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A076310
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a(n) = floor(n/10) + 4*(n mod 10).
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7
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0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 1, 5, 9, 13, 17, 21, 25, 29, 33, 37, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 5, 9, 13, 17, 21, 25, 29, 33, 37, 41, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 7, 11, 15, 19, 23
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OFFSET
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0,2
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COMMENTS
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(n==0 modulo 13) iff (a(n)==0 modulo 13); applied recursively, this property provides a divisibility test for numbers given in base 10 notation.
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REFERENCES
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Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.
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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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a(n) = +a(n-1) +a(n-10) -a(n-11). G.f.: -x*(-4-4*x-4*x^2-4*x^3-4*x^4-4*x^5-4*x^6-4*x^7-4*x^8+35*x^9) / ( (1+x) *(x^4+x^3+x^2+x+1) *(x^4-x^3+x^2-x+1) *(x-1)^2 ). - R. J. Mathar, Feb 20 2011
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EXAMPLE
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435598 is not a multiple of 13, as 435598 -> 43559+4*8=43591 -> 4359+4*1=4363 -> 436+4*3=448 -> 44+4*8=76 -> 7+4*6=29=13*2+3, therefore the answer is NO.
Is 8424 divisible by 13? 8424 -> 842+4*4=858 -> 85+4*8=117 -> 11+4*7=39=13*3, therefore the answer is YES.
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MAPLE
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MATHEMATICA
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Table[Floor[n/10] + 4*Mod[n, 10], {n, 0, 100}] (* Wesley Ivan Hurt, Jan 30 2014 *)
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 1}, 80] (* Harvey P. Dale, Sep 30 2015 *)
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PROG
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(Haskell)
a076310 n = n' + 4 * m where (n', m) = divMod n 10
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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