Search: id:A033940
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%I A033940
%S A033940 1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,
               5,
%T A033940 1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,
%U A033940 5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6
%N A033940 10^n mod 7.
%C A033940 This sequence can be employed in a test for divisibility by seven. Given 
               the decimal expansion of some natural number, it is easily shown 
               that the following sum has the same remainder under division by seven 
               as the original number and that this sum is stricly smaller than 
               the original number: Successively take the digits of the number in 
               reverse order and multiply each of them by the respective term of 
               the sequence A033940, then sum the products. By repeating this process, 
               since the sums decrease in size, one ends up with seven if and only 
               if the initial number is divisible by seven. Example: 43638 is divisible 
               by seven since 8*1 + 3*3 + 6*2 + 3*6 + 4*4 = 63 and 3*1 + 6*3 = 21 
               and 1*1 + 2*3 = 7. - Peter C. Heinig (algorithms(AT)gmx.de), Apr 
               16 2007
%C A033940 Contribution from Eric Desbiaux (moongerms(AT)wanadoo.fr), Feb 15 2009: 
               Representation of (3^n) in the circle with seven equidistant points, 
               (10^n) mod 7=(3^n) mod 7,
%C A033940 Representation of multiples of 3 in the circle (with seven equidistant 
               points), see the Chryzodes links. - Eric Desbiaux (moongerms(AT)wanadoo.fr), 
               Feb 14 2009
%C A033940 Equivalently 3^n mod 7. [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), 
               Nov 24 2009]
%H A033940 Author?, <a href="http://www.chryzode.org/fr/ligne.htm">Chryzodes</a>
%H A033940 Author?, <a href="http://www.chryzode.org/fr/ligne.htm">Chryzodes "3in7"</
               a>
%H A033940 Author?, <a href="http://www.chryzode.org/">Chryzodes</a>
%F A033940 a(n)=a(n-1)-a(n-3)+a(n-4) = a(n-6). G.f.: (1+2x-x^2+5^x3)/((1-x)(1+x)(1-x+x^2)). 
               a(n)=7/2 -7*(-1)^n/6 -4*A010892(n)/3-A010892(n-1)/3. [From R. J. 
               Mathar (mathar(AT)strw.leidenuniv.nl), Feb 13 2009]
%o A033940 (Other) 1.)sage: [power_mod(10,n, 7)for n in xrange(0,106)] # 2.)sage: 
               [power_mod(3, n, 7)for n in xrange(0,106)] # [From Zerinvary Lajos 
               (zerinvarylajos(AT)yahoo.com), Nov 24 2009]
%Y A033940 Sequence in context: A057050 A123042 A121647 this_sequence A106409 A115510 
               A070264
%Y A033940 Adjacent sequences: A033937 A033938 A033939 this_sequence A033941 A033942 
               A033943
%K A033940 nonn,new
%O A033940 0,2
%A A033940 Jeff Burch (gburch(AT)erols.com)

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