cp's OEIS Frontend

Row sums are A011782. Inverse is A065547.

Triangle T(n,k), 0 <= k <= n, read by rows, given by [0, 1, -1, 0, 0, 0, 0, 0, ...] DELTA [1, 0, 1, 0, 0, 0, 0, 0, ...] where DELTA is the operator defined in A084938. - Philippe Deléham, Jul 29 2006

Sum of entries in column k is A001519(k+1) (the odd-indexed Fibonacci numbers). - Philippe Deléham, Dec 02 2008

Number of permutations of length n avoiding simultaneously the patterns 123 and 132 with k left-to-right minima. A left-to-right minimum in a permutation a(1)a(2)...a(n) is position i such that a(j) > a(i) for all j < i. - Tian Han, Nov 16 2023

The A_n lattice consists of all vectors v = (x_1,...,x_(n+1)) in Z^(n+1) such that x_1 + ... + x_(n+1) = 0. The lattice is equipped with the norm ||v|| = 1/2*(|x_1| + ... + |x_(n+1)|). Pairs of lattice points (v,w) in the product lattice A_n x A_n have norm ||(v,w)|| = ||v|| + ||w||. Then the k-th term in the crystal ball sequence for the A_n x A_n lattice gives the number of such pairs (v,w) for which ||(v,w)|| is less than or equal to k.

This array has a remarkable relationship with Apery's constant zeta(3). The row (or column) and main diagonal entries of the array occur in series acceleration formulas for zeta(3). For row n entries there holds zeta(3) = (1+1/2^3+...+1/n^3) + Sum_{k >= 1} 1/(k^3*T(n,k-1)*T(n,k)). Also, as consequence of Apery's proof of the irrationality of zeta(3), we have a series acceleration formula along the main diagonal of the table: zeta(3) = 6 * sum {n >= 1} 1/(n^3*T(n-1,n-1)*T(n,n)). Apery's result appears to generalize to the other diagonals of the table. Calculation suggests the following result may hold: zeta(3) = 1 + 1/2^3 + ... + 1/k^3 + Sum_{n >= 1} (2*n+k)*(3*n^2 +3*n*k +k^2)/(n^3*(n+k)^3*T(n-1,n+k-1)*T(n,n+k)).

For the corresponding results for the constant zeta(2), related to the crystal ball sequences of the lattices A_n, see A108625. For corresponding results for log(2), coming from either the crystal ball sequences of the hypercubic lattices A_1 x ... x A_1 or the lattices of type C_n, see A008288 and A142992 respectively.

Multiplicative suborder of 2 (mod 2n+1) (or sord(2, 2n+1)).

This is called quasi-order of 2 mod b, with b = 2*n+1, for n >= 1, in the Hilton/Pederson reference.

For the complexity of computing this, see A002326.

Also, the order of the so-called "milk shuffle" of a deck of n cards, which maps cards (1,2,...,n) to (1,n,2,n-1,3,n-2,...). See the paper of Lévy. - Jeffrey Shallit, Jun 09 2019

It appears that under iteration of the base-n Kaprekar map, for even n > 2 (A165012, A165051, A165090, A151949 in bases 4, 6, 8, 10), almost all cycles are of length a(n/2 - 1); proved under the additional constraint that the cycle contains at least one element satisfying "number of digits (n-1) - number of digits 0 = o(total number of digits)". - Joseph Myers, Sep 05 2009

From Gary W. Adamson, Sep 20 2011: (Start)

a(n) can be determined by the cycle lengths of iterates using x^2 - 2, seed 2*cos(2*Pi/N); as shown in the A065941 comment of Sep 06 2011. The iterative map of the logistic equation 4x*(1-x) is likewise chaotic with the same cycle lengths but initiating the trajectory with sin^2(2*Pi/N), N = 2n+1 [Kappraff & Adamson, 2004]. Chaotic terms with the identical cycle lengths can be obtained by applying Newton's method to i = sqrt(-1) [Strang, also Kappraff and Adamson, 2003], resulting in the morphism for the cot(2*Pi/N) trajectory: (x^2-1)/2x. (End)

From Gary W. Adamson, Sep 11 2019: (Start)

Using x^2 - 2 with seed 2*cos(Pi/7), we obtain the period-three trajectory 1.8019377...-> 1.24697...-> -0.445041... For an odd prime N, the trajectory terms represent diagonal lengths of regular star 2N-gons, with edge the shortest value (0.445... in this case.) (Cf. "Polygons and Chaos", p. 9, Fig 4.) We can normalize such lengths by dividing through with the lowest value, giving 3 diagonals of the 14-gon: (1, 2.801937..., 4.048917...). Label the terms ranked in magnitude with odd integers (1, 3, 5), and we find that the diagonal lengths are in agreement with the diagonal formula (sin(j*Pi)/14)/(sin(Pi/14)), with j = (1,3,5). (End)

Roots of signed n-th row A054142 polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths a(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697...); we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...); the roots to the polynomial with cycle length 3 matching a(3) = 3. - Gary W. Adamson, Sep 21 2011

From Juhani Heino, Oct 26 2015: (Start)

Start a sequence with numbers 1 and n. For next numbers, add previous numbers going backwards until the sum is even. Then the new number is sum/2. I conjecture that the sequence returns to 1,n and a(n) is the cycle length.

For example:

1,7,4,2,1,7,... so a(7) = 4.

1,6,3,5,4,2,1,6,... so a(6) = 6. (End)

From Juhani Heino, Nov 06 2015: (Start)

Proof of the above conjecture: Let n = -1/2; thus 2n + 1 = 0, so operations are performed mod (2n + 1). When the member is even, it is divided by 2. When it is odd, multiply by n, so effectively divide by -2. This is all well-defined in the sense that new members m are 1 <= m <= n. Now see what happens starting from an odd member m. The next member is -m/2. As long as there are even members, divide by 2 and end up with an odd -m/(2^k). Now add all the members starting with m. The sum is m/(2^k). It's divided by 2, so the next member is m/(2^(k+1)). That is the same as (-m/(2^k))/(-2), as with the definition.

So actually start from 1 and always divide by 2, although the sign sometimes changes. Eventually 1 is reached again. The chain can be traversed backwards and then 2^(cycle length) == +-1 (mod 2n + 1).

To conclude, we take care of a(0): sequence 1,0 continues with zeros and never returns to 1. So let us declare that cycle length 0 means unavailable. (End)

From Gary W. Adamson, Aug 20 2019: (Start)

Terms in the sequence can be obtained by applying the doubling sequence mod (2n + 1), then counting the terms until the next term is == +1 (mod 2n + 1). Example: given 25, the trajectory is (1, 2, 4, 8, 16, 7, 14, 3, 6, 12).

The cycle ends since the next term is 24 == -1 (mod 25) and has a period of 10. (End)

From Gary W. Adamson, Sep 04 2019: (Start)

Conjecture of Kappraff and Adamson in "Polygons and Chaos", p. 13 Section 7, "Chaos and Number": Given the cycle length for N = 2n + 1, the same cycle length is present in bases 4, 9, 16, 25, ..., m^2, for the expansion of 1/N.

Examples: The cycle length for 7 is 3, likewise for 1/7 in base 4: 0.021021021.... In base 9 the expansion of 1/7 is 0.125125125... Check: The first few terms are 1/9 + 2/81 + 5/729 = 104/279 = 0.1426611... (close to 1/7 = 0.142857...). (End)

From Gary W. Adamson, Sep 24 2019: (Start)

An exception to the rule for 1/N in bases m^2: (when N divides m^2 as in 1/7 in base 49, = 7/49, rational). When all terms in the cycle are the same, the identity reduces to 1/N in (some bases) = .a, a, a, .... The minimal values of "a" for 1/N are provided as examples, with the generalization 1/N in base (N-1)^2 = .a, a, a, ... for N odd:

1/3 in base 4 = .1, 1, 1, ...

1/5 in base 16 = .3, 3, 3, ...

1/7 in base 36 = .5, 5, 5, ...

1/9 in base 64 = .7, 7, 7, ...

1/11 in base 100 = .9, 9, 9, ... (Check: the first three terms are 9/100 +9/(100^2) + 9/(100^3) = 0.090909 where 1/11 = 0.09090909...). (End)

For N = 2n+1, the corresponding entry is equal to the degree of the polynomial for N shown in (Lang, Table 2, p. 46). As shown, x^3 - 3x - 1 is the minimal polynomial for N = 9, with roots (1.87938..., -1.53208..., 0.347296...); matching the (abs) values of the 2*cos(Pi/9) trajectory using x^2 - 2. Thus, a(4) = 3. If N is prime, the polynomials shown in Table 2 are the same as those for the same N in A065941. If different, the minimal polynomials shown in Table 2 are factors of those in A065941. - Gary W. Adamson, Oct 01 2019

The terms in the 2*cos(Pi/N) trajectory (roots to the minimal polynomials in A187360 and (Lang)), are quickly obtained from the doubling trajectory (mod N) by using the operation L(m) 2*cos(x)--> 2*cos(m*x), where L(2), the second degree Lucas polynomial (A034807) is x^2 - 2. Relating to the heptagon and using seed 2*cos(Pi/7), we obtain the trajectory 1.8019..., 1.24697..., and 0.445041....; cyclic with period 3. All such roots can be derived from the N-th roots of Unity and can be mapped on the Vesica Piscis. Given the roots of Unity (Polar 1Angle(k*2*Pi/N), k = 1, 2, ..., (N-1)/2) the Vesica Piscis maps these points on the left (L) circle to the (R) circle by adding 1A(0) or (a + b*I) = (1 + 0i). But this operation is the same as vector addition in which the resultant vector is 1 + 1A(k*(2*Pi/N)). Example: given the radius at 2*Pi/7 on the left circle, this maps to (1 + 1A(2*Pi/7)) on the right circle; or 1A(2*Pi/7) --> (1.8019377...A(Pi/7). Similarly, 1 + 1A((2)*2*Pi/7)) maps to (1.24697...A (2*Pi/7); and 1 + 1A(3*2*Pi/7) maps to (.0445041...A(3*Pi/7). - Gary W. Adamson, Oct 23 2019

From Gary W. Adamson, Dec 01 2021: (Start)

As to segregating the two sets: (A014659 terms are those N = (2*n+1), N divides (2^m - 1), and (A014657 terms are those N that divide (2^m + 1)); it appears that the following criteria apply: Given IcoS(N, 1) (cf. Lang link "On the Equivalence...", p. 16, Definition 20), if the number of odd terms is odd, then N belongs to A014659, otherwise A014657. In IcoS(11, 1): (1, 2, 4, 3, 5), three odd terms indicate that 11 is a term in A014657. IcoS(15, 1) has the orbit (1, 2, 4, 7) with two odd terms indicating that 15 is a term in A014659.

It appears that if sin(2^m * Pi/N) has a negative sign, then N is in A014659; otherwise N is in A014657. With N = 15, m is 4 and sin(16 * Pi/15) is -0.2079116... If N is 11, m is 5 and sin(32 * Pi/11) is 0.2817325. (End)

On the iterative map using x^2 - 2, (Devaney, p. 126) states that we must find the function that takes 2*cos(Pi) -> 2*cos(2*Pi). "However, we may write 2*cos(2*Pi) = 2*(2*cos^2(Pi) - 1) = (2*cos(Pi))^2 - 2. So the required function is x^2 - 2." On the period 3 implies chaos theorem of James Yorke and T.Y. Li, proved in 1975; Devaney (p. 133) states that if F is continuous and we find a cycle of period 3, there are infinitely many other cycles for this map with every possible period. Check: The x^2 - 2 orbit for 7 has a period of 3, so this entry has periodic points of all other periods. - Gary W. Adamson, Jan 04 2023

It appears that a(n) is the length of the cycle starting at 2/(2*n+1) for the map x->1 - abs(2*x-1). - Michel Marcus, Jul 16 2025

The row polynomials with exponents in increasing order (e.g., third row: 1+3x+3x^2) are Grosswald's y_{n}(x) polynomials, p. 18, Eq. (7).

Also called Bessel numbers of first kind.

The triangle a(n,k) has factorization [C(n,k)][C(k,n-k)]Diag((2n-1)!!) The triangle a(n-k,k) is A100861, which gives coefficients of scaled Hermite polynomials. - Paul Barry, May 21 2005

Related to k-matchings of the complete graph K_n by a(n,k)=A100861(n+k,k). Related to the Morgan-Voyce polynomials by a(n,k)=(2k-1)!!*A085478(n,k). - Paul Barry, Aug 17 2005

Related to Hermite polynomials by a(n,k)=(-1)^k*A060821(n+k, n-k)/2^n. - Paul Barry, Aug 28 2005

The row polynomials, the Bessel polynomials y(n,x):=Sum_{m=0..n} (a(n,m)*x^m) (called y_{n}(x) in the Grosswald reference) satisfy (x^2)*(d^2/dx^2)y(n,x) + 2*(x+1)*(d/dx)y(n,x) - n*(n+1)*y(n,x) = 0.

a(n-1, m-1), n >= m >= 1, enumerates unordered n-vertex forests composed of m plane (aka ordered) increasing (rooted) trees. Proof from the e.g.f. of the first column Y(z):=1-sqrt(1-2*z) (offset 1) and the Bergeron et al. eq. (8) Y'(z)= phi(Y(z)), Y(0)=0, with out-degree o.g.f. phi(w)=1/(1-w). See their remark on p. 28 on plane recursive trees. For m=1 see the D. Callan comment on A001147 from Oct 26 2006. - Wolfdieter Lang, Sep 14 2007

The asymptotic expansions of the higher order exponential integrals E(x,m,n), see A163931 for information, lead to the Bessel numbers of the first kind in an intriguing way. For the first four values of m these asymptotic expansions lead to the triangles A130534 (m=1), A028421 (m=2), A163932 (m=3) and A163934 (m=4). The o.g.f.s. of the right hand columns of these triangles in their turn lead to the triangles A163936 (m=1), A163937 (m=2), A163938 (m=3) and A163939 (m=4). The row sums of these four triangles lead to A001147, A001147 (minus a(0)), A001879 and A000457 which are the first four right hand columns of A001498. We checked this phenomenon for a few more values of m and found that this pattern persists: m = 5 leads to A001880, m=6 to A001881, m=7 to A038121 and m=8 to A130563 which are the next four right hand columns of A001498. So one by one all columns of the triangle of coefficients of Bessel polynomials appear. - Johannes W. Meijer, Oct 07 2009

a(n,k) also appear as coefficients of (n+1)st degree of the differential operator D:=1/t d/dt, namely D^{n+1}= Sum_{k=0..n} a(n,k) (-1)^{n-k} t^{1-(n+k)} (d^{n+1-k}/dt^{n+1-k}. - Leonid Bedratyuk, Aug 06 2010

a(n-1,k) are the coefficients when expanding (xI)^n in terms of powers of I. Let I(f)(x) := Integral_{a..x} f(t) dt, and (xI)^n := x Integral_{a..x} [ x_{n-1} Integral_{a..x_{n-1}} [ x_{n-2} Integral_{a..x_{n-2}} ... [ x_1 Integral_{a..x_1} f(t) dt ] dx_1 ] .. dx_{n-2} ] dx_{n-1}. Then: (xI)^n = Sum_{k=0..n-1} (-1)^k * a(n-1,k) * x^(n-k) * I^(n+k)(f)(x) where I^(n) denotes iterated integration. - Abdelhay Benmoussa, Apr 11 2025

Warning: formulas and programs sometimes refer to offset 0 and sometimes to offset 1.

Apart from signs, identical to A053122.

Coefficient array for Morgan-Voyce polynomial B(n,x); see A085478 for references. - Philippe Deléham, Feb 16 2004

T(n,k) is the number of compositions of n having k parts when there are q kinds of part q (q=1,2,...). Example: T(4,2) = 10 because we have (1,3),(1,3'),(1,3"), (3,1),(3',1),(3",1),(2,2),(2,2'),(2',2) and (2',2'). - Emeric Deutsch, Apr 09 2005

T(n, k) is also the number of idempotent order-preserving full transformations (of an n-chain) of height k (height(alpha) = |Im(alpha)|). - Abdullahi Umar, Oct 02 2008

This sequence is jointly generated with A085478 as a triangular array of coefficients of polynomials v(n,x): initially, u(1,x) = v(1,x) = 1; for n > 1, u(n,x) = u(n-1,x) + x*v(n-1)x and v(n,x) = u(n-1,x) + (x+1)*v(n-1,x). See the Mathematica section. - Clark Kimberling, Feb 25 2012

Concerning Kimberling's recursion relations, see A102426. - Tom Copeland, Jan 19 2016

Subtriangle of the triangle T(n,k), 0 <= k <= n, read by rows, given by (0, 2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 27 2012

From Wolfdieter Lang, Aug 30 2012: (Start)

With offset [0,0] the triangle with entries R(n,k) = T(n+1,k+1):= binomial(n+k+1, 2*k+1), n >= k >= 0, and zero otherwise, becomes the Riordan lower triangular convolution matrix R = (G(x)/x, G(x)) with G(x):=x/(1-x)^2 (o.g.f. of A000027). This means that the o.g.f. of column number k of R is (G(x)^(k+1))/x. This matrix R is the inverse of the signed Riordan lower triangular matrix A039598, called in a comment there S.

The Riordan matrix with entries R(n,k), just defined, provides the transition matrix between the sequence entry F(4*m*(n+1))/L(2*l), with m >= 0, for n=0,1,... and the sequence entries 5^k*F(2*m)^(2*k+1) for k = 0,1,...,n, with F=A000045 (Fibonacci) and L=A000032 (Lucas). Proof: from the inverse of the signed triangle Riordan matrix S used in a comment on A039598.

For the transition matrix R (T with offset [0,0]) defined above, row n=2: F(12*m) /L(2*m) = 3*5^0*F(2*m)^1 + 4*5^1*F(2*m)^3 + 1*5^2*F(2*m)^5, m >= 0. (End)

From R. Bagula's comment in A053122 (cf. Damianou link p. 10), this array gives the coefficients (mod sign) of the characteristic polynomials for the Cartan matrix of the root system A_n. - Tom Copeland, Oct 11 2014

For 1 <= k <= n, T(n,k) equals the number of (n-1)-length ternary words containing k-1 letters equal 2 and avoiding 01. - Milan Janjic, Dec 20 2016

The infinite sum (Sum_{i >= 0} (T(s+i,1+i) / 2^(s+2*i)) * zeta(s+1+2*i)) = 1 allows any zeta(s+1) to be expressed as a sum of rational multiples of zeta(s+1+2*i) having higher arguments. For example, zeta(3) can be expressed as a sum involving zeta(5), zeta(7), etc. The summation for each s >= 1 uses the s-th diagonal of the triangle. - Robert B Fowler, Feb 23 2022

The convolution triangle of the nonnegative integers. - Peter Luschny, Oct 07 2022

Riordan array ((1+x)/(1-x)^2, x/(1-x)^2). Row sums are A002878. Diagonal sums are A003945. Inverse is A113187. An interesting factorization is (1/(1-x), x/(1-x))(1+2*x, 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(k-1,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(n-1,x) +x*(x+1)*v(n-1) and v(n,x)= u(n-1,x) +x*v(n-1,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 F-numbers. 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_{s=0..k} ( T(k,s)*(-1)^((k+s)*n)*5^s*F_{n}^(2*s+1) ), 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_{s=0..k} ( T(k,s)*(x-x^(-1))^(2*s+1) ), 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 Z-sequence of this Riordan array is A217477, and the A-sequence is (-1)^n*A115141(n). For the notion of A- and Z-sequences for Riordan triangles see a W. Lang link under A006232. (End)

The signed triangle ((-1)^(k-s))*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 T-polynomials whose coefficients are given in A127672 (C is there called R). See the odd numbered rows there. This signed triangle is the Riordan array ((1-x)/(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)^(k-s))*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=0..k} ( S(k,s)*rho(4*(k+1))^(2*s+1) ).

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 n-gon, 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_{s=0..k} ( (-1)^(k-s)*T(k,s)*x^s ) = S(k, x-2) - S(k-1, x-2), k >= 0, with the Chebyshev S-polynomials (see their coefficient triangle (A049310) and S(-1, x) = 0), satisfy the recurrence todd(k, x) = (-1)^(k-1)*((x-4)/2)*todd(k-1, 4-x) + ((x-2)/2)*todd(k-1, 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)^(k-s))*S(k-1,s-1) + (2*(s+1)*(-1)^(k-s) - 1)*S(k-1,s) + (1/2)*(-1)^(k-s)*Sum_{j=0..k-s-2} ( binomial(j+s+2,s)*4^(j+2)* S(k-1, s+1+j) ) 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 A-sequence A115141 is similar but has simpler coefficients: S(k,s) = sum(A115141(j)*S(k-1,s-1+j), j=0..k-s), 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 Damianou and Evripidou, the signed row polynomials of this entry are given by (p(n,x))^2 = (A(2*n+1, x) + 2)/x = (F(2*n+1, (2-x), 1, 0, 0, ... ) + 2)/x = F(2*n+1, -x, 2*x, -3*x, ..., (-1)^n n*x)/x = -F(2*n+1, x, 2*x, 3*x, ..., 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)

The row polynomials P(k, x) of the signed triangle S(k, s) = ((-1)^(k-s))*T(k, s) are given from the row polynomials R(2*k+1, x) of triangle A127672 by

P(k, x) = R(2*k+1, sqrt(x))/sqrt(x). - Wolfdieter Lang, May 02 2021

Row sums: A006318 (Schroeder numbers). Essentially same as triangle A060693 transposed.

T(n,k) is number of Schroeder paths (i.e., consisting of steps U=(1,1), D=(1,-1), H=(2,0) and never going below the x-axis) from (0,0) to (2n,0), having k U's. E.g., T(2,1)=3 because we have UHD, UDH and HUD. - Emeric Deutsch, Dec 06 2003

Little Schroeder numbers A001003 have a(n) = Sum_{k=0..n} A088617(n,k)*(-1)^(n-k)*2^k. - Paul Barry, May 24 2005

Conjecture: The expected number of U's in a Schroeder n-path is asymptotically Sqrt[1/2]*n for large n. - David Callan, Jul 25 2008

T(n, k) is also the number of order-preserving and order-decreasing partial transformations (of an n-chain) of width k (width(alpha) = |Dom(alpha)|). - Abdullahi Umar, Oct 02 2008

The antidiagonals of this lower triangular matrix are the rows of A055151. - Tom Copeland, Jun 17 2015

Row sums are odd-indexed Fibonacci numbers.

T(n,k) is the number of nondecreasing Dyck paths of semilength n+1, having k double rises. Mirror image of A085478. - Emeric Deutsch, May 31 2004

Diagonal sums are A052535. - Paul Barry, Jan 21 2005

Matrix inverse is the triangle of Salie numbers A098435. - Paul Barry, Jan 21 2005

Coefficients of Morgan-Voyce polynomial b(n,x); e.g., b(3,x)=x^3+5x^2+6x+1. See A172431 for coefficients of Morgan-Voyce polynomial B(n,x). - Clark Kimberling, Feb 13 2010

T(n,k) is the number of stack polyominoes of perimeter 2n+4 with k+1 columns. - Emanuele Munarini, Apr 07 2011

Roots of signed n-th polynomials are chaotic with respect to the operation (-2, x^2), with cycle lengths A003558(n). Example: starting with a root to x^3 - 5x^2 + 6x - 1 = 0; (2 + 2*cos(2*Pi/N) = 3.24697... = A116415; we obtain the trajectory (3.24697...-> 1.55495...-> 0.198062...; the 3 roots to the polynomial with cycle length 3 matching A003558(3) = 3. The operation (-2, x^2) is the reversal of the well known chaotic operation (x^2 - 2) [Kappraff, Adamson, 2004] starting with seed 2*cos(2*Pi/N). Check: given 2*cos(2*Pi/7) = 1.24697..., we obtain the 3-cycle using (x^2 - 2): (1.24697...-> -0.445041...-> 1.801937...; where the terms in either set are intermediate terms in the other, irrespective of sign. - Gary W. Adamson, Sep 22 2011

A054142 is jointly generated with A172431 as an array of coefficients of polynomials u(n,x): initially, u(1,x)=v(1,x)=1; for n>1, u(n,x)=x*u(n-1,x)+v(n-1,x) and v(n,x)=x*u(n-1,x)+(x+1)*v(n-1,x). See the Mathematica section of A172431. - Clark Kimberling, Mar 09 2012

Subtriangle of the triangle given by (0, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Apr 01 2012

The o.g.f. for row n of the array A(n, k) = binomial(2*n-k,k), k >= 0, n >= 0 is G(n,x) = Sum_{k=0..n} T(n, k)*x^k + (-x)^(2*n+1) * c(-x)^(2*n+1) / sqrt(1-4*(-x)), for n >= 0. Here c(x) is the o.g.f. of A000108 (Catalan). For powers of c(x) see the W. Lang link in A115139. For the alternating sign case replace x by -x. - Wolfdieter Lang, Sep 12 2016

Multiplying the n-th diagonal by A001147(n) generates A001497. - Tom Copeland, Oct 04 2016

Row sums: A000045(n+1), Fibonacci numbers.

A168561*A007318 = A037027, as lower triangular matrices. Diagonal sums : A077957. - Philippe Deléham, Dec 02 2009

T(n,k) is the number of compositions of n+1 into k+1 odd parts. Example: T(4,2)=3 because we have 5 = 1+1+3 = 1+3+1 = 3+1+1.

Coefficients of monic Fibonacci polynomials (rising powers of x). Ftilde(n, x) = x*Ftilde(n-1, x) + Ftilde(n-2, x), n >=0, Ftilde(-1,x) = 0, Ftilde(0, x) = 1. G.f.: 1/(1 - x*z - z^2). Compare with Chebyshev S-polynomials (A049310). - Wolfdieter Lang, Jul 29 2014

The row polynomials Pe(n, x) := Sum_{m=0..n} a(n, m)*x^m appear as numerators of the generating functions for the even numbered column sequences of array A034870.

Elements have the same parity as those of Pascal's triangle.

All zeros of polynomial Pe(n, x) are negative. They are -tan^2(Pi/2*n+1), -tan^2(2*Pi/2*n+1), ..., -tan^2(n*Pi/2*n+1). Moreover, for m >= 1, Pe(m, -x^2) is the characteristic polynomial of the linear difference equation with constant coefficients for differences between multiples of 2*m+1 with even and odd digit sum in base 2*m in the interval [0,(2*m)^n). - Vladimir Shevelev and Peter J. C. Moses, May 22 2012

Row reverse of A103327. - Peter Bala, Jul 29 2013

The row polynomial Pe(d, x), multiplied by (2*d)!/d! = A001813(d), gives the numerator polynomial of the o.g.f. of the sequence of the diagonal d, for d >= 0, of the Sheffer triangle Lah[4,1] given in A048854. - Wolfdieter Lang, Oct 12 2017