

A065941


T(n,k) = binomial(nfloor((k+1)/2), floor(k/2)). Triangle read by rows, for 0 <= k <= n.


71



1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 4, 3, 3, 1, 1, 1, 5, 4, 6, 3, 1, 1, 1, 6, 5, 10, 6, 4, 1, 1, 1, 7, 6, 15, 10, 10, 4, 1, 1, 1, 8, 7, 21, 15, 20, 10, 5, 1, 1, 1, 9, 8, 28, 21, 35, 20, 15, 5, 1, 1, 1, 10, 9, 36, 28, 56, 35, 35, 15, 6, 1, 1, 1, 11, 10, 45, 36, 84, 56, 70, 35, 21, 6, 1
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OFFSET

0,9


COMMENTS

Also the qStirling2 numbers at q = 1.  Peter Luschny, Mar 09 2020
Row sums give the Fibonacci sequence. So do the alternating row sums.
Triangle of coefficients of polynomials defined by p(1,x) = p(0,x) = 1, p(n, x) = x*p(n1, x) + p(n2, x), for n >= 1.  Benoit Cloitre, May 08 2005 [rewritten with correct offset.  Wolfdieter Lang, Feb 18 2020]
The T(n,k) coefficients appear in appendix 2 of Parks's remarkable article "A new proof of the RouthHurwitz stability criterion using the second method of Liapunov" if we assume that the b(n) coefficients are all equal to 1 and ignore the first column. The complete version of this triangle including the first column is A103631.  Johannes W. Meijer, Aug 11 2011
Signed ++++..., the roots are chaotic using f(x) > x^2  2 with cycle lengths shown in A003558 by nth rows. Example: given row 3, x^3 + x^2  2x  1; the roots are (a = 1.24697, ...; b = 0.445041, ...; c = 1.802937, ...). Then (say using seed b with x^2  2) we obtain the trajectory 0.445041, ... > 1.80193, ... > 1.24697, ...; matching the entry "3" in A003558(3).  Gary W. Adamson, Sep 06 2011
Roots to the polynomials and terms in A003558 can all be obtained from the numbers below using a doubling series mod N procedure as follows: (more than one row may result). Any row ends when the trajectory produces a term already used. Then try the next higher odd term not used as the leftmost term, then repeat.
For example, for N = 11, we get: (1, 2, 4, 3, 5), showing that when confronted with two choices after the 4: (8 and 3), pick the smaller (abs) term, = 3. Then for the next row pick 7 (not used) and repeat the algorithm; succeeding only if the trajectory produces new terms. But 7 is also (4) mod 11 and 4 was used. Therefore what I call the "rt table" (for roots trajectory) has only one row: (1, 2, 4, 3, 5). Conjecture: The numbers of terms in the first row is equal to A003558 corresponding to N, i.e., 5 in this case with period 2.
Now for the roots to the polynomials. Pick N = 7. The polynomial is x^3  x^2  2x + 1 = 0, with roots 1.8019..., 1.2469... and 0.445... corresponding to 2*cos(j*Pi/N), N = 7, and j = (1, 2, and 3). The terms (1, 2, 3) are the rt terms for N = 7. For 11, the rt terms are (1, 2, 4, 3, 5). This implies that given any roots of the corresponding polynomial, they are cyclic using f(x) > x^2  2 with cycle lengths shown in A003558. The terms thus generated are 2*cos(j*Pi), with j = (1, 2, 4, 3, 5). Check: Begin with 2*j*Pi/N, with j = 1 (1.9189...). The other trajectory terms are: > 1.6825..., > 0.83083..., 1.3097...; 545...; (a 5 period and cyclic since we can begin with any of the constants). The rt table for odd N begins as follows:
3...............1
5...............1, 2
7...............1, 2, 3
9...............1, 2, 4
...............3 (singleton terms reduce to "1") (9 has two rows)
11...............1, 2, 4, 3, 5
13...............1, 2, 4, 5, 3, 6
15...............1, 2, 4, 7
................3, 6 (dividing through by the gcd gives (1, 2))
................5. (singleton terms reduce to "1")
The result is that 15 has 3 factors (since 3 rows), and the values of those factors are the previous terms "N", corresponding to the rt terms in each row. Thus, the first row is new, the second (1, 2), corresponds to N = 5, and the "1" in row 3 corresponds to N = 3. The factors are those values apart from 15 and 1. Note that all of the unreduced rt terms in all rows for N form a complete set of the terms 1 through (N1)/2 without duplication. (End)
The 3 factors of the 7th degree polynomial for 15: (x^7  x^6  6x^5 + 5x^4 + 10x^3  6x^2  4x + 1) can be determined by getting the roots for 2*cos(j*Pi/1), j = (1, 2, 4, 7) and finding the corresponding polynomial, which is x^4 + x^3  4x^2  4x + 1. This is the minimal polynomial for N = 15 as shown in Table 2, p. 46 of (Lang). The degree of this polynomial is 4, corresponding to the entry in A003558 for 15, = 4. The trajectories (3, 6) and (5) are j values for 2*cos(j*Pi/15) which are roots to x^2  x  1 (relating to the pentagon), and (x  1), relating to the triangle. (End)
Matrices M of the form: (1's in the main diagonal, 1's in the subdiagonal, and the rest zeros) are chaotic if we replace (f(x) > x^2  2) with f(x) > M^2  2I, where I is the Identity matrix (1, 0, 0; 0, 1, 0; 0, 0, 1]. For example, with the 3 X 3 matrix M: [0, 0, 1; 0, 1, 1; 1, 1, 0]; the f(x) trajectory is:
....M^2  2I: [1, 1, 0; 1, 0, 1; 0, 1, 0], then for the latter,
....M^2  2I: [0, 1, 1; 1, 0, 0; 1, 0, 1]. The cycle ends with period 3 since the next matrix is (1) * the seed matrix. As in the case with f(x) > x^2  2, the eigenvalues of the 3 chaotic matrices are (abs) 1.24697, 0.44504... and 1.80193, ... Also, the characteristic equations of the 3 matrices are the same as or variants of row 4 of the triangle below: (x^3 + x  2x  1) with different signs. (End)
Let x = 2*cos(2A) (A = Angle); then
sin(A)/sin A = 1
sin(3A)/sin A = x + 1
sin(5A)/sin A = x^2 + x  1
sin(7A)/sin A = x^3 + x  2x  1
sin(9A)/sin A = x^4 + x^3  3x^2  2x + 1
... (signed ++++...). (End)
Or Pascal's triangle (A007318) with duplicated diagonals. Also triangle of coefficients of polynomials defined by P_0(x) = 1 and for n>=1, P_n(x) = F_n(x) + F_(n+1)(x), where F_n(x) is Fibonacci polynomial (cf. A049310): F_n(x) = Sum_{i=0..floor((n1)/2)} C(ni1,i)*x^(n2*i1).  Vladimir Shevelev, Apr 12 2012
The matrix inverse is given by
1;
1, 1;
0, 1, 1;
0, 1, 2, 1;
0, 0, 1, 2, 1;
0, 0, 1, 3, 3, 1;
0, 0, 0, 1, 3, 3, 1;
0, 0, 0, 1, 4, 6, 4, 1;
0, 0, 0, 0, 1, 4, 6, 4, 1;


LINKS

Jay Kappraff, S. Jablan, G. W. Adamson, and R. Sazdanovich, Golden Fields, Generalized Fibonacci Sequences, and Chaotic Matrices, FORMA, Vol. 19, No. 4, 2005.


FORMULA

T(n, k) = binomial(nfloor((k+1)/2), floor(k/2)).
As a square array read by antidiagonals, this is given by T1(n, k) = binomial(floor(n/2) + k, k).  Paul Barry, Mar 11 2003
Recurrences: T(k, 0) = 1, T(k, n) = T(k1, n) + T(k2, n2), or T(k, n) = T(k1, n) + T(k1, n1) if n even, T(k1, n1) if n odd.  Ralf Stephan, May 17 2004
G.f.: sum[n, sum[k, T(k, n)x^ky^n]] = (1+xy)/(1yx^2y^2). sum[n>=0, T(k, n)y^n] = y^k/(1y)^[k/2].  Ralf Stephan, May 17 2004
T(n, k) = sum(T(j, k2), j=k2..n2), 2 <= k <= n, n>=2;
T(n, 0) =1, T(n+1, 1) = 1, n >= 0. (End)


EXAMPLE

Triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 ...

[0] 1,
[1] 1, 1,
[2] 1, 1, 1,
[3] 1, 1, 2, 1,
[4] 1, 1, 3, 2, 1,
[5] 1, 1, 4, 3, 3, 1,
[6] 1, 1, 5, 4, 6, 3, 1,
[7] 1, 1, 6, 5, 10, 6, 4, 1,
[8] 1, 1, 7, 6, 15, 10, 10, 4, 1,
[9] 1, 1, 8, 7, 21, 15, 20, 10, 5, 1,

Consider the roots of the polynomials corresponding to odd N such that for N=7 the polynomial is (x^3 + x^2  2x  1) and the roots (a, b, c) are (1.8019377..., 1.247697..., and 0.445041...). The discriminant of a polynomial derived from the roots is the square of the product of successive differences: ((ab), (bc), (ca))^2 in this case, resulting in 49, matching the method derived from the coefficients of a cubic. For our purposes we use the product of the differences, not the square, resulting in (3.048...) * (1.69202...) * (1.35689...) = 7.0. Conjecture: for all polynomials in the set, the product of the differences of the roots = the corresponding N. For N = 7, we get x^3  7x + 7. It appears that for all prime N's, these resulting companion polynomials are monic (left coefficient is 1), and all other coefficients are N or multiples thereof, with the rightmost term = N. The companion polynomials for the first few primes are:
N = 5: x^2  5;
N = 7: x^3  7x + 7;
N = 11: x^5  11x^3 + 11x^2 + 11x  11;
N = 13: x^6  13x^4 + 13x^3 + 26x^2  39x + 13;
N = 17: x^8  17x^6 + 17x^5 + 68x^4  119x^3 + 17x^2 + 51x  17;
N = 19: x^9  19x^7 + 19x^6 + 95x^5  171x^4  19x^3 + 190x^2  114x + 19. (End)


MAPLE

A065941 := proc(n, k) option remember: local j: if k=0 then 1 elif k=1 then 1: elif k>=2 then add(procname(j, k2), j=k2..n2) fi: end: seq(seq(A065941(n, k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
# The function qStirling2 is defined in A333143.
seq(print(seq(qStirling2(n, k, 1), k=0..n)), n=0..9);


MATHEMATICA

Flatten[Table[Binomial[nFloor[(k+1)/2], Floor[k/2]], {n, 0, 15}, {k, 0, n}]] (* Harvey P. Dale, Dec 11 2011 *)


PROG

(Haskell)
a065941 n k = a065941_tabl !! n !! k
a065941_row n = a065941_tabl !! n
a065941_tabl = iterate (\row >
zipWith (+) ([0] ++ row) (zipWith (*) (row ++ [0]) a059841_list)) [1]
(PARI) T065941(n, k) = binomial(n(k+1)\2, k\2); \\ Michel Marcus, Apr 28 2014
(Magma) [Binomial(n  Floor((k+1)/2), Floor(k/2)): k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 10 2019
(Sage) [[binomial(n  floor((k+1)/2), floor(k/2)) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Jul 10 2019


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



