%I A119910
%S A119910 1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,
%T A119910 2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,3,2,1,
%U A119910 3,2,1,3,2,1,3,2,1,3,2,1,3,2
%V A119910 1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,
               -3,-2,1,3,2,-1,
%W A119910 -3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,
               2,-1,-3,-2,1,3,
%X A119910 2,-1,-3,-2,1,3,2,-1,-3,-2,1,3,2,-1,-3,-2
%N A119910 Simple periodic sequence with period 1, 3, 2, -1, -3, -2.
%C A119910 Take any of term, multiply it to units place digit of any taken no. then 
               save the product, then take the next term of this sequence, multiply 
               it to the next place digit of the taken no., add the product to previous 
               one and save it, then take the next term of the sequence, multiply 
               it to the next place digit of the taken no. and add it to the previous 
               sum, keep on doing this until all the digits of the taken no. are 
               done, now if the calculated sum is divisible by `7`, then the initial 
               number taken must also be comletely divisible by seven, otherwise 
               not.
%C A119910 Can be converted into the sequence "10^n mod 7", 1) 1,3,2,6,4,5,1,3,2,
               6,4,5,1,3,2,6,4,5,1,3,2,6,4,5 .... 2) -6,-4,-5,6,4,5,-6,-4,-5,6,4,
               5,-6,-4,-5,6,4,5 ... 3) -6,-4,-5,-1,-3,-2,-6,-4,-5,-1,-3,-2,-6,-4,
               -5,-1,-3,-2 ... Many variations can be made by adding or subtracting 
               7 from any term of the previous sequences. Still the divisibility 
               rule will be valid.
%H A119910 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A119910 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%F A119910 a(n)=(1/6)*{-3*(n mod 6)-[(n+1) mod 6]+2*[(n+2) mod 6]+3*[(n+3) mod 4]+[(n+4) 
               mod 4]-2*[(n+5) mod 4]} - Paolo P. Lava (ppl(AT)spl.at), Nov 21 2006
%F A119910 O.g.f.: 2+(3*x-2)/(x^2-x+1) . a(n) = 3*A010892(n-1)-2*A010892(n). a(n) 
               = -a(n-3). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Feb 08 
               2008
%F A119910 a(n)=a(n-1)-a(n-2)for n>2, a(1)=1, a(2)=3. [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), 
               Nov 16 2008]
%F A119910 Closed form. a(n)=(1/2)*[(1/2)-(1/2)*I*sqrt(3)]^n+(1/2)*[(1/2)+(1/2)*I*sqrt(3)]^n+(5/
               6)*I*[(1/2)-(1/2)*I *sqrt(3)]^n*sqrt(3)-(5/6)*I*[(1/2)+(1/2)*I*sqrt(3)]^n*sqrt(3), 
               with n>=0 [From Paolo P. Lava (ppl(AT)spl.at), Nov 19 2008]
%e A119910 a(32)=?: 32%7=4, therefore a(32)=-1.
%e A119910 Let us test the divisibility of 342 with the series:
%e A119910 Take 1 from the sequence, multiply it by 2, the product is 2,
%e A119910 take 3 from the sequence, multiply it by 4, the product is 12,
%e A119910 take 2 from the sequence, multiply it by 3, the product is 6,
%e A119910 the sum of the products is 2 + 12 + 6 = 20,
%e A119910 because 20 is not divisible by 7, therefore 342 will also not be.
%Y A119910 Cf. A033940.
%Y A119910 Sequence in context: A130827 A070309 A130784 this_sequence A138034 A087818 
               A112746
%Y A119910 Adjacent sequences: A119907 A119908 A119909 this_sequence A119911 A119912 
               A119913
%K A119910 sign
%O A119910 1,2
%A A119910 Kartikeya Shandilya (kartikeya.shandilya(AT)gmail.com), May 28 2006