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%I #19 Mar 27 2015 21:20:09
%S 1,2,4,1,7,2,2,12,4,5,2,1,20,7,10,4,2,2,33,12,18,8,4,4,2,1,54,20,31,
%T 14,7,8,4,2,2,88,33,52,24,12,14,8,4,4,2,1,143,54,86,40,20,24,14,7,8,4,
%U 2,2,232,88,141,66,33,40,24,12,14,8,4,4,2,1,376,143,230,108,54,66
%N Irregular triangle read by rows, T(n,k) is the number of pentagrams on the k layers at n iterations of a pentagram fractal (see comment).
%C Refer to A255870, the number of pentagrams on one side of the outer layer (including pentagrams on two vertices) at n iterations would be T(n,0), the next layers k >= 1 T(n,k) are the number of pentagrams toward the center. For k >= 2, the row length is A032766. The first differences of A255870 = 5*(rows sum - 1). T(n,k) = A000071 with a shift for k = 0 or k mod 3 = 1. T(n,2) = A006327 with a shift. For k >= 3, T(n,k) = 2*A000071 with a shift for k mod 3 = 0 or 2. See illustration in the links.
%H Kival Ngaokrajang, <a href="/A256107/a256107_1.pdf">Illustration of initial terms</a>, <a href="/A256107/a256107_2.pdf">T(n,k) for n = 0..10, k = 0..13</a>
%e Irreuglar triangle begins:
%e n/k 0 1 2 3 4 5 6 7 8 ...
%e 0 1
%e 1 2
%e 2 4 1
%e 3 7 2 2
%e 4 12 4 5 2 1
%e 5 20 7 10 4 2 2
%e 6 33 12 18 8 4 4 2 1
%e 7 54 20 31 14 7 8 4 2 2
%e 8 88 33 52 24 12 14 8 4 4
%e ...
%o (PARI)
%o {for(n=0, 20, if(n<2, lk=0, lk=floor(3*(n-2)/2)+1); for (k=0, lk, if(k<>0, if(k<>2, if(Mod(k,3)==1, t=fibonacci(n+1-2*(k-1)/3)-1, t=2*(fibonacci(n+2-ceil((2*k+1)/3))-1)), t=fibonacci(n+2)-3), t=fibonacci(n+3-2*k/3)-1); print1(t, ", ")))}
%Y Cf. A255870, A032766, A000071, A006327.
%K nonn,tabf
%O 0,2
%A _Kival Ngaokrajang_, Mar 14 2015
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