

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
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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

Table of n, a(n) for n=1..84.
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
O.g.f.: 2+(3*x2)/(x^2x+1). a(n) = 3*A010892(n1)2*A010892(n). a(n) = a(n3) for n>3.  R. J. Mathar, Feb 08 2008
a(n) = a(n1)  a(n2) 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: A287556 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

Corrected modular formula by Bruno Berselli, Sep 27 2010
New name from Omar E. Pol, Oct 31 2013


STATUS

approved



