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A256107
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).
1
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, 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, 2, 2, 232, 88, 141, 66, 33, 40, 24, 12, 14, 8, 4, 4, 2, 1, 376, 143, 230, 108, 54, 66
OFFSET
0,2
COMMENTS
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.
EXAMPLE
Irreuglar triangle begins:
n/k 0 1 2 3 4 5 6 7 8 ...
0 1
1 2
2 4 1
3 7 2 2
4 12 4 5 2 1
5 20 7 10 4 2 2
6 33 12 18 8 4 4 2 1
7 54 20 31 14 7 8 4 2 2
8 88 33 52 24 12 14 8 4 4
...
PROG
(PARI)
{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, ", ")))}
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Kival Ngaokrajang, Mar 14 2015
STATUS
approved