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%I #38 Jul 15 2023 05:41:13
%S 1,1,1,1,0,1,1,-1,-1,1,1,-2,-2,-2,1,1,-3,-2,-2,-3,1,1,-4,-1,0,-1,-4,1,
%T 1,-5,1,3,3,1,-5,1,1,-6,4,6,6,6,4,-6,1,1,-7,8,8,6,6,8,8,-7,1,1,-8,13,
%U 8,2,0,2,8,13,-8,1,1,-9,19,5,-6,-10,-10,-6,5,19,-9,1,1,-10,26,-2,-17,-20,-20,-20,-17,-2,26,-10,1,1,-11,34,-14,-29,-25
%N A triangle of Krawtchouk coefficients.
%C Row sums are A009545(n+1), with e.g.f. exp(x)(cos(x)+sin(x)). Diagonal sums are A077948.
%C The rows are the diagonals of the Krawtchouk matrices. Coincides with the Riordan array (1/(1-x),(1-2x)/(1-x)). - _Paul Barry_, Sep 24 2004
%C Corresponds to Pascal-(1,-2,1) array, read by antidiagonals. The Pascal-(1,-2,1) array has n-th row generated by (1-2x)^n/(1-x)^(n+1). - _Paul Barry_, Sep 24 2004
%C A modified version (different signs) of this triangle is given by T(n,k) = Sum_{j=0..n} C(n-k,j)*C(k,j)*cos(Pi*(k-j)). - _Paul Barry_, Jun 14 2007
%D P. Feinsilver and J. Kocik, Krawtchouk matrices from classical and quantum walks, Contemporary Mathematics, 287 2001, pp. 83-96.
%H G. C. Greubel, <a href="/A098593/b098593.txt">Table of n, a(n) for the first 50 rows, flattened</a>
%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL11/Barry/barry594.html">A note on Krawtchouk Polynomials and Riordan Arrays</a>, JIS 11 (2008) 08.2.2
%H P. Feinsilver and J. Kocik, <a href="http://arxiv.org/abs/quant-ph/0702073">Krawtchouk polynomials and Krawtchouk matrices</a>, arxiv:quant-ph/0702073, 2007.
%F T(n, k) = Sum_{i=0..k} binomial(n-k, k-i)*binomial(k, i)*(-1)^(k-i), k<=n.
%F T(n, k) = T(n-1, k) + T(n-1, k-1) - 2*T(n-2, k-1) (n>0). - _Paul Barry_, Sep 24 2004
%F T(n, k) = [k<=n]*Hypergeometric2F1(-k,k-n;1;-1). - _Paul Barry_, Jan 24 2011
%F E.g.f. for the n-th subdiagonal: exp(x)*P(n,x), where P(n,x) is the polynomial Sum_{k = 0..n} (-1)^k*binomial(n,k)* x^k/k!. For example, the e.g.f. for the second subdiagonal is exp(x)*(1 - 2*x + x^2/2) = 1 - x - 2*x^2/2! - 2*x^3/3! - x^4/4! + x^5/5! + .... - _Peter Bala_, Mar 05 2017
%e Rows begin {1}, {1,1}, {1,0,1}, {1,-1,-1,1}, {1,-2,-2,-2,1}, ...
%e From _Paul Barry_, Oct 05 2010: (Start)
%e Triangle begins
%e 1,
%e 1, 1,
%e 1, 0, 1,
%e 1, -1, -1, 1,
%e 1, -2, -2, -2, 1,
%e 1, -3, -2, -2, -3, 1,
%e 1, -4, -1, 0, -1, -4, 1,
%e 1, -5, 1, 3, 3, 1, -5, 1,
%e 1, -6, 4, 6, 6, 6, 4, -6, 1,
%e 1, -7, 8, 8, 6, 6, 8, 8, -7, 1,
%e 1, -8, 13, 8, 2, 0, 2, 8, 13, -8, 1
%e Production matrix (related to large Schroeder numbers A006318) begins
%e 1, 1,
%e 0, -1, 1,
%e 0, -2, -1, 1,
%e 0, -6, -2, -1, 1,
%e 0, -22, -6, -2, -1, 1,
%e 0, -90, -22, -6, -2, -1, 1,
%e 0, -394, -90, -22, -6, -2, -1, 1,
%e 0, -1806, -394, -90, -22, -6, -2, -1, 1,
%e 0, -8558, -1806, -394, -90, -22, -6, -2, -1, 1
%e Production matrix of inverse is
%e -1, 1,
%e -2, 1, 1,
%e -4, 2, 1, 1,
%e -8, 4, 2, 1, 1,
%e -16, 8, 4, 2, 1, 1,
%e -32, 16, 8, 4, 2, 1, 1,
%e -64, 32, 16, 8, 4, 2, 1, 1,
%e -128, 64, 32, 16, 8, 4, 2, 1, 1,
%e -256, 128, 64, 32, 16, 8, 4, 2, 1, 1 (End)
%t T[n_, k_] := Sum[Binomial[n - k, k - j]*Binomial[k, j]*(-1)^(k - j), {j, 0, n}]; Table[T[n, k], {n, 0, 49}, {k, 0, n}] // Flatten (* _G. C. Greubel_, Oct 15 2017 *)
%o (PARI) for(n=0,10, for(k=0,n, print1(sum(i=0,k, binomial(n-k, k-i) *binomial(k, i)*(-1)^(k-i)), ", "))) \\ _G. C. Greubel_, Oct 15 2017
%Y Cf. Pascal (1,m,1) array: A123562 (m = -3), A000012 (m = -1), A007318 (m = 0), A008288 (m = 1), A081577 (m = 2), A081578 (m = 3), A081579 (m = 4), A081580 (m = 5), A081581 (m = 6), A081582 (m = 7), A143683 (m = 8).
%K easy,sign,tabl
%O 0,12
%A _Paul Barry_, Sep 17 2004
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