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A059259
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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), ...
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26
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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, 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, 148, 166, 130, 70, 25, 5, 1, 1, 10, 46, 128, 239, 314, 296, 200
(list;
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graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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0,8
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COMMENTS
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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
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
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
See the Edwards-Allen 2020 paper, page 14, for proof of Englander's conjecture. - Michael De Vlieger, Dec 10 2020
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LINKS
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FORMULA
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G.f.: 1/(1 - x - x*y - y^2).
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
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
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
T(n,k) + T(n-1,k-1) = binomial(n,k) if n >= k > 0.
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.
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.
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EXAMPLE
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Triangle begins:
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, 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, 148, 166, 130, 70, 25, 5, 1;
1, 10, 46, 128, 239, 314, 296, 200, 95, 30, 6, 0;
...
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MAPLE
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read transforms; 1/(1-x-x*y-y^2); SERIES2(%, x, y, 12); SERIES2TOLIST(%, x, y, 12);
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MATHEMATICA
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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 *)
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PROG
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(Sage)
@cached_function
def prec(n, k):
if k==n: return (-1)^n
if k==0: return 0
return prec(n-1, k-1)-sum(prec(n, k+i-1) for i in (2..n-k+1))
return [(-1)^(n-k+1)*prec(n+1, k) for k in (1..n)]
(PARI) {T(n, k) = if(k==0, 1, if(k==n, (1+(-1)^n)/2, T(n-1, k) +T(n-1, k-1)) )};
for(n=0, 10, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Apr 29 2019
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CROSSREFS
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See A059260 for an explicit formula.
Diagonals of this triangle are given by A006498.
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KEYWORD
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AUTHOR
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STATUS
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approved
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