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A137277 Triangle of the coefficients [x^k] P_n(x) of the polynomials P_n(x) = 1/n * sum(j=0..floor(n/2), (-1)^j * binomial(n,j) * (n-4*j) * x^(n-2*j) ).

%I A137277 #7 Mar 30 2012 17:34:25
%S A137277 1,0,1,2,0,1,0,1,0,1,-6,0,0,0,1,0,-6,0,-1,0,1,20,0,-5,0,-2,0,1,0,25,0,
%T A137277 -3,0,-3,0,1,-70,0,28,0,0,0,-4,0,1,0,-98,0,28,0,4,0,-5,0,1,252,0,-126,
%U A137277 0,24,0,9,0,-6,0,1,0,378,0,-150,0,15,0,15,0,-7,0,1,-924,0,528,0,-165,0,0,0,22,0,-8,0,1,0,-1452
%N A137277 Triangle of the coefficients [x^k] P_n(x) of the polynomials P_n(x) = 1/n * sum(j=0..floor(n/2), (-1)^j * binomial(n,j) * (n-4*j) * x^(n-2*j) ).
%C A137277 The first four P_n(x) are the same as in A137276.
%C A137277 Row sums are 1, 1, 3, 2, -5, -6, 14, 20, -45, -70, 154, a signed variant of A047074.
%F A137277 P(0,n)=1. P_n(x) = 1/n*sum(j=0..floor(n/2), (-1)^j*binomial(n,j)*(n-4*j)*x^(n-2*j)).
%e A137277 {1}, = 1
%e A137277 {0, 1}, = x
%e A137277 {2, 0, 1}, = 2+x^2
%e A137277 {0, 1, 0, 1}, = x+x^3
%e A137277 {-6, 0, 0, 0, 1}, = -6+x^4
%e A137277 {0, -6, 0, -1, 0, 1},
%e A137277 {20, 0, -5, 0, -2, 0, 1},
%e A137277 {0, 25, 0, -3,0, -3, 0, 1},
%e A137277 {-70, 0, 28, 0, 0, 0, -4, 0, 1},
%e A137277 {0, -98, 0, 28, 0,4, 0, -5, 0, 1},
%e A137277 {252, 0, -126, 0, 24, 0, 9, 0, -6, 0, 1}
%p A137277 A137277 := proc(n,k) if n = 0 then 1; else add( (-1)^j*binomial(n,j)*(n-4*j)*x^(n-2*j),j=0..n/2)/n ; coeftayl(%,x=0,k) ; fi; end:
%p A137277 seq( seq(A137277(n,k),k=0..n),n=0..15) ;
%t A137277 B[x_, n_] = If[n > 0, Sum[(-1)^p*Binomial[n,p]*(n - 4*p)*x^(n - 2*p)/ n, {p, 0, Floor[n/2]}], 1]; a = Table[CoefficientList[B[x, n], x], {n, 0, 10}]; Flatten[a]
%Y A137277 Cf. A138034.
%K A137277 sign,easy,tabl
%O A137277 0,4
%A A137277 _Roger L. Bagula_, Mar 13 2008
%E A137277 Edited by the Associate Editors of the OEIS, Aug 27 2009