1,2

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.

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.

Index entries for sequences related to linear recurrences with constant coefficients

Tanya Khovanova, Recursive Sequences

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

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

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]

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]

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.

Cf. A033940.

Sequence in context: A130827 A070309 A130784 this_sequence A138034 A087818 A112746

Adjacent sequences: A119907 A119908 A119909 this_sequence A119911 A119912 A119913

sign

Kartikeya Shandilya (kartikeya.shandilya(AT)gmail.com), May 28 2006