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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. 10
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, 8, 24, 80, 80, 192, 64, 128, 1, 8, 40, 80, 240, 192, 448, 128, 256, 1, 10, 40, 160, 240, 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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Conjecture 1. If n>1 is even then F(n,x) has no real roots.

Conjecture 2. If n>0 is odd then F(n,x) has exactly one real root, r,

and if n>4 then 0 < -r < n.

Conjectures 1 and 2 are true. [From Alain Thiery (Alain.Thiery(AT)math.u-bordeaux1.fr), May 14 2010]

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

Appears to be the unsigned row reverse of A180870 and A228565. - Peter Bala, Feb 17 2014

From Wolfdieter Lang, Jul 29 2014: (Start)

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.

(End)

REFERENCES

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.

LINKS

G. C. Greubel, Rows n=0..100 of triangle, flattened

FORMULA

Count the top row as row 0 and let C(n,k) denote the usual binomial

coefficient. For row 2n, define p(0)=C(n,0), p(1)=C(n,1), p(2)=C(n+1,2),

p(3)=C(n+1,3), p(4)=C(n+2,4), p(5)=c(n+2,5),..., until reaching two final

1's: p(2n-1)=C(2n-1,2n-1) and p(2n)=C(2n,2n). Then the k-th number in row

2n is p(k)*2^k. For row 2n+1, define q(0)=C(n,0), q(1)=C(n+1,1),

q(2)=C(n+1,2), q(3)=C(n+2,3),..., until reaching q(2n+1)=C(2n+1,2n+1).

Then the k-th number in row 2n+1 is q(k)*2^k.

From Peter Bala, Jan 17 2014: (Start)

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 + ....

Recurrence equation: T(n,k) = 2*T(n-1,k-1) + T(n-2,k-2) with T(n,0) = 1.

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)

From Wolfdieter Lang, Jul 29 2014: (Start)

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.

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.

R(n,x) = Ftilde(n,2*x) + Ftilde(n-1,2*x) with the monic Fibonacci polynomials Ftilde(n,x) given in A168561.

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.

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, ...

Row sums A078057. Alternating row sums A123335.

(End)

EXAMPLE

Rows 0 to 8:

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 8 24 80 80 192 64 128

1 8 40 80 240 192 448 128 256

(Row 8) = (1, 4*2, 10*4, 10*8, 15*16, 6*32, 7*64, 1*128, 1*256).

First few polynomials:

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.

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

MATHEMATICA

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 *)

PROG

(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)));

for(n=0, 15, for(k=0, n, print1(t(n, k), ", "))) \\ G. C. Greubel, Sep 24 2018

CROSSREFS

Cf. A140070, A140071, A180870, A228565, A078057, A123335, A000012, 2*A004526, 4*A008805, 8*A058187, 16*A189976, 32*A189980.

Sequence in context: A261358 A261356 A244419 * A177701 A119765 A303325

Adjacent sequences:  A157748 A157749 A157750 * A157752 A157753 A157754

KEYWORD

nonn,tabl,easy

AUTHOR

Clark Kimberling, Mar 05 2009

EXTENSIONS

Offset corrected to 0. Cf.s added, keyword easy added by Wolfdieter Lang, Jul 29 2014

STATUS

approved

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