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A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.

%I A157751 #20 Sep 24 2018 08:56:34
%S A157751 1,1,2,1,2,4,1,4,4,8,1,4,12,8,16,1,6,12,32,16,32,1,6,24,32,80,32,64,1,
%T A157751 8,24,80,80,192,64,128,1,8,40,80,240,192,448,128,256,1,10,40,160,240,
%U A157751 672,448,1024,256,512,1,10,60,160,560,672,1792,1024,2304,512,1024,1,12,60,280,560,1792,1792,4608,2304,5120,1024,2048
%N A157751 Triangle of coefficients of polynomials F(n,x) in descending powers of x generated by F(n,x)=(x+1)*F(n-1,x)+F(n-1,-x), with initial F(0,x)=1.
%C A157751 Conjecture 1. If n>1 is even then F(n,x) has no real roots.
%C A157751 Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,
%C A157751 and if n>4 then 0 < -r < n.
%C A157751 Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]
%C A157751 Cayley (1876) states "We, in fact, find  1 + sin u = 1 + x,  1 - sin 3u = (1 + x)(1 - 2x)^2,  1 + sin 5u = (1 + x)(1 + 2x - 4x^2)^2,  1 - sin 7u = (1 + x)(1 - 4x - 4x^2 + 8x^3)^2, &c.". - _Michael Somos_, Jun 19 2012
%C A157751 Appears to be the unsigned row reverse of A180870 and A228565. - _Peter Bala_, Feb 17 2014
%C A157751 From _Wolfdieter Lang_, Jul 29 2014: (Start)
%C A157751 This triangle is the Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). For Riordan triangles see the W. Lang link 'Sheffer a-and z-sequences' under A006232, also for references. The o.g.f. given by Peter Bala in the formula section refers to the row reversed triangle. The usual information on this triangle, like o.g.f. for the columns, the row sums, the alternating row sums, the recurrences using A- and Z-sequences, etc. follows from this Riordan property. The Riordan proof follows from the given o.g.f. by Peter Bala, call it Grev(x,z), by row reversion: G(x,z) = Grev(1/x,x*z) = (1+z)/(1- 2*x*z - z^2) = G(z)*(1/(1 - x*F(z))) with G(z) = (1+z)/(1-z^2) and F(z) = z*2/(1-z^2). See A244419 for the discussion for a signed version of this triangle.
%C A157751 (End)
%D A157751 A. Cayley, On an Expression for 1 +- sin(2p+1)u in Terms of sin u, Messenger of Mathematics, 5 (1876), pp. 7-8 = Mathematical Papers Vol. 10, n. 630, pp. 1-2.
%H A157751 G. C. Greubel, <a href="/A157751/b157751.txt">Rows n=0..100 of triangle, flattened</a>
%F A157751 Count the top row as row 0 and let C(n,k) denote the usual binomial
%F A157751 coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),
%F A157751 p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final
%F A157751 1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row
%F A157751 2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),
%F A157751 q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).
%F A157751 Then the k-th number in row 2n+1 is q(k)*2^k.
%F A157751 From _Peter Bala_, Jan 17 2014: (Start)
%F A157751 Working with an offset of 0, the o.g.f. is (1 + x*z)/(1 - 2*z - x^2*z^2) = 1 + (x + 2)*z + (x^2 + 2*x + 4)*z^2 + ....
%F A157751 Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.
%F A157751 The polynomials G(n,x) defined by G(0,x) = 1 and G(n,x) = x*F(n-1,x) for n >= 1 satisfy G(n,x) = (x + 1)*G(n-1,x) - G(n-1,-x). Cf. A140070 and A140071. (End)
%F A157751 From _Wolfdieter Lang_, Jul 29 2014: (Start)
%F A157751 O.g.f. for the row polynomials (rising powers of x) R(n,x) = x^n*F(n,1/x): (1+z)/(1 - 2*x*z - z^2). Riordan triangle ((1+z)/(1-z^2), 2*z/(1-z^2)). See a comment above.
%F A157751 Recurrence for the row polynomials R(n,x) = (1+x)*R(n-1,x) - (-1)^n*x*R(n-1,-x), n >= 1, R(0,x) = 1.
%F A157751 R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.
%F A157751 Recurrence for the triangle: R(n,m) = R(n-1,m) + (1 + (-1)^(n-m))*R(n-1,m-1), n >= m >= 1, R(n,m) = 0 if n < m, R(n,0) = 1.
%F A157751 O.g.f. column sequences ((1+x)/(1-x^2))*(2*x/(1-x^2))^m, m >= 0. See A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980, ...
%F A157751 Row sums A078057. Alternating row sums A123335.
%F A157751 (End)
%e A157751 Rows 0 to 8:
%e A157751 1
%e A157751 1 2
%e A157751 1 2 4
%e A157751 1 4 4 8
%e A157751 1 4 12 8 16
%e A157751 1 6 12 32 16 32
%e A157751 1 6 24 32 80 32 64
%e A157751 1 8 24 80 80 192 64 128
%e A157751 1 8 40 80 240 192 448 128 256
%e A157751 (Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).
%e A157751 First few polynomials:
%e A157751 F(0,x)=1, F(1,x)=x+2, F(2,x)=x^2+2*x+4, F(3,x)=x^3+4*x^2+4*x+8.
%e A157751 The row polynomials R(n,x) start: 1, 1 + 2*x = x*F(1,1/x), 1 + 2*x + 4^x^2 = x^2*F(2,1/x), ...  - _Wolfdieter Lang_, Jul 29 2014
%t A157751 T[n_, 0]:= 1; T[n_, n_]:= 2^n; T[n_, k_]:= T[n, k] = T[n-1, k] + (1 + (-1)^(n-k))*T[n-1, k-1]; Table[T[n, k], {n, 0, 15}, {k, 0, n}] (* _G. C. Greubel_, Sep 24 2018 *)
%o A157751 (PARI) t(n,k) = if(k==0, 1, if(k==n, 2^n, t(n-1,k) + (1+(-1)^(n-k))*t(n-1,k-1)));
%o A157751 for(n=0,15, for(k=0,n, print1(t(n,k), ", "))) \\ _G. C. Greubel_, Sep 24 2018
%Y A157751 Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.
%K A157751 nonn,tabl,easy
%O A157751 0,3
%A A157751 _Clark Kimberling_, Mar 05 2009
%E A157751 Offset corrected to 0. Cf.s added, keyword easy added by _Wolfdieter Lang_, Jul 29 2014