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A059259 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ... 26

%I #70 Dec 10 2020 12:58:13

%S 1,1,0,1,1,1,1,2,2,0,1,3,4,2,1,1,4,7,6,3,0,1,5,11,13,9,3,1,1,6,16,24,

%T 22,12,4,0,1,7,22,40,46,34,16,4,1,1,8,29,62,86,80,50,20,5,0,1,9,37,91,

%U 148,166,130,70,25,5,1,1,10,46,128,239,314,296,200

%N Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-x-x*y-y^2) = 1/((1+y)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

%C This sequence provides the general solution to the recurrence a(n) = a(n-1) + k*(k+1)*a(n-2), a(0)=a(1)=1. The solution is (1, 1, k^2 + k + 1, 2*k^2 + 2*k + 1, ...) whose coefficients can be read from the rows of the triangle. The row sums of the triangle are given by the case k=1. These are the Jacobsthal numbers, A001045. Viewed as a square array, its first row is (1,0,1,0,1,...) with e.g.f. cosh(x), g.f. 1/(1-x^2) and subsequent rows are successive partial sums given by 1/((1-x)^n * (1-x^2)). - _Paul Barry_, Mar 17 2003

%C Conjecture: every second column of this triangle is identical to a column in the square array A071921. For example, column 4 of A059259 (1, 3, 9, 22, 46, ...) appears to be the same as column 3 of A071921; column 6 of A059259 (1, 4, 16, 50, 130, 296, ...) appears to be the same as column 4 of A071921; and in general column 2k of A059259 appears to be the same as column k+1 of A071921. Furthermore, since A225010 is a transposition of A071921 (ignoring the latter's top row and two leftmost columns), there appears to be a correspondence between column 2k of A059259 and row k of A225010. - _Mathew Englander_, May 17 2014

%C T(n,k) is the number of n-tilings of a (one-dimensional) board that use k (1,1)-fence tiles and n-k squares. A (1,1)-fence is a tile composed of two pieces of width 1 separated by a gap of width 1. - _Michael A. Allen_, Jun 25 2020

%C See the Edwards-Allen 2020 paper, page 14, for proof of Englander's conjecture. - _Michael De Vlieger_, Dec 10 2020

%H G. C. Greubel, <a href="/A059259/b059259.txt">Table of n, a(n) for the first 50 rows, flattened</a>

%H Kenneth Edwards and Michael A. Allen, <a href="https://arxiv.org/abs/2009.04649">New Combinatorial Interpretations of the Fibonacci Numbers Squared, Golden Rectangle Numbers, and Jacobsthal Numbers Using Two Types of Tile</a>, arXiv:2009.04649 [math.CO], 2020.

%F G.f.: 1/(1 - x - x*y - y^2).

%F As a square array read by antidiagonals, this is T(n, k) = Sum_{i=0..n} (-1)^(n-i)*C(i+k, k). - _Paul Barry_, Jul 01 2003

%F T(2*n,n) = A026641(n). - _Philippe Deléham_, Mar 08 2007

%F T(n,k) = T(n-1,k) + T(n-2,k-1) + T(n-2,k-2), T(0,0) = T(1,0) = T(2,0) = T(2,1) = T(2,2)=1, T(1,1)=0, T(n,k)=0 if k < 0 or if k > n. - _Philippe Deléham_, Nov 24 2013

%F T(n,0) = 1, T(n,n) = (1+(-1)^n)/2, and T(n,k) = T(n-1,k) + T(n-1,k-1) for 0 < k < n. - _Mathew Englander_, May 24 2014

%F From _Michael A. Allen_, Jun 25 2020: (Start)

%F T(n,k) + T(n-1,k-1) = binomial(n,k) if n >= k > 0.

%F T(2*n-1,2*n-2) = T(2*n,2*n-1) = n, T(2*n,2*n-2) = n^2, T(2*n+1,2*n-1) = n*(n+1) for n > 0.

%F T(n,2) = binomial(n-2,2) + n - 1 for n > 1 and T(n,3) = binomial(n-3,3) + 2*binomial(n-2,2) for n > 2.

%F T(2*n-k,k) = A123521(n,k). (End)

%e Triangle begins:

%e 1;

%e 1, 0;

%e 1, 1, 1;

%e 1, 2, 2, 0;

%e 1, 3, 4, 2, 1;

%e 1, 4, 7, 6, 3, 0;

%e 1, 5, 11, 13, 9, 3, 1;

%e 1, 6, 16, 24, 22, 12, 4, 0;

%e 1, 7, 22, 40, 46, 34, 16, 4, 1;

%e 1, 8, 29, 62, 86, 80, 50, 20, 5, 0;

%e 1, 9, 37, 91, 148, 166, 130, 70, 25, 5, 1;

%e 1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0;

%e ...

%p read transforms; 1/(1-x-x*y-y^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);

%t T[n_, 0]:= 1; T[n_, n_]:= (1+(-1)^n)/2; T[n_, k_]:= T[n, k] = T[n-1, k] + T[n-1, k-1]; Table[T[n, k], {n, 0, 10} , {k, 0, n}]//Flatten (* _G. C. Greubel_, Jan 03 2017 *)

%o (Sage)

%o def A059259_row(n):

%o @cached_function

%o def prec(n, k):

%o if k==n: return (-1)^n

%o if k==0: return 0

%o return prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))

%o return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n)]

%o for n in (1..12): print(A059259_row(n)) # _Peter Luschny_, Mar 16 2016

%o (PARI) {T(n,k) = if(k==0, 1, if(k==n, (1+(-1)^n)/2, T(n-1,k) +T(n-1,k-1)) )};

%o for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ _G. C. Greubel_, Apr 29 2019

%Y See A059260 for an explicit formula.

%Y Diagonals of this triangle are given by A006498.

%Y Similar to the triangles A035317, A080242, A108561, A112555.

%Y Cf. A123521, A157897.

%K nonn,tabl

%O 0,8

%A _N. J. A. Sloane_, Jan 23 2001

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Last modified March 29 06:57 EDT 2024. Contains 371265 sequences. (Running on oeis4.)