7, 0; (End)

%p T := (n, k) -> simplify(GegenbauerC(k, n-k, 1)):

%p for n from 0 to 10 do seq(T(n,k), k=0..n-1) od; # _Peter Luschny_, May 10 2016

%t u[1, x_] := 1; v[1, x_] := 1; z = 16;

%t u[n_, x_] := x*u[n - 1, x] + v[n - 1, x];

%t v[n_, x_] := x*u[n - 1, x] + (x + 1)*v[n - 1, x];

%t Table[Expand[u[n, x]], {n, 1, z/2}]

%t Table[Expand[v[n, x]], {n, 1, z/2}]

%t cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];

%t TableForm[cu]

%t Flatten[%] (* A054142 *)

%t Table[Expand[v[n, x]], {n, 1, z}]

%t cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];

%t TableForm[cv]

%t Flatten[%] (* A172431 *)

%t (* _Clark Kimberling_, Mar 09 2012 *)

%t Table[GegenbauerC[k-1, n-k+1, 1], {n, 15}, {k, n}]//Flatten (* _G. C. Greubel_, Dec 15 2019 *)

%o (PARI) T(n,k) = sum(j=0, (k-1)\2, (-1)^j*(n-j-1)!*2^(k-2*j-1)/(j!*(n-k)!*(k-2*j-1)!) );

%o for(n=1, 10, for(k=1, n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Dec 15 2019

%o (Magma) F:=Factorial; [ &+[(-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)): j in [0..Floor((k-1)/2)]]: k in [1..n], n in [1..15]]; // _G. C. Greubel_, Dec 15 2019

%o (Sage) [[gegenbauer(k-1, n-k+1, 1) for k in (1..n)] for n in (1..15)] # _G. C. Greubel_, Dec 15 2019

%o (GAP) F:=Factorial;; Flat(List([1..15], n-> List([1..n], k-> Sum([0..Int((k-1)/2)], j-> (-1)^j*F(n-j-1)*2^(k-2*j-1)/(F(j)*F(n-k)*F(k-2*j-1)) )))); # _G. C. Greubel_, Dec 15 2019

%Y Cf. A078812, A053123, A007318, A001906 (antidiagonals sums), A007685.

%Y Cf. also A054142, A082985.

%K nonn,tabl

%O 1,3

%A _Mark Dols_, Feb 02 2010