

A111125


Triangle read by rows: T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.


42



1, 3, 1, 5, 5, 1, 7, 14, 7, 1, 9, 30, 27, 9, 1, 11, 55, 77, 44, 11, 1, 13, 91, 182, 156, 65, 13, 1, 15, 140, 378, 450, 275, 90, 15, 1, 17, 204, 714, 1122, 935, 442, 119, 17, 1, 19, 285, 1254, 2508, 2717, 1729, 665, 152, 19, 1, 21, 385, 2079, 5148, 7007, 5733, 2940, 952
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OFFSET

0,2


COMMENTS

Riordan array ((1+x)/(1x)^2,x/(1x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1x),x/(1x))(1+2x,x(1+x)).  Paul Barry, Oct 17 2005
Central coefficients of rows with odd numbers of term are A052227.
From Wolfdieter Lang, Jun 26 2011: (Start)
T(k,s) appears as T_s(k) in the Knuth reference, p. 285.
This triangle is related to triangle A156308(n,m), appearing in this reference as U_m(n) on p. 285, by T(k,s)T(k1,s) = A156308(k,s), k>=s>=1 (identity on p. 286).
T(k,s) = A156308(k+1,s+1)  A156308(k,s+1), k>=s>=0 (identity on p. 286).
(End)
A111125 is jointly generated with A208513 as an array of coefficients of polynomials v(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=u(n1,x)+(x+1)*v(n1)x and v(n,x)=u(n1,x)+x*v(n1,x)+1. See the Mathematica section. The columns of A111125 are identical to those of A208508. Here, however, the alternating row sums are periodic (with period 1,2,1,1,2,1).  Clark Kimberling, Feb 28 2012
This triangle T(k,s) (with signs and columns scaled with powers of 5) appears in the expansion of Fibonacci numbers F=A000045 with multiples of odd numbers as indices in terms of odd powers of Fnumbers. See the Jennings reference, p. 108, Theorem 1. Quoted as Lemma 3 in the Ozeki reference given in A111418. The formula is: F_{(2*k+1)*n} = sum(T(k,s)*(1)^((k+s)*n)*5^s*F_{n}^(2*s+1),s=0..k), k >= 0, n >= 0.  Wolfdieter Lang, Aug 24 2012
From Wolfdieter Lang, Oct 18 2012: (Start)
This triangle T(k,s) appears in the formula x^(2*k+1)  x^((2*k+1)) = sum(T(k,s)*(xx^(1))^(2*s+1),s=0..k), k>=0. Prove the inverse formula (due to the Riordan property this will suffice) with the binomial theorem. Motivated to look into this by the quoted paper of Wang and Zhang, eq. (1.4).
Alternating row sums are A057079.
The Zsequence of this Riordan array is A217477, and the Asequence is (1)^n*A115141(n). For the notion of A and Zsequences for Riordan triangles see a W. Lang link under A006232. (End)
The signed triangle ((1)^(ks))*T(k,s) gives the coefficients of (x^2)^s of the polynomials C(2*k+1,x)/x, with C the monic integer Chebyshev Tpolynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1x)/(1+x)^2, x/(1+x)^2). Proof by comparing the o.g.f. of the row polynomials where x is replaced by x^2 with the odd part of the bisection of the o.g.f. for C(n,x)/x.  Wolfdieter Lang, Oct 23 2012
From Wolfdieter Lang, Oct 04 2013: (Start)
The signed triangle S(k,s) := ((1)^(ks))*T(k,s) (see the preceding comment) is used to express in a (4*(k+1))gon the length ratio s(4*(k+1)) = 2*sin(Pi/4*(k+1)) = 2*cos((2*k+1)*Pi/(4*(k+1))) of a side/radius as a polynomial in rho(4*(k+1)) = 2*cos(Pi/4*(k+1)), the length ratio (smallest diagonal)/side:
s(4*(k+1)) = sum(S(k,s)*rho(4*(k+1))^(2*s+1), s=0..k).
This is to be computed modulo C(4*(k+1),rho(4*(k+1)) = 0, the minimal polynomial (see A187360) in order to obtain s(4*(k+1)) as an integer in the algebraic number field Q(rho(4*(k+1))) of degree delta(4*(k+1)) (see A055034). Thanks go to Seppo Mustonen for asking me to look into the problem of the square of the total length in a regular ngon, where this formula is used in the even n case. See A127677 for the formula in the (4*k+2)gon. (End)
From Wolfdieter Lang, Aug 14 2014: (Start)
The row polynomials for the signed triangle (see the Oct 23 2012 comment above), call them todd(k,x) = sum(((1)^(ks))*T(k,s)*x^s) = S(k, x2)  S(k1, x2), k >= 0, with the Chebyshev Spolynomials (see their coefficient triangle (A049310) and S(1, x) = 0), satisfy the recurrence todd(k, x) = (1)^(k1)*((x4)/2)*todd(k1, 4x) + ((x2)/2)*todd(k1, x), k >= 1, todd(0, x) = 1. From the Aug 03 2014 comment on A130777.
This leads to a recurrence for the signed triangle, call it S(k,s) as in the Oct 04 2013 comment: S(k,s) = (1/2)*(1 + (1)^(ks))*S(k1,s1) + (2*(s+1)*(1)^(ks)  1)*S(k1,s) + (1/2)*(1)^(ks)*sum(binomial(j+s+2,s)*4^(j+2)* S(k1,s+1+j), j=0..ks2) for k >= s >= 1, and S(k,s) = 0 if k < s and S(k,0) = (1)^k*(2*k+1). Note that the recurrence derived from the Riordan Asequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k1,s1+j), j=0..ks), k >= s >=1.
(End)
From Tom Copeland, Nov 07 2015: (Start)
Rephrasing notes here: Append an initial column of zeros, except for a 1 at the top, to A111125 here. Then the partial sums of the columns of this modified entry are contained in A208513. Append an initial row of zeros to A208513. Then the difference of consecutive pairs of rows of the modified A208513 generates the modified A111125. Cf. A034807 and A127677.
For relations among the characteristic polynomials of Cartan matrices of the Coxeter root groups, Chebyshev polynomials, cyclotomic polynomials, and the polynomials of this entry, see Damianou (p. 20 and 21) and Damianou and Evripidou (p. 7).
As suggested by the equations on p. 7 of Damainou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = [A(2n+1,x)+2]/x = [F(2n+1,(2x),1,0,0,..)+2]/x = F(2n+1,x,2x,3x,..,(1)^n n*x)/x= F(2n+1,x,2x,3x,..,n*x)/x, where A(n,x) are the polynomials of A127677 and F(n,..) are the Faber polynomials of A263196. Cf. A127672 and A127677.
(End)


LINKS

Table of n, a(n) for n=0..62.
R. AndreJeannin, A generalization of MorganVoyce polynomials, The Fibonacci Quarterly 32.3 (1994): 22831.
K. Dilcher and K. B. Stolarsky, A Pascaltype triangle characterizing twin primes, Amer. Math. Monthly, 112 (2005), 673681.
P. Damianou , On the characteristic polynomials of Cartan matrices and Chebyshev polynomials, arXiv preprint arXiv:1110.6620 [math.RT], 2014.
P. Damianou and C. Evripidou, Characteristic and Coxeter polynomials for affine Lie algebras, arXiv preprint arXiv:1409.3956 [math.RT], 2014.
D. Jennings, Some Polynomial Identities for the Fibonacci and Lucas Numbers, Fib. Quart., 31(2) (1993), 134137.
D. E. Knuth, Johann Faulhaber and sums of powers, Math. Comp. 61 (1993), no. 203, 277294.
Yidong Sun, Numerical triangles and several classical sequences, Fib. Quart., Nov. 2005, pp. 359370.
T. Wang and W. Zhang, Some identities involving Fibonacci, Lucas polynomials and their applications, Bull. Math. Soc. Sci. Math. Roumanie, Tome 55(103), No.1, (2012) 95103.
Eric Weisstein's World of Mathematics, MorganVoyce polynomials


FORMULA

T(k,s) = ((2*k+1)/(2*s+1))*binomial(k+s,2*s), 0 <= s <= k.
From Peter Bala, Apr 30 2012: (Start)
T(n,k) = binomial(n+k,2*k) + 2*binomial(n+k,2*k+1).
The row generating polynomials P(n,x) are a generalization of the MorganVoyce polynomials b(n,x) and B(n,x). They satisfy the recurrence equation P(n,x) = (x+2)*P(n1,x)  P(n2,x) for n >= 2, with initial conditions P(0,x) = 1, P(1,x) = x+r+1 and with r = 2. The cases r = 0 and r = 1 give the MorganVoyce polynomials A085478 and A078812 respectively. AndreJeannin has considered the case of general r.
P(n,x) = U(n+1,1+x/2) + U(n,1+x/2), where U(n,x) denotes the Chebyshev polynomial of the second kind  see A053117. P(n,x) = 2/x*{T(2*n+2,u)T(2*n,u)), where u = sqrt((x+4)/4) and T(n,x) denotes the Chebyshev polynomial of the first kind  see A053120. P(n,x) = product {k = 1..n} ( x + 4*(sin(k*Pi/(2*n+1)))^2 ). P(n,x) = 1/x*(b(n+1,x)  b(n1,x)) and P(n,x) = 1/x*{(b(2*n+2,x)+1)/b(n+1,x)  (b(2*n,x)+1)/b(n,x)}, where b(n,x) := sum {k = 0..n} binomial(n+k,2*k)*x^k are the MorganVoyce polynomials of A085478. Cf. A211957.
(End)
From Wolfdieter Lang, Oct 18, 2012 (Start)
O.g.f. column No. s: ((1+x)/(1x)^2)*(x/(1x)^2)^s, s >= 0. (from the Riordan data given in a comment above).
O.g.f. of the row polynomials R(k,x):=sum(T(k,s)*x^s,s=0..k), k>=0: (1+z)/(1(2+x)*z+z^2) (from the Riordan property).
(End)
T(n,k) = 2*T(n1,k) + T(n1,k1)  T(n2,k), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n.  Philippe Deléham, Nov 12 2013


EXAMPLE

Triangle T(k,s) begins:
k\s 0 1 2 3 4 5 6 7 8 9 10
0: 1
1: 3 1
2: 5 5 1
3: 7 14 7 1
4: 9 30 27 9 1
5: 11 55 77 44 11 1
6: 13 91 182 156 65 13 1
7: 15 140 378 450 275 90 15 1
8: 17 204 714 1122 935 442 119 17 1
9: 19 285 1254 2508 2717 1729 665 152 19 1
10: 21 385 2079 5148 7007 5733 2940 952 189 21 1
... Extended and reformatted by Wolfdieter Lang, Oct 18 2012
Application for Fibonacci numbers F_{(2*k+1)*n}, row k=3:
F_{7*n} = 7*(1)^(3*n)*F_n + 14*(1)^(4*n)*5*F_n^3 + 7*(1)^(5*n)*5^2*F_n^5 + 1*(1)^(6*n)*5^3*F_n^7, n>=0.  Wolfdieter Lang, Aug 24 2012
Example for the Z and Asequence recurrences of this Riordan triangle: Z = A217477 = [3,4,12,40,...]; T(4,0) = 3*7 4*14 +12*7 40*1 = 9. A = [1, 2, 1, 2, 5, 14, ..]; T(5,2) = 1*30 + 2*27  1*9 + 2*1= 77. Wolfdieter Lang, Oct 18 2012
Example for the (4*(k+1))gon length ratio s(4*(k+1))(side/radius) as polynomial in the ratio rho(4*(k+1)) ((smallest diagonal)/side): k=0, s(4) = 1*rho(4) = sqrt(2); k=1, s(8) = 3*rho(8) + rho(8)^3 = sqrt(2sqrt(2)); k=2, s(12) = 5*rho(12)  5*rho(12)^3 + rho(12)^5, and C(12,x) = x^4  4*x^2 + 1, hence rho(12)^5 = 4*rho(12)^3  rho(12), and s(12) = 4*rho(12)  rho(12)^3 = sqrt(2  sqrt(3)).  Wolfdieter Lang, Oct 04 2013
Example for the recurrence for the signed triangle S(k,s)= ((1)^(ks))*T(k,s) (see the Aug 14 2014 comment above):
S(4,1) = 0 + (2*2  1)*S(3,1)  (1/2)*(3*4^2*S(3,2) + 4*4^3*S(3,3)) =  5*14  3*8*(7)  128*1 = 30. The recurrence from the Riordan Asequence A115141 is S(4,1) = 7 2*14 (7) 2*1 = 30.  Wolfdieter Lang, Aug 14 2014


MATHEMATICA

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n  1, x] + x*v[n  1, x];
v[n_, x_] := u[n  1, x] + (x + 1)*v[n  1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%] (* A208513 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%] (* A111125 *)
(* Clark Kimberling, Feb 28 2012 *)


PROG

(Sage)
@CachedFunction
def T(n, k):
if n< 0: return 0
if n==0: return 1 if k == 0 else 0
h = 3*T(n1, k) if n==1 else 2*T(n1, k)
return T(n1, k1)  T(n2, k)  h
A111125 = lambda n, k: (1)^(nk)*T(n, k)
for n in (0..9): [A111125(n, k) for k in (0..n)] # Peter Luschny, Nov 20 2012


CROSSREFS

Mirror image of A082985, which see for further references, etc.
Also closely related to triangles in A098599 and A100218.
A052227, A208513, A208508. A085478, A211957
Cf. A034807, A127672, A127677, A263196.
Sequence in context: A232632 A084533 A082985 * A209159 A182397 A209560
Adjacent sequences: A111122 A111123 A111124 * A111126 A111127 A111128


KEYWORD

nonn,tabl,easy,changed


AUTHOR

N. J. A. Sloane, Oct 16 2005


EXTENSIONS

More terms from Paul Barry, Oct 17 2005


STATUS

approved



