cp's OEIS Frontend

A006943 Rows of Sierpiński's triangle (Pascal's triangle mod 2).

%I A006943 M4802 #119 Aug 01 2025 01:54:45
%S A006943 1,11,101,1111,10001,110011,1010101,11111111,100000001,1100000011,
%T A006943 10100000101,111100001111,1000100010001,11001100110011,
%U A006943 101010101010101,1111111111111111,10000000000000001,110000000000000011
%N A006943 Rows of Sierpiński's triangle (Pascal's triangle mod 2).
%C A006943 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 A006943 Rows 1 to 31 are the binary representation of the 31 (2^5-1) nonempty products of distinct Fermat primes, giving the number of sides of constructible (with straightedge and compass) odd-sided polygons. - _Daniel Forgues_, Jun 21 2011
%C A006943 Sierpiński's triangles typically refer to any finite triangle with rows 0 to 2^n-1 so as to get complete triangles, with n at least 4 so as to show the fractal-like 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
%C A006943 Also, binary representation of the n-th iteration of the "Rule 60" elementary cellular automaton starting with a single ON (black) cell. - _Robert Price_, Feb 21 2016
%C A006943 a(n) is the concatenation of the coefficients of (x+1)^n in GF(2)[x]. - _Thomas Anton_, Oct 04 2022
%C A006943 Subsequence of A057148. - _Chai Wah Wu_, Jul 30 2025
%D A006943 C. Pickover, Mazes for the Mind, St. Martin's Press, NY, 1992, p. 353.
%D A006943 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A006943 S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 55.
%H A006943 N. J. A. Sloane, <a href="/A006943/b006943.txt">Table of n, a(n) for n = 0..200</a>
%H A006943 Antti Karttunen, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Papers1/42-1/quartkarttunen01_2004.pdf">On Pascal's Triangle Modulo 2 in Fibonacci Representation</a>, Fibonacci Quarterly, 42 (2004), 38-46. (For Denton Hewgill's identity)
%H A006943 OEIS Wiki, <a href="/wiki/Sierpinski&#39;s_triangle">Sierpinski's triangle</a>
%H A006943 Vladimir Shevelev, <a href="http://arxiv.org/abs/1011.6083">On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization</a>, arXiv:1011.6083 [math.NT], 2010-2012; J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29.
%H A006943 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/ElementaryCellularAutomaton.html">Elementary Cellular Automaton</a>
%H A006943 Stephen Wolfram, <a href="http://wolframscience.com/">A New Kind of Science</a>
%H A006943 <a href="https://oeis.org/wiki/Index_to_Elementary_Cellular_Automata">Index to Elementary Cellular Automata</a>
%H A006943 <a href="/index/Ce#cell">Index entries for sequences related to cellular automata</a>
%F A006943 From _Daniel Forgues_, Jun 20-21 2011: (Start)
%F A006943 In the following formulas, [...]_2 means converted to base 2.
%F A006943 a(n) = [Sum_{i=0..n} (binomial(n,i) mod 2) 2^i]_2, n >= 0.
%F A006943 From row n, 0 <= n <= 2^k - 1, k >= 0, being
%F A006943   a(n) = [Product_{i=0..k-1} (F_i)^(alpha_i)]_2, alpha_i in {0, 1},
%F A006943 where for k = 0, we get the empty product, i.e., 1, giving a(0) = 1,
%F A006943 we induce from the triangle that row 2^k + n, 0 <= n <= 2^k - 1, is
%F A006943   a(2^k + n) = a(n)*[F_k]_2, k >= 0.
%F A006943 Denton Hewgill's identity (cf. links):
%F A006943 a(n) = [Product_{i>=0} (F_i)^(floor(n/2^i) mod 2)]_2, F_i = 2^(2^i)+1.
%F A006943 a(0) = 1; a(n) = [Product_{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 A006943 From _Vladimir Shevelev_, Dec 26-27 2013: (Start)
%F A006943 Sum_{n>=0} 1/a(n)^r = Product_{k>=0} (1 + 1/(10^(2^k)+1)^r),
%F A006943 Sum_{n>=0} (-1)^A000120(n)/a(n)^r = Product_{k>=0} (1 - 1/(10^(2^k)+1)^r), where r > 0 is a real number.
%F A006943 In particular,
%F A006943 Sum_{n>=0} 1/a(n) = Product_{k>=0} (1 + 1/(10^(2^k)+1)) = 1.10182034...;
%F A006943 Sum_{n>=0} (-1)^A000120(n)/a(n) = 0.9;
%F A006943 a(2^n) = 10^(2^n) + 1, n >= 0.
%F A006943 Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations a(2^t*n+2^(t-1)) = 99*(10^(2^(t-1)+1))/(10^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t >= 2. In particular, for t=2,3,4, we have the following formulas:
%F A006943   a(4*n+2)  = 101*a(4*n);
%F A006943   a(8*n+4)  = (10001/101)*a(8*n+2);
%F A006943   a(16*n+8) = (100000001/1010101)*(16*n+6), etc. (End)
%F A006943 From _Tom Edgar_, Oct 11 2015: (Start)
%F A006943 a(2*n+1) = 11*a(2*n).
%F A006943 a(n) = Product_{b_j != 0} a(2^j) where n = Sum_{j>=0} b_j*2^j is the binary representation of n.
%F A006943 (End)
%e A006943 From _Daniel Forgues_, Jun 20 2011: (Start)
%e A006943 Terms as products of distinct Fermat numbers in binary representation (Cf. A080176 comment) (Cf. Sierpiński's triangle on OEIS Wiki):
%e A006943 a(0) = 1 = (empty product);
%e A006943 a(1) = 11 = F_0;
%e A006943 a(2) = 101 = F_1;
%e A006943 a(3) = 1111 = 11*101 = F_0*F_1;
%e A006943 a(4) = 10001 = F_2;
%e A006943 a(5) = 110011 = 11*10001 = F_0*F_2;
%e A006943 a(6) = 1010101 = 101*10001 = F_1*F_2;
%e A006943 a(7) = 11111111 = 11*101*10001 = F_0*F_1*F_2. (End)
%p A006943 A006943 := proc(n) local k; add((binomial(n,k) mod 2)*10^k, k=0..n); end;
%t A006943 f[n_] := FromDigits@ Mod[Binomial[n, Range[0, n]], 2]; Array[f, 17, 0] (* _Robert G. Wilson v_, Jun 26 2011 *)
%o A006943 (Python)
%o A006943 def A006943(n): return sum((bool(~n&n-k)^1)*10**k for k in range(n+1)) # _Chai Wah Wu_, May 03 2023
%Y A006943 Cf. A001317 (decimal representation).
%Y A006943 Cf. A080176 (Fermat numbers in binary).
%Y A006943 Cf. A249183, A057148.
%K A006943 nonn,easy,base
%O A006943 0,2
%A A006943 _N. J. A. Sloane_
%E A006943 More terms from _James Sellers_, Aug 21 2000
%E A006943 Edited by _Daniel Forgues_, Jun 20 2011