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%I #77 Dec 27 2021 11:08:35
%S 1,11,1101,111111,11001101001,1101101010101011,111011111010011111101,
%T 111110101100011100011101011111,
%U 110001110011100010001101110001110010001,11001100100101010011111101111110101010010010010011
%N Concatenation of binary expansion of n-th row of Pascal's triangle.
%C Binary analog of A003590. More generally, this sequence is the 2nd row of the matrix whose k-th row is the concatenation of the base-k expression of n-th row of Pascal's triangle. The 10th row of that array is A003590.
%C a(n) is a repunit for n = 0, 1, 3, but otherwise not since C(n,1) = n has a 0-bit apart from at n=2^k-1 and there C(n,2) = n*(n-1)/2 == 1 (mod 4) has a 0-bit. - _Kevin Ryde_ and _Bernard Schott_, Nov 11 2021
%H Harvey P. Dale, <a href="/A133342/b133342.txt">Table of n, a(n) for n = 0..38</a>
%F a(n) = Concatenate[k=1..n] binomial(n, k) (base 2).
%F a(n) = Concatenate[i=A000217(n)..A000217(n+1)] A007088(A007318(i)).
%e a(0) = 1 because the 0th row of Pascal's triangle is 1.
%e a(1) = 11 because the 1st row of Pascal's triangle is 1,1 which concatenates to 11.
%e 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.
%e 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.
%e 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.
%e 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.
%e a(6) = 111011111010011111101 because the 6th 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.
%e The array of base k concatenations begins:
%e k/n 0 1 2 3 4
%e 1.| 1 11 1111 11111111 1111111111111111 2^(n-1) repetitions of 1
%e 2.| 1 11 1101 111111 11001101001
%e 3.| 1 11 121 110101 11120111
%e 4.| 1 11 121 1331 11012101
%e 5.| 1 11 121 1331 141141
%e 6.| 1 11 121 1331 141041
%p 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
%t FromDigits[Flatten[IntegerDigits[#,2]]]&/@Table[Binomial[n,k],{n,0,10},{k,0,n}] (* _Harvey P. Dale_, Apr 12 2020 *)
%Y Cf. A000217, A002275, A003590, A007318, A349304.
%K base,easy,nonn
%O 0,2
%A _Jonathan Vos Post_, Oct 20 2007
%E Corrected and extended by _R. J. Mathar_, Jan 08 2008
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