|
%I #29 Apr 17 2020 12:54:32
%S 1,1,1,1,1,1,1,1,1,2,1,2,1,3,1,1,3,2,1,4,4,1,4,6,1,5,9,1,1,5,12,2,1,6,
%T 16,6,1,6,20,10,1,7,25,19,1,1,7,30,28,3,1,8,36,44,9,1,8,42,60,19,1,9,
%U 49,85,38,1,1,9,56,110,66,3
%N Triangle read by rows, formed from antidiagonals of triangle A228570: T(n,k) = A034851(n-3*k, k) for n >= 0 and 0 <= k <= floor(n/4).
%C The row sums of this triangle are A192928.
%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. Observe A102541 and A228570 for the same phenomenom. The number of zeros in the columns for these three triangles are multiples of 2, 3 and 4 respectively.
%C Also the number of equivalence classes of ways of placing k 4 X 4 tiles in an n X 4 rectangle under all symmetry operations of the rectangle. - _Christopher Hunt Gribble_, Apr 24 2015
%H Andrew Howroyd, <a href="/A228572/b228572.txt">Table of n, a(n) for n = 0..989</a>
%F T(n, k) = A034851(n-3*k, k) for n >= 0 and 0 <= k <= floor(n/4).
%F T(n, k) = T(n-1, k) + T(n-4, k-1) - C((n+3)/2 - 3*(k+1)/2-1, (k+1)/2-1), where the last term is present only if n odd and k odd; T(0, 0) = 1, T(1, 0) = 1, T(2, 0) = 1, T(3, 0) = 1, T(n, k) = 0 for n < 0 and T(n, k) = 0 for k < 0 and k > floor(n/4).
%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
%e 4: 1, 1
%e 5: 1, 1
%e 6: 1, 2
%e 7: 1, 2
%e 8: 1, 3, 1
%e 9: 1, 3, 2
%e 10: 1, 4, 4
%e 11: 1, 4, 6
%e 12: 1, 5, 9, 1
%p T := proc(n, k) option remember: if n <0 then return(0) fi: if k < 0 or k > floor(n/4) then return(0) fi: A034851(n-3*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/4)), n=0..21); # 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 or n=3 and k=0 then return(1) fi: if k <0 or k > floor(n/4) then return(0) fi: if type(n, odd) and type(k, odd) then procname(n-1,k) + procname(n-4, k-1) - binomial((n+3)/2 - 3*(k+1)/2 - 1,(k+1)/2-1) else procname(n-1, k) + procname(n-4, k-1) fi: end: seq(seq(T(n, k), k=0..floor(n/4)), n=0..21); # End second program
%t T[n_, k_] := (Binomial[n - 3k, k] + Boole[EvenQ[k] || EvenQ[n]]* Binomial[(n - 3k - Mod[k, 2] - Mod[n, 2])/2, Quotient[k, 2]])/2; Table[T[n, k], {n, 0, 20}, {k, 0, Quotient[n, 4]}] // Flatten (* _Jean-François Alcover_, Oct 06 2017, after _Andrew Howroyd_ *)
%o (PARI)
%o T(n,k)={(binomial(n-3*k,k) + (k%2==0||n%2==0)*binomial((n-3*k-k%2-n%2)/2,k\2))/2}
%o for(n=1,20,for(k=0,(n\4), print1(T(n,k), ", "));print) \\ _Andrew Howroyd_, May 29 2017
%Y Cf. A034851, A102541, A228570, A192928.
%K nonn,easy,tabf
%O 0,10
%A _Johannes W. Meijer_, Aug 26 2013
|
