

A011973


Irregular triangle of numbers read by rows: {binomial(nk, k), n >= 0, 0 <= k <= floor(n/2)}; or, triangle of coefficients of (one version of) Fibonacci polynomials.


81



1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 4, 3, 1, 5, 6, 1, 1, 6, 10, 4, 1, 7, 15, 10, 1, 1, 8, 21, 20, 5, 1, 9, 28, 35, 15, 1, 1, 10, 36, 56, 35, 6, 1, 11, 45, 84, 70, 21, 1, 1, 12, 55, 120, 126, 56, 7, 1, 13, 66, 165, 210, 126, 28, 1, 1, 14, 78, 220, 330, 252, 84, 8, 1, 15, 91, 286, 495, 462
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OFFSET

0,6


COMMENTS

T(n,k) is the number of subsets of {1,2,...,n1} of size k and containing no consecutive integers. Example: T(6,2)=6 because the subsets of size 2 of {1,2,3,4,5} with no consecutive integers are {1,3},{1,4},{1,5},{2,4},{2,5} and {3,5}. Equivalently, T(n,k) is the number of kmatchings of the path graph P_n.  Emeric Deutsch, Dec 10 2003
T(n,k)= number of compositions of n+2 into k+1 parts, all >= 2. Example: T(6,2)=6 because we have (2,2,4),(2,4,2),(4,2,2),(2,3,3),(3,2,3) and (3,3,2).  Emeric Deutsch, Apr 09 2005
Given any recurrence sequence S(k) = x*a(k1) + a(k2), starting (1, x, x^2+1, ...); the (k+1)th term of the series = f(x) in the kth degree polynomial: (1, (x), (x^2 + 1), (x^3 + 2x), (x^4 + 3x^2 + 1), (x^5 + 4x^3 + 3x), (x^6 + 5x^4 + 6x^2 + 1), ... Example: let x = 2, then S(k) = 1, 2, 5, 12, 29, 70, 169, ... such that A000129(7) = 169 = f(x), x^6 + 5x^4 + 6x^2 + 1 = (64 + 80 + 24 + 1).  Gary W. Adamson, Apr 16 2008
Row k gives the nonzero coefficients of U(k,x/2) where U is the Chebyshev polynomial of the second kind. For example, row 6 is 1,5,6,1 and U(6,x/2) = x^6  5x^4 + 6x^2  1.  David Callan, Jul 22 2008
T(n,k) is the number of nodes at level k in the Fibonacci tree f(k1). The Fibonacci trees f(k) of order k are defined as follows: 1. f(1) and f(0) each consist of a single node. 2. For k >= 1, to the root of f(k1), taken as the root of f(k), we attach with a rightmost edge the tree f(k2). See the Iyer and Reddy references. These trees are not the same as the Fibonacci trees in A180566. Example: T(3,0)=1 and T(3,1)=2 because in f(2) = /\ we have 1 node at level 0 and 2 nodes at level 1.  Emeric Deutsch, Jun 21 2011
Triangle, with zeros omitted, given by (1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.  Philippe Deléham, Dec 12 2011
This sequence is the elements on the rising diagonals of the Pascal triangle, where the sum of the elements in each rising diagonal represents a Fibonacci number.  Mohammad K. Azarian, Mar 08 2012
If we set F(0;x) = 0, F(1;x) = 1, F(n+1;x) = x*F(n;x) + F(n1;x), then we obtain the sequence of VietaFibonacci polynomials discussed by Gary W. Adamson above. We note that F(n;x) = (i)^n * U(n;i*x/2), where U denotes the respective Chebyshev polynomial of the second kind (see David Callan's remark above). Let us fix a,b,f(0),f(1) in C, b is not the zero, and set f(n) = a*f(n1) + b*f(n2). Then we deduce the relation: f(n) = b^((n1)/2) * F(n;a/sqrt(b))*f(1) + b^(n/2) * F(n1;a/sqrt(b))*f(0), where for a given value of the complex root sqrt(b) we set b^(n/2) = (sqrt(b))^n. Moreover, if b=1 then we get f(n+k) + (1)^k * f(nk) = L(k;a)*f(n), for every k=0,1,...,n, and where L(0;a)=2, L(1;a)=a, L(n+1;a)=a*L(n;a) + L(n1;a) are the VietaLucas polynomials. Let us observe that L(n+2;a) = F(n+2;a) + F(n;a), L(m+n;a) = L(m;a)*F(n;a) + L(m1;a)*F(n1;a), which implies also L(n+1;a) = a*F(n;a) + 2*F(n1;a). Further we have L(n;a) = 2*(i)^n * T(n;i*x/2), where T(n;x) denotes the nth Chebyshev polynomial of the first kind. For the proofs, other relations and facts  see WitulaSlota's papers.  Roman Witula, Oct 12 2012
Aside from signs and index shift, the coefficients of the characteristic polynomial of the Coxeter adjacency matrix for the Coxeter group A_n related to the Chebyshev polynomial of the second kind (cf. Damianou link p. 19).  Tom Copeland, Oct 11 2014
For a mirrored, shifted version showing the relation of these coefficients to the Pascal triangle, Fibonacci, and other number triangles, see A030528. See also A053122 for a relation to Cartan matrices.  Tom Copeland, Nov 04 2014
For a relation to a formulation for a universal Lie Weyl algebra for su(1,1), see page 16 of Durov et al.  Tom Copeland, Nov 29 2014
A reversed, signed and aerated version is given by A049310, related to Chebyshev polynomials.  Tom Copeland, Dec 06 2015
For n >= 3, the nth row gives the coefficients of the independence polynomial of the (n2)path graph P_{n2}.  Eric W. Weisstein, Apr 07 2017
For n >= 2, the nth row gives the coefficients of the matchinggenerating polynomial of the (n1)path graph P_{n1}.  Eric W. Weisstein, Apr 10 2017
Antidiagonals of the Pascal matrix A007318 read bottom to top. These are also the antidiagonals read from top to bottom of the numerical coefficients of the MaurerCartan form matrix of the Leibniz group L^(n)(1,1) presented on p. 9 of the Olver paper), which is generated as exp[c. * M] with (c.)^n = c_n and M the Lie infinitesimal generator A218272. Reverse is A102426.  Tom Copeland, Jul 02 2018
T(n,k) is the number of Markov equivalence classes with skeleton the path on n+1 nodes having exactly k immoralities. See Theorem 2.1 in the article by A. Radhakrishnan et al. below.  Liam Solus, Aug 23 2018
T(n, k) = number of compositions of n+1 into n+12*k odd parts. For example, T(6,2) = 6 because 7 = 5+1+1 = 3+3+1 = 3+1+3 = 1+1+5 = 1+3+3 = 1+1+5.  Michael Somos, Sep 19 2019
Alternate rows can be parsed into those with odd integer coefficients to the right of the leftmost 1, and those with even integer coefficients to the right of the leftmost 1. The first set is shown in A054142 and are characteristic polynomials of submatrices of an infinite tridiagonal matrix (A332602) with all 1's in the super and subdiagonals and (1,2,2,2,...) as the main diagonal. For example, the characteristic equation of the 3 X 3 submatrix (1,1,0; 1,2,1; 0,1,2) is x^3  5x^2 + 6x  1. The roots are the Beraha constants B(7,1) = 3.24697...; B(7,2) = 1.55495...; and B(7,3) = 0.198062.... For n X n matrices of this form, the largest eigenvalue is B(2n+1, 1). The 3 X 3 matrix has an eigenvalue of 3.24697... = B(7,1).
Polynomials with even integer coefficients to the right of the leftmost 1 are in A053123 with roots being the even indexed Beraha constants. The generating Cartan matrices are those with (2,2,2,...) as the main diagonal and 1's as the sub and super diagonals. The largest eigenvalue of n X n matrices of this form are B(2n+2,1). For example, the largest eigenvalue of (2,1,0; 1,2,1; 0,1,2) is 3.414... = B(8,1) = a root to x^3  6x^2 + 10x  4. (End)
T(n,k) is the number of edge covers of P_(n+2) with (nk) edges. For example, T(6,2)=6 because among edges 1, 2, ..., 7 of P_8, we can eliminate any two nonconsecutive edges among 26. These numbers can be found using the recurrence relation for the edge cover polynomial of P_n, which is E(P_n,x)= xE(P_(n1),x)+xE(P_(n2),x) and E(P_1,x)=0, E(P_2,x)=x (ref. Akbari and Oboudi).  Feryal Alayont, Jun 03 2022
T(n,k) is the number of ways to tile an nboard (an n X 1 array of 1 X 1 cells) using k dominoes and n2*k squares.  Michael A. Allen, Dec 28 2022
T(n,k) is the number of positive integer sequences (s(1),s(2),...,s(n2k)) such that s(i) < s(i+1), s(1) is odd, s(n2k) <= n, and s(i) and s(i+1) have opposite parity (ref. Donnelly, Dunkum, and McCoy). Example: T(6,0)=1 corresponds to 123456; T(6,1)=5 corresponds to 1234, 1236, 1256, 1456, 3456; T(6,2)=6 corresponds to 12, 14, 16, 34, 36; and T(6,3)=1 corresponds to the empty sequence () with length 0.  Molly W. Dunkum, Jun 27 2023


REFERENCES

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 141ff.
C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.
I. Kaplansky and J. Riordan, The problème des ménages, Scripta Math. 12, (1946), 113124. See p. 117.
J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 182183.


LINKS

D. Merlini, R. Sprugnoli and M. C. Verri, The tennis ball problem, J. Combin. Theory, A 99 (2002), 307344 (Table 3).


FORMULA

Let F(n, x) be the nth Fibonacci polynomial in x; the g.f. for F(n, x) is Sum_{n>=0} F(n, x)*y^n = (1 + x*y)/(1  y  x*y^2).  Paul D. Hanna
T(m, n) = 0 for n != 0 and m <= 1 T(0, 0) = T(1, 0) = 1 T(m, n) = T(m  1, n) + T(m2, n1) for m >= 2 (i.e., like the recurrence for Pascal's triangle A007318, but going up one row as well as left one column for the second summand). E.g., T(7, 2) = 10 = T(6, 2) + T(5, 1) = 6 + 4.  Rob Arthan, Sep 22 2003
G.f. for kth column: x^(2*k1)/(1x)^(k+1).
Identities for the Fibonacci polynomials F(n, x):
F(m+n+1, x) = F(m+1, x)*F(n+1, x) + x*F(m, x)F(n, x).
F(n, x)^2F(n1, x)*F(n+1, x) = (x)^(n1).
The degree of F(n, x) is floor((n1)/2) and F(2p, x) = F(p, x) times a polynomial of equal degree which is 1 mod p.
p(x,n) = Sum_{m=0..floor((n+1)/2)} binomial(nm+1, m)*x^m;
p(x,n) = p(x, n  1) + x*p(x, n  2). (End)
G.f.: 1/(1xy*x^2) = R(0)/2, where R(k) = 1 + 1/(1  (2*k+1+ x*y)*x/((2*k+2+ x*y)*x + 1/R(k+1) )); (continued fraction).  Sergei N. Gladkovskii, Nov 09 2013
O.g.f. G(x,t) = x/(1xtx^2) = x + x^2 + (1+t) x^3 + (1+2t) x^4 + ... has the inverse Ginv(x,t) = [1+xsqrt[(1+x)^2 + 4tx^2]]/(2tx) = x  x^2 + (1t) x^3 + (1+3t) x^4 + ..., an o.g.f. for the signed Motzkin polynomials of A055151, consistent with A134264 with h_0 = 1, h_1 = 1, h_2 = t, and h_n = 0 otherwise.  Tom Copeland, Jan 21 2016
O.g.f. H(x,t) = x (1+tx)/ [1x(1+tx)] = x + (1+t) x^2 + (1+2t) x^3 + ... = L[Cinv(tx)/t], where L(x) = x/(1+x) with inverse Linv(x) = x/(1x) and Cinv(x) = x (1x) is the inverse of C(x) = (1sqrt(14x))/2, the o.g.f. of the shifted Catalan numbers A000108. Then Hinv(x,t) = C[t Linv(x)]/t = [1 + sqrt(1+4tx/(1+x))]/2t = x  (1+t) x^2 + (1+2t+2t^2) x^3  (1+3t+6t^2+5t^3) x^4 + ..., which is signed A098474, reverse of A124644.  Tom Copeland, Jan 25 2016
T(n, k) = GegenbauerC(k, (n+1)/2k, 1).  Peter Luschny, May 10 2016


EXAMPLE

The first few Fibonacci polynomials (defined here by F(0,x) = 0, F(1,x) = 1; F(n+1, x) = F(n, x) + x*F(n1, x)) are:
0: 0
1: 1
2: 1
3: 1 + x
4: 1 + 2*x
5: 1 + 3*x + x^2
6: (1 + x)*(1 + 3*x)
7: 1 + 5*x + 6*x^2 + x^3
8: (1 + 2*x)*(1 + 4*x + 2*x^2)
9: (1 + x)*(1 + 6*x + 9*x^2 + x^3)
10: (1 + 3*x + x^2 )*(1 + 5*x + 5*x^2)
11: 1 + 9*x + 28*x^2 + 35*x^3 + 15*x^4 + x^5
1
1
1 1
1 2
1 3 1
1 4 3
1 5 6 1
1 6 10 4
1 7 15 10 1
1 8 21 20 5
1 9 28 35 15 1
1 10 36 56 35 6
1 11 45 84 70 21 1
1 12 55 120 126 56 7 (End)
For n=9 and k=4, T(9,4) = C(5,4) = 5 since there are exactly five size4 subsets of {1,2,...,8} that contain no consecutive integers, namely, {1,3,5,7}, {1,3,5,8}, {1,3,6,8}, {1,4,6,8}, and {2,4,6,8}.  Dennis P. Walsh, Mar 31 2011
When the rows of the triangle are displayed as centered text, the falling diagonal sums are A005314. The first few terms are row1 = 1 = 1; row2 = 1+1 = 2; row3 = 2+1 = 3; row4 = 1+3+1 = 5; row5 = 1+3+4+1 = 9; row6 = 4+6+5+1 = 16; row7 = 1+10+10+6+1 = 28; row8 = 1+5+20+15+7+1 = 49; row9 = 6+15+35+21+8+1 = 86; row10 = 1+21+35+56+28+9+1 = 151.  John Molokach, Jul 08 2013
In the example, you can see that the nth row of Pascal's triangle is given by T(n, 0), T(n+1, 1), ..., T(2n1, n1), T(2n, n).  Daniel Forgues, Jul 07 2018


MAPLE

a := proc(n) local k; [ seq(binomial(nk, k), k=0..floor(n/2)) ]; end;
T := proc(n, k): if k<0 or k>floor(n/2) then return(0) fi: binomial(nk, k) end: seq(seq(T(n, k), k=0..floor(n/2)), n=0..15); # Johannes W. Meijer, Aug 26 2013


MATHEMATICA

(* first: sum method *) Table[CoefficientList[Sum[Binomial[n  m + 1, m]*x^m, {m, 0, Floor[(n + 1)/2]}], x], {n, 0, 12}] (* Roger L. Bagula, Feb 20 2009 *)
(* second: polynomial recursion method *) Clear[L, p, x, n, m]; L[x, 0] = 1; L[x, 1] = 1 + x; L[x_, n_] := L[x, n  1] + x*L[x, n  2]; Table[ExpandAll[L[x, n]], {n, 0, 10}]; Table[CoefficientList[ExpandAll[L[x, n]], x], {n, 0, 12}]; Flatten[%] (* Roger L. Bagula, Feb 20 2009 *)
(* Center option shows falling diagonals are A224838 *) Column[Table[Binomial[n  m, m], {n, 0, 25}, {m, 0, Floor[n/2]}], Center] (* John Molokach, Jul 26 2013 *)
Table[ Select[ CoefficientList[ Fibonacci[n, x], x], Positive] // Reverse, {n, 1, 18} ] // Flatten (* JeanFrançois Alcover, Oct 21 2013 *)
CoefficientList[LinearRecurrence[{1, x}, {1 + x, 1 + 2 x}, {1, 10}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)
CoefficientList[Table[x^((n  1)/2) Fibonacci[n, 1/Sqrt[x]], {n, 15}], x] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)


PROG

(PARI) {T(n, k) = if( k<0  2*k>n, 0, binomial(nk, k))};
(Sage) # Prints the table; cf. A145574.
for n in (2..20): [Compositions(n, length=m, min_part=2).cardinality() for m in (1..n//2)] # Peter Luschny, Oct 18 2012
(Haskell)
a011973 n k = a011973_tabf !! n !! k
a011973_row n = a011973_tabf !! n
a011973_tabf = zipWith (zipWith a007318) a025581_tabl a055087_tabf


CROSSREFS



KEYWORD

tabf,easy,nonn,nice


AUTHOR



STATUS

approved



