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A076309
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a(n) = floor(n/10) - 2*(n mod 10).
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8
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0, -2, -4, -6, -8, -10, -12, -14, -16, -18, 1, -1, -3, -5, -7, -9, -11, -13, -15, -17, 2, 0, -2, -4, -6, -8, -10, -12, -14, -16, 3, 1, -1, -3, -5, -7, -9, -11, -13, -15, 4, 2, 0, -2, -4, -6, -8, -10, -12, -14, 5, 3, 1, -1, -3, -5, -7, -9, -11, -13, 6, 4, 2, 0, -2, -4, -6, -8, -10
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OFFSET
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0,2
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COMMENTS
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Delete the last digit from n and subtract twice this digit from the shortened number. - N. J. A. Sloane, May 25 2019
(n==0 modulo 7) iff (a(n)==0 modulo 7); applied recursively, this property provides a useful test for divisibility by 7.
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REFERENCES
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Erdős, Paul, and János Surányi. Topics in the Theory of Numbers. New York: Springer, 2003. Problem 6, page 3.
Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.
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LINKS
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FORMULA
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a(n) = a(n-1) + a(n-10) - a(n-11).
G.f.: x*(-2 -2*x -2*x^2 -2*x^3 -2*x^4 -2*x^5 -2*x^6 -2*x^7 -2*x^8 +19*x^9)/((1+x)*(x^4-x^3+x^2-x+1)*(x^4+x^3+x^2+x+1)*(x-1)^2). (End)
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EXAMPLE
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695591 is not a multiple of 7, as 695591 -> 69559-2*1=69557 -> 6955-2*7=6941 -> 694-2*1=692 -> 69-2*2=65=7*9+2, therefore the answer is NO.
Is 3206 divisible by 7? 3206 -> 320-2*6=308 -> 30-2*8=14=7*2, therefore the answer is YES, indeed 3206=2*7*229.
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MATHEMATICA
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Table[Floor[n/10] - 2*Mod[n, 10], {n, 0, 100}] (* G. C. Greubel, Apr 07 2016 *)
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PROG
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(Haskell)
a076309 n = n' - 2 * m where (n', m) = divMod n 10
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CROSSREFS
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KEYWORD
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sign
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AUTHOR
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STATUS
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approved
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