
EXAMPLE

a(0) = 1 because the 0th row of Pascal's triangle is 1.
a(1) = 11 because the 1st row of Pascal's triangle is 1,1 which concatenates to 11.
a(2) = 1101 because the 2nd row of Pascal's triangle is 1,2,1 which in binary is 1,10,1 which concatenates to 1101.
a(3) = 111111 because the 3rd row of Pascal's triangle is 1,3,3,1 which in binary is 1,11,11,1 which concatenates to 111111.
a(4) = 110010101001 because the 4th row of Pascal's triangle is 1,4,6,4,1 which in binary is 1,100,110,100,1 which concatenates to 11001101001.
a(5) = 1101101010101011 because the 5th row of Pascal's triangle is 1,5,10,10,5,1 which in binary is 1,101,1010,1010,101,1 which concatenates to 1101101010101011.
a(6) = 111011111010011111101 because the 5th row of Pascal's triangle is 1,6,15,20,15,6,1 which in binary is 1,110,1111,10100,1111,110,1 which concatenates to 111011111010011111101.


MAPLE

catL := proc(L) local resul, a ; resul:=0 ; for a in L do resul := resul*10^(max(ilog10(a)+1, 1))+a ; od: RETURN(resul) ; end: A133342 := proc(n) local prow, k ; prow := [1] ; for k from 1 to n do prow := [op(prow), convert(binomial(n, k), binary) ] ; od: catL(prow) ; end: seq(A133342(n), n=0..11) ; # R. J. Mathar, Jan 08 2008
