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A133342
Concatenation of binary expansion of n-th row of Pascal's triangle.
2
1, 11, 1101, 111111, 11001101001, 1101101010101011, 111011111010011111101, 111110101100011100011101011111, 110001110011100010001101110001110010001, 11001100100101010011111101111110101010010010010011
OFFSET
0,2
COMMENTS
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.
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
LINKS
FORMULA
a(n) = Concatenate[k=1..n] binomial(n, k) (base 2).
a(n) = Concatenate[i=A000217(n)..A000217(n+1)] A007088(A007318(i)).
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 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.
The array of base k concatenations begins:
k/n 0 1 2 3 4
1.| 1 11 1111 11111111 1111111111111111 2^(n-1) repetitions of 1
2.| 1 11 1101 111111 11001101001
3.| 1 11 121 110101 11120111
4.| 1 11 121 1331 11012101
5.| 1 11 121 1331 141141
6.| 1 11 121 1331 141041
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
MATHEMATICA
FromDigits[Flatten[IntegerDigits[#, 2]]]&/@Table[Binomial[n, k], {n, 0, 10}, {k, 0, n}] (* Harvey P. Dale, Apr 12 2020 *)
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Jonathan Vos Post, Oct 20 2007
EXTENSIONS
Corrected and extended by R. J. Mathar, Jan 08 2008
STATUS
approved