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
From R. J. Mathar, Feb 08 2008: (Start)
O.g.f.: 2 + (3*x-2)/(x^2-x+1).
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) = (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
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