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%I #14 Apr 21 2014 22:58:54
%S 1,0,1,2,0,1,0,1,0,1,-2,0,0,0,1,0,-3,0,-1,0,1,2,0,-3,0,-2,0,1,0,5,0,
%T -2,0,-3,0,1,-2,0,8,0,0,0,-4,0,1,0,-7,0,10,0,3,0,-5,0,1,2,0,-15,0,10,
%U 0,7,0,-6,0,1,0,9,0,-25,0,7,0,12,0,-7,0,1,-2,0,24,0,-35,0,0,0,18,0,-8,0,1,0,-11,0,49,0,-42,0,-12,0
%N Triangle T(n,k), read by rows: T(n,k)= 0 if n-k odd. T(n,k)= 2*(-1)^((n-k)/2)*(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.
%C Polynomial coefficients of P(n,x) in increasing powers, read by rows, where P(0,x)=1, P(1,x)=x, P(2,x)=2+x^2, P(3,x)=x+x^3, P(4,x)=-2+x^4, and P(n,x) = x*P(n-1,x) - P(n-2,x) for n>=5.
%C The row-reversed version of A135929.
%C Row sums are repeating 1, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1..., see A138034 and A119910.
%H P. Steinbach, <a href="http://www.jstor.org/stable/2691048">Golden fields: a case for the heptagon</a>, Math. Mag. 70 (1997), no. 1, 22-31, <a href="http://www.ams.org/mathscinet-getitem?mr=1439165">MR 1439165</a>
%F T(n,k)= 0 if n-k odd. T(n,k)= 2*(-1)^((n-k)/2)*(2k-n)/(n+k)*binomial((n+k)/2,(n-k)/2) if n-k even.
%F P(n,x) = x*P(n-1,x)-P(n-2,x), n>=5.
%F P(n,2*x) = -2*T(n,x)+4*x*U(n-1,x), where T(n,x) is A053120 and U(n,x) is A053117.
%e {1}, = 1
%e {0, 1}, = x
%e {2, 0, 1}, = 2+x^2
%e {0, 1, 0, 1}, = x+x^3
%e {-2, 0, 0, 0, 1}, = -2+x^4
%e {0, -3, 0, -1, 0, 1}, = -3x-x^3+x^5
%e {2, 0, -3, 0, -2, 0, 1},
%e {0, 5, 0, -2, 0, -3, 0, 1},
%e {-2, 0, 8, 0, 0, 0, -4, 0, 1},
%e {0, -7, 0, 10, 0, 3, 0, -5, 0, 1},
%e {2, 0, -15, 0, 10, 0, 7, 0, -6, 0, 1},
%e {0, 9, 0, -25, 0, 7, 0, 12, 0, -7, 0, 1}
%p A137276 := proc(n,k) local nmk,npk; if n = 0 then 1; elif (n-k) mod 2 <> 0 then 0; else nmk := (n-k)/2 ; npk := (n+k)/2 ; (-1)^nmk*(2*k-n)/npk*binomial(npk,nmk) ; fi; end:
%p seq( seq(A137276(n,k),k=0..n),n=0..13) ;
%Y Cf. A123956, A137289.
%K sign,tabl
%O 0,4
%A _Roger L. Bagula_ and _Gary W. Adamson_, Mar 13 2008
%E Fourth row inserted by the Associate Editors of the OEIS, Aug 27 2009
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