%I M4802
%S 1,11,101,1111,10001,110011,1010101,11111111,100000001,1100000011,
%T 10100000101,111100001111,1000100010001,11001100110011,
%U 101010101010101,1111111111111111,10000000000000001,110000000000000011
%N Rows of Sierpiński's triangle (Pascal's triangle mod 2).
%C The rows of Sierpiński's triangle, read as numbers in binary representation, are products of distinct Fermat numbers, row 0 being the empty product. (See also the comment in A080176.)
%C Rows 1 to 31 are the binary representation of the 31 (2^51) nonempty products of distinct Fermat primes, giving the number of sides of constructible (with straightedge and compass) oddsided polygons.  _Daniel Forgues_, Jun 21 2011
%C Sierpiński's triangles typically refer to any finite triangle with rows 0 to 2^n1 so as to get complete triangles, with n at least 4 so as to show the fractallike pattern of nested triangles. We may consider these finite Sierpiński's triangles as finite parts of "the" infinite Sierpiński's triangle, so to speak.  _Daniel Forgues_, Jun 22 2011
%D C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 353.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H N. J. A. Sloane, <a href="/A006943/b006943.txt">Table of n, a(n) for n = 0..200</a>
%H Antti Karttunen, <a href="http://ndirty.cute.fi/~karttu/matikka/A048757/A048757.htm">On Pascal's Triangle Modulo 2 in Fibonacci Representation</a> (Abstract). (For Denton Hewgill's identity)
%H OEIS Wiki, <a href="/wiki/Sierpinski's_triangle">Sierpinski's triangle</a>
%H V. Shevelev, <a href="http://arxiv.org/abs/1011.6083">On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization</a>, J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 1129.
%F From _Daniel Forgues_, Jun 2021 2011: (Start)
%F In the following formulae, [...]_2 means converted to base 2.
%F a(n) = [sum_(i=0..n) (binom(n,i) mod 2) 2^i]_2, n >= 0.
%F From row n, 0 <= n <= 2^k  1, k >= 0, being
%F a(n) = [prod_(i=0..k1) (F_i)^(alpha_i)]_2, alpha_i in {0, 1},
%F where for k = 0, we get the empty product, i.e., 1, giving a(0) = 1,
%F we induce from the triangle that row 2^k + n, 0 <= n <= 2^k  1, is
%F a(2^k + n) = a(n)*[F_k]_2, k >= 0.
%F Denton Hewgill's identity: (Cf. links)
%F a(n) = [prod_{i>=0} (F_i)^(floor(n/2^i) mod 2)]_2, F_i = 2^(2^i)+1.
%F a(0) = 1; a(n) = [prod_{i=0..floor(log_2(n))} (F_i)^(floor(n/2^i) mod 2)]_2, F_i = 2^(2^i)+1, n >= 1. (End)
%F From _Vladimir Shevelev_, Dec 2627 2013: (Start)
%F sum_{n>=0} 1/a(n)^r = prod_{k>=0} (1 + 1/(10^(2^k)+1)^r),
%F sum_{n>=0} (1)^A000120(n)/a(n)^r = prod_{k>=0} (1  1/(10^(2^k)+1)^r), where r>0 is a real number.
%F In particular,
%F sum_{n>=0} 1/a(n) = prod{k>=0} (1 + 1/(10^(2^k)+1)) = 1.10182034...;
%F sum_{n>=0} (1)^A000120(n)/a(n) = 0.9
%F a(2^n) = 10^(2^n)+1, n>=0.
%F Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations a(2^t*n+2^(t1)) = 99*(10^(2^(t1)+1))/(10^(2^(t1))1) * a(2^t*n+2^(t1)2), t>=2. In particular, for t=2,3,4, we have the following formulas:
%F a(4*n+2) = 101*a(4*n);
%F a(8*n+4) = 10001/101*a(8*n+2);
%F a(16*n+8)= 100000001/1010101*(16*n+6), etc. (End)
%e From _Daniel Forgues_, Jun 20 2011: (Start)
%e Terms as products of distinct Fermat numbers in binary representation (Cf. A080176 comment) (Cf. Sierpiński's triangle on OEIS Wiki):
%e a(0) = 1 = (empty product);
%e a(1) = 11 = F_0;
%e a(2) = 101 = F_1;
%e a(3) = 1111 = 11*101 = F_0*F_1;
%e a(4) = 10001 = F_2;
%e a(5) = 110011 = 11*10001 = F_0*F_2;
%e a(6) = 1010101 = 101*10001 = F_1*F_2;
%e a(7) = 11111111 = 11*101*10001 = F_0*F_1*F_2. (End)
%p A006943 := proc(n) local k; add((binomial(n,k) mod 2)*10^k, k=0..n); end;
%t f[n_] := FromDigits@ Mod[Binomial[n, Range[0, n]], 2]; Array[f, 17, 0] (* _Robert G. Wilson v_, Jun 26 2011 *)
%Y Cf. A080176 for Fermat numbers in binary representation.
%Y Cf. A001317 for the decimal representation of A006943.
%Y Cf. A249183.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_
%E More terms from _James A. Sellers_, Aug 21 2000
%E Edited by _Daniel Forgues_, Jun 20 2011
