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 A046854 Triangle in which k-th entry of row n is binomial(floor(n/2+k/2),k), n>=0, n >= k >= 0. 50
 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 3, 4, 1, 1, 1, 3, 6, 4, 5, 1, 1, 1, 4, 6, 10, 5, 6, 1, 1, 1, 4, 10, 10, 15, 6, 7, 1, 1, 1, 5, 10, 20, 15, 21, 7, 8, 1, 1, 1, 5, 15, 20, 35, 21, 28, 8, 9, 1, 1, 1, 6, 15, 35, 35, 56, 28, 36, 9, 10, 1, 1, 1, 6, 21 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,8 COMMENTS Row sums are fibonacci(n+2). Diagonal sums are A016116. - Paul Barry, Jul 07 2004 Riordan array (1/(1-x), x/(1-x^2)). Matrix inverse is A106180. - Paul Barry, Apr 24 2005 As an infinite lower triangular matrix * [1,2,3,...] = A055244: (1, 1, 3, 6, 12, 23,...). [Gary W. Adamson, Dec 23 2008] From Emeric Deutsch, Jun 18 2010: (Start) T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n} of size k, starting with an odd number (Terquem's problem, see the Riordan reference, p. 17). Example: T(8,5)=6 because we have 12345, 12347, 12367, 12567, 14567, and 34567. T(n,k) is the number of alternating parity increasing subsequences of {1,2,...,n,n+1} of size k, starting with an even number. Example: T(7,4)=5 because we have 2345, 2347, 2367, 2567, and 4567. (End) From L. Edson Jeffery, Mar 01 2011: (Start) This triangle can be constructed as follows. Interlace two copies of the table of binomial coefficients to get the preliminary table 1 1 1  1 1  1 1  2  1 1  2  1 1  3  3  1 1  3  3  1 ..., then shift each entire r-th column up r rows, r=0,1,2,.... Also, a signed version of this sequence (A187660 in tabular form) begins with {1} {1,-1} {1,-1,-1} {1,-2,-1,1} {1,-2,-3,1,1}, ... (compare with A066170, A130777). Let T(N,k) denote the k-th entry in row N of the signed table. Then, for N>1, row N gives the coefficients of the characteristic function p_N(x)=Sum[k=0..N, T(N,k)x^(N-k)]=0 of the N X N matrix U_N=[(0 ... 0 1);(0 ... 0 1 1);...;(0 1 ... 1);(1 ... 1)]. Now let Q_r(t) be a polynomial with recurrence relation Q_r(t)=t*Q_(r-1)(t)-Q_(r-2)(t)  (r>1), with Q_0(t)=1 and Q_1(t)=t. Then p_N(x)=0 has solutions Q_(N-1)(phi_j), where phi_j=2*(-1)^(j-1)*cos(j*Pi/(2*N+1)), j=1,2,...,N. For example, row N=3 is {1,-2,-1,1}, giving the coefficients of the characteristic function p_3(x)=x^3-2*x^2-x+1=0 for the 3 X 3 matrix U_3=[(0 0 1);(0 1 1);(1 1 1)], with eigenvalues Q_2(phi_j)=[2*(-1)^(j-1)*cos(j*Pi/7)]^2-1, j=1,2,3. (End) Given the signed polynomials (+--++--,...) of the triangle, the largest root of the n-th row polynomial is the longest (2n+1) regular polygon diagonal length, with edge = 1. Example: the largest root to x^3 - 2x^2 - x + 1 = 0 is 2.24697...; the longest heptagon diagonal, ((Sin 3*Pi/7)/Sin Pi/7)) - Gary W. Adamson, Sep 06 2011. Given the signed polynomials from Gary W. Adamson's comment, the largest root of the n-th polynomial also equals the length from the center to a corner (vertex) of a regular 2*(2*n+1)-sided polygon with side (edge) length = 1. - L. Edson Jeffery, Jan 01 2012 Put f(x,0) = 1 and f(x,n) = x + 1/f(x,n-1).  Then f(x,n) = u(x,n)/v(x,n), where u(x,n) and v(x,n) are polynomials.  The flattened triangles of coefficients of u and v are both essentially A046854, as indicated by the Mathematica program headed "Polynomials".  - Clark Kimberling, Oct 12 2014 From Jeremy Dover, Jun 07 2016: (Start) T(n,k) is the number of binary strings of length n+1 starting with 0 that have exactly k pairs of consecutive 0s and no pairs of consecutive 1s. T(n,k) is the number of binary strings of length n+2 starting with 1 that have exactly k pairs of consecutive 0s and no pairs of consecutive 1s. (End) REFERENCES J. Riordan, An Introduction to Combinatorial Analysis, Princeton University Press, 1978. [Emeric Deutsch, Jun 18 2010] LINKS Nathaniel Johnston, Rows 0..100, flattened Jeremy M. Dover, Some Notes on Pairs in Binary Strings, arXiv:1609.00980 [math.CO], 2016. FORMULA T(n,k) = binomial(floor(n/2+k/2),k) G.f.: (1+x)/(1-xy-x^2). - Ralf Stephan, Feb 13 2005 Triangle = A097806 * A049310, as infinite lower triangular matrices. - Gary W. Adamson, Oct 28 2007 T(n,k) = A065941(n,n-k) = abs(A130777(n,k)) = abs(A066170(n,k)) = abs(A187660(n,k)) [Johannes W. Meijer, Aug 08 2011] For n > 1: T(n, k) = T(n-1, k-1) + T(n-2, k), 0 < k < n-1. - Reinhard Zumkeller, Apr 24 2013 EXAMPLE Triangle begins: 1 1 1 1 1 1 1 2 1 1 1 2 3 1 1 1 3 3 4 1 1 ... MAPLE A046854:= proc(n, k): binomial(floor(n/2+k/2), k) end: seq(seq(A046854(n, k), k=0..n), n=0..11); # Nathaniel Johnston, Jun 30 2011 MATHEMATICA Table[ Binomial[ Floor[ n/2 +k/2 ], k ], {n, 0, 16}, {k, 0, n} ] (* next program: Polynomials *) z = 12; f[x_, n_] := x + 1/f[x, n - 1]; f[x_, 1] = 1; t = Table[Factor[f[x, n]], {n, 1, z}] u = Flatten[CoefficientList[Numerator[t], x]] (* A046854 *) v = Flatten[CoefficientList[Denominator[t], x]] (* Clark Kimberling, Oct 13 2014 *) PROG (Haskell) a046854 n k = a046854_tabl !! n !! k a046854_row n = a046854_tabl !! n a046854_tabl = [1] : f [1] [1, 1] where    f us vs = vs : f vs  (zipWith (+) (us ++ [0, 0]) ([0] ++ vs)) -- Reinhard Zumkeller, Apr 24 2013 CROSSREFS Reflected version of A065941, which is considered the main entry. A deficient version is in A030111. Cf. A066170, A097806, A049310, A187660. Cf. A055244 [From Gary W. Adamson, Dec 23 2008] Sequence in context: A267482 A130777 A187660 * A066170 A184957 A228349 Adjacent sequences:  A046851 A046852 A046853 * A046855 A046856 A046857 KEYWORD nonn,tabl,easy AUTHOR STATUS approved

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Last modified December 14 17:37 EST 2018. Contains 318103 sequences. (Running on oeis4.)