

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]
Another version of triangle in A103631.  Philippe Deléham, Jan 01 2009
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
From Gary W. Adamson, Aug 25 2019: (Start)
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)
From Gary W. Adamson, Sep 30 2019: (Start)
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)
From Gary W. Adamson, Aug 21 2019: (Start)
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)
Received from Herb Conn, Jan 2004: (Start)
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;
... apart from signs the same as A124645.  R. J. Mathar, Mar 12 2013


LINKS

Nathaniel Johnston, Rows 0..100, flattened
Henry W. Gould, A Variant of Pascal's Triangle , The Fibonacci Quarterly, Vol. 3, Nr. 4, Dec. 1965, p. 257271, with corrections.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonaccitype sequences, pp. 5564 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. See Table 4.
A. F. Horadam, R. P. Loh and A. G. Shannon, Divisibility properties of some Fibonaccitype sequences, pp. 5564 of Combinatorial Mathematics VI (Armidale 1978), Lect. Notes Math. 748, 1979. [Annotated scanned copy]
Jay Kappraff, Beyond Measure, A Guided Tour Through Nature, Myth and Number, World Scientific, 2002; p. 490.
Jay Kappraff, S. Jablan, G. W. Adamson, and R. Sazdanovich, Golden Fields, Generalized Fibonacci Sequences, and Chaotic Matrices, FORMA, Vol. 19, No. 4, 2005.
Jay Kappraff and Gary W. Adamson, Polygons and Chaos, Journal of Dynamical Systems and Geometric Theories, Vol 2 (2004), p 65.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, 2001 (Chapter 14)
E. Munarini and N. Z. Salvi, Binary strings without zigzags, Séminaire Lotharingien de Combinatoire, B49h (2004), 15 pp.
Wolfdieter Lang, "The field Q((2Cos Pi/N), its Galois group, and length
ratios in the regular ngon" https://arxiv.org/abs/1210.1018
P.C. Parks, A new proof of the RouthHurwitz stability criterion using the second method of Liapunov , Math. Proc. of the Cambridge Philosophical Society, Vol. 58, Issue 04 (1962) p. 694702.
P. Steinbach, Golden fields: a case for the heptagon, Math. Mag. 70 (1997), no. 1, 2231.
Index entries for triangles and arrays related to Pascal's triangle


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
Triangle is a reflection of that in A066170 (absolute values).  Gary W. Adamson, Feb 16 2004
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) = A108299(n, k)*A087960(k) = abs(A108299(n, k)).  Reinhard Zumkeller, Jun 01 2005
From Johannes W. Meijer, Aug 11 2011: (Start)
T(n,k) = A046854(n, nk) = abs(A066170(n, nk)).
T(n+k, nk) = A109223(n,k).
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)
For n > 1: T(n, k) = T(n2, k) + T(n1, k), 1 < k < n.  Reinhard Zumkeller, Apr 24 2013


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,

From Gary W. Adamson, Oct 23 2019: (Start)
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): binomial(nfloor((k+1)/2), floor(k/2)) end: seq(seq(A065941(n, k), k=0..n), n=0..15); # Johannes W. Meijer, Aug 11 2011
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);
# Peter Luschny, Mar 09 2020


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]
 Reinhard Zumkeller, May 07 2012
(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

Cf. A065942 (central stalk sequence), A000045 (row sums), A108299.
Reflected version of A046854.
Some triangle sums (see A180662): A000045 (Fi1), A016116 (Kn21), A000295 (Kn23), A094967 (Fi2), A000931 (Ca2), A001519 (Gi3), A000930 (Ze3).
Cf. A003558, A182579, A059841, A333143.
Sequence in context: A152157 A039961 A108299 * A123320 A054123 A119269
Adjacent sequences: A065938 A065939 A065940 * A065942 A065943 A065944


KEYWORD

nonn,tabl,easy


AUTHOR

Len Smiley, Nov 29 2001


STATUS

approved



