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%I #22 Oct 06 2017 04:27:44
%S 1,1,1,1,1,1,1,1,2,1,2,1,1,3,2,1,3,4,1,4,6,1,1,4,9,2,1,5,12,6,1,5,16,
%T 10,1,1,6,20,19,3,1,6,25,28,9,1,7,30,44,19,1,1,7,36,60,38,3,1,8,42,85,
%U 66,12,1,8,49,110,110,28,1
%N Triangle read by rows, formed from antidiagonals of triangle A102541. T(n, k) = A034851(n-2*k, k), n>= 0 and 0 <= k <= floor(n/3).
%C The row sums of this triangle are A102543. The antidiagonal sums are given by A192928 and the backwards antidiagonal sums are given by A228571.
%C Moving the terms in each column of this triangle, see the example, upwards to row 0 gives Losanitsch’s triangle A034851 as a square array.
%C Also the number of equivalence classes of ways of placing k 3 X 3 tiles in an n X 3 rectangle under all symmetry operations of the rectangle. - _Christopher Hunt Gribble_, Feb 16 2014
%H Andrew Howroyd, <a href="/A228570/b228570.txt">Table of n, a(n) for n = 0..1000</a>
%F T(n, k) = A034851(n-2*k, k), n >= 0 and 0 <= k <= floor(n/3).
%F T(n, k) = T(n-1, k) + T(n-3, k-1) - C((n-4)/2 - 2*(k-1)/2, (k-1)/2) where the last term is present only if n even and k odd; T(0, 0) = 1, T(1, 0) = 1, T(2, 0) = 1, T(n, k) = 0 for n < 0 and T(n, k) = 0 for k < 0 and k > floor(n/3).
%e The first few rows of triangle T(n, k) are:
%e n/k: 0, 1, 2, 3
%e 0: 1
%e 1: 1
%e 2: 1
%e 3: 1, 1
%e 4: 1, 1
%e 5: 1, 2
%e 6: 1, 2, 1
%e 7: 1, 3, 2
%e 8: 1, 3, 4
%e 9: 1, 4, 6, 1
%e 10: 1, 4, 9, 2
%e 11: 1, 5, 12, 6
%p T := proc(n,k) option remember: if n <0 then return(0) fi: if k < 0 or k > floor(n/3) then return(0) fi: A034851(n-2*k, k) end: A034851 := proc(n, k) option remember; local t; if k = 0 or k = n then return(1) fi; if n mod 2 = 0 and k mod 2 = 1 then t := binomial(n/2-1, (k-1)/2) else t := 0; fi; A034851(n-1, k-1) + A034851(n-1, k) - t; end: seq(seq(T(n, k), k=0..floor(n/3)), n=0..18); # End first program
%p T := proc(n,k) option remember: if n=0 and k=0 or n=1 and k=0 or n=2 and k=0 then return(1) fi: if k <0 or k > floor(n/3) then return(0) fi: if type(n, even) and type(k, odd) then procname(n-1, k) + procname(n-3, k-1) - binomial((n-4)/2-2*(k-1)/2, (k-1)/2) else procname(n-1, k) + procname(n-3, k-1) fi: end: seq(seq(T(n,k), k=0..floor(n/3)), n=0..18); # End second program
%t T[n_, k_] := (Binomial[n - 2k, k] + Boole[EvenQ[k] || OddQ[n]] Binomial[(n - 2k - Mod[n, 2])/2, Quotient[k, 2]])/2; Table[T[n, k], {n, 0, 20}, {k, 0, Quotient[n, 3]}] // Flatten (* _Jean-François Alcover_, Oct 06 2017, after _Andrew Howroyd_ *)
%o (PARI)
%o T(n,k)={(binomial(n-2*k,k) + (k%2==0||n%2==1)*binomial((n-2*k-n%2)/2,k\2))/2}
%o for(n=1,20,for(k=0,(n\3), print1(T(n,k), ", "));print) \\ _Andrew Howroyd_, May 30 2017
%Y Cf. A034851, A102541, A228572, A102543, A192928, A228571, A005691.
%K nonn,easy,tabf
%O 0,9
%A _Johannes W. Meijer_, Aug 26 2013
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