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 A119910 Period 6: repeat [1, 3, 2, -1, -3, -2]. 9
 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, -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, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS 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 completely divisible by seven, otherwise not. 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. Nonsimple continued fraction of (6+2*sqrt(2))/7 = 1.26120387... - R. J. Mathar, Mar 08 2012 LINKS Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (1,-1). FORMULA a(n) = (1/6)*(-3*(n mod 6) - ((n+1) mod 6) + 2*((n+2) mod 6) + 3*((n+3) mod 6) + ((n+4) mod 6) - 2*((n+5) mod 6)). - Paolo P. Lava, Nov 21 2006; corrected by Bruno Berselli, Sep 27 2010 From R. J. Mathar, Feb 08 2008: (Start) 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) for n>3. (End) a(n) = a(n-1) - a(n-2) for n>2. - Philippe Deléham, Nov 16 2008 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. - Paolo P. Lava, Nov 19 2008 a(n) = (4*sqrt(3)*sin(n*Pi/3) - 6*cos(n*Pi/3))/3. - Wesley Ivan Hurt, Jun 19 2016 EXAMPLE a(32)=?: 32%7=4, therefore a(32)=-1. Let us test the divisibility of 342 with the series: Take 1 from the sequence, multiply it by 2, the product is 2, take 3 from the sequence, multiply it by 4, the product is 12, take 2 from the sequence, multiply it by 3, the product is 6, the sum of the products is 2 + 12 + 6 = 20, because 20 is not divisible by 7, therefore 342 will also not be. MAPLE A119910:=n->[1, 3, 2, -1, -3, -2][(n mod 6)+1]: seq(A119910(n), n=0..100); # Wesley Ivan Hurt, Jun 19 2016 MATHEMATICA PadRight[{}, 100, {1, 3, 2, -1, -3, -2}] (* Wesley Ivan Hurt, Jun 19 2016 *) PROG (Magma) &cat[[1, 3, 2, -1, -3, -2]^^20]; // Wesley Ivan Hurt, Jun 19 2016 CROSSREFS Cf. A010892, A033940. Sequence in context: A353748 A236966 A280048 * A130784 A138034 A229216 Adjacent sequences:  A119907 A119908 A119909 * A119911 A119912 A119913 KEYWORD sign,easy AUTHOR Kartikeya Shandilya (kartikeya.shandilya(AT)gmail.com), May 28 2006 EXTENSIONS New name from Omar E. Pol, Oct 31 2013 STATUS approved

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Last modified September 26 18:11 EDT 2022. Contains 357002 sequences. (Running on oeis4.)