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%I #33 Sep 08 2022 08:45:16
%S 1,1,1,1,3,1,1,5,5,1,1,7,12,7,1,1,9,22,22,9,1,1,11,35,50,35,11,1,1,13,
%T 51,95,95,51,13,1,1,15,70,161,210,161,70,15,1,1,17,92,252,406,406,252,
%U 92,17,1,1,19,117,372,714,882,714,372,117,19,1,1,21,145,525,1170,1722,1722,1170,525,145,21,1
%N A figurate number triangle read by rows.
%C Row coefficients are the absolute values of the coefficients of the characteristic polynomials of the n X n matrices A(n) with A(n)_{i,i} = 2, i>0, A(n)_{i,j} = 1, otherwise (starts with (0,0) position).
%C The triangle can be generated by the matrix multiplication A007318 * A114219s, where A114219s = 0; 0,1; 0,1,1; 0,-1,2,1; 0,1,-2,3,1; 0,-1,2,-3,4,1; ... = A097807 * A128229 is a signed variant of A114219. - _Gary W. Adamson_, Feb 20 2007
%H G. C. Greubel, <a href="/A103450/b103450.txt">Rows n = 0..100 of the triangle, flattened</a>
%H G. Chiaselotti, W. Keith, and P. A. Oliverio, <a href="https://www.researchgate.net/publication/275617885_Two_Self-Dual_Lattices_of_Signed_Integer_Partitions">Two Self-Dual Lattices of Signed Integer Partitions</a>, Appl. Math. Inf. Sci. 8, No. 6, 3191-3199 (2014), via ResearchGate.
%F T(n, k) = binomial(n-1, k-1)*(k*(n-k) + n)/k with T(n, 0) = 1.
%F T(n, k) = T(n-1, k-1) + T(n-1, k) + binomial(n-2, k-1) with T(n, 0) = 1.
%F Column k is generated by (1+k*x)*x^k/(1-x)^(k+1).
%F Rows are coefficients of the polynomials P(0, x) = 1, P(n, x) = (1+x)^(n-2)*(1 +(n+1)*x + x^2) for n>0.
%F T(n,k) = Sum_{j=0..n} binomial(k, k-j)*binomial(n-k, j)*(j+1). - _Paul Barry_, Oct 28 2006
%F A signed version arises from the coefficients of the polynomials defined by: p(x, 0) = 1, p(x, 1) = (-1 +x), p(x, 2) = (1 -3*x +x^2), p(x,n) = (-1 +x)^(n-2)*(1 - (n + 1)*x + x^2); T(n, k) = (-1)^(n+k)*coefficient of x^k of ( p(x,n) ). - _Roger L. Bagula_ and _Gary W. Adamson_, Oct 21 2008
%F T(2*n+1, n) = A141222(n). - _Emanuele Munarini_, Jun 01 2012 [corrected by _Werner Schulte_, Nov 27 2021]
%F G.f.: is 1 / ( (1-q*x/(1-x)) * (1-x/(1-q*x)) ). - _Joerg Arndt_, Aug 27 2013
%F Sum_{k=0..floor(n/2)} T(n-k, k) = (1/5)*((-n+5)*Fibonacci(n+1) + (3*n- 2)*Fibonacci(n)) = A208354(n). - _G. C. Greubel_, Jun 17 2021
%F T(2*n, n) = A000984(n) * (n + 2) / 2 for n >= 0. - _Werner Schulte_, Nov 27 2021
%e From _Roger L. Bagula_, Oct 21 2008: (Start)
%e The triangle begins:
%e 1;
%e 1, 1;
%e 1, 3, 1;
%e 1, 5, 5, 1;
%e 1, 7, 12, 7, 1;
%e 1, 9, 22, 22, 9, 1;
%e 1, 11, 35, 50, 35, 11, 1;
%e 1, 13, 51, 95, 95, 51, 13, 1;
%e 1, 15, 70, 161, 210, 161, 70, 15, 1;
%e 1, 17, 92, 252, 406, 406, 252, 92, 17, 1;
%e 1, 19, 117, 372, 714, 882, 714, 372, 117, 19, 1; ... (End)
%t (* First program *)
%t p[x_, n_]:= p[x, n]= If[n==0, 1, (-1+x)^(n-2)*(1 -(n+1)*x +x^2)];
%t T[n_, k_]:= T[n,k]= (-1)^(n+k)*SeriesCoefficient[p[x, n], {x, 0, k}];
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _Roger L. Bagula_ and _Gary W. Adamson_, Oct 21 2008 *)(* corrected by _G. C. Greubel_, Jun 17 2021 *)
%t (* Second program *)
%t T[n_, k_]:= If[k==0, 1, Binomial[n, k]*(n*(k+1) -k^2)/n];
%t Table[T[n, k], {n,0,16}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jun 17 2021 *)
%o (Magma)
%o A103450:= func< n,k | k eq 0 select 1 else Binomial(n, k)*(k*(n-k) + n)/n >;
%o [A103450(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jun 17 2021
%o (Sage)
%o def A103450(n, k): return 1 if (k==0) else binomial(n, k)*(k*(n-k) + n)/n
%o flatten([[A103450(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jun 17 2021
%Y Row sums are A045623.
%Y Columns include: A000326, A002412, A002418, A005408.
%Y Cf. A000045, A208354.
%K easy,nonn,tabl
%O 0,5
%A _Paul Barry_, Feb 06 2005
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