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A196279
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Let r= (7n) mod 10 and x=floor(7n/10) be the last digit and leading part of 7n. Then a(n) = (x-2r)/7.
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1
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0, -2, -1, 0, -2, -1, 0, -2, -1, 0, 1, -1, 0, 1, -1, 0, 1, -1, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, 4, 2, 3, 4, 2, 3, 4, 2, 3, 4, 5, 3, 4, 5, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 8, 9, 10, 8, 9
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OFFSET
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0,2
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COMMENTS
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).
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FORMULA
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Conjecture: a(n)= +a(n-1) +a(n-10) -a(n-11) with G.f. x*(-2 +x -2*x^3 +x^4 -2*x^6 +x^7 +x^9 +x^2 +x^5 +x^8) / ( (1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) *(x-1)^2 ). - R. J. Mathar, Oct 04 2011
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EXAMPLE
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Check to see if 273 is divisible by 7 : double the last digit 3*2=6 ; subtract that from the rest of the number 27-6=21 ; check to see if the difference is divisible by 7: 21/7 is divisible by 7, therefore 273 is also divisible by 7. 273=7*39 and 21=7*3 so a(39)=3.
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MAPLE
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r := (7*n) mod 10 ;
x := floor(7*n/10) ;
(x-2*r)/7 ;
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PROG
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CROSSREFS
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KEYWORD
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sign,base,easy
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AUTHOR
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STATUS
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approved
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