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A076312
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a(n) = floor(n/10) + 2*(n mod 10).
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7
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0, 2, 4, 6, 8, 10, 12, 14, 16, 18, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 7, 9, 11, 13, 15, 17, 19, 21
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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 add twice this digit to the shortened number. - N. J. A. Sloane, May 25 2019
(n==0 modulo 19) iff (a(n)==0 modulo 19); applied recursively, this property provides a useful test for divisibility by 19.
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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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G.f.: -x(17x^9-2-2x-2x^2-2x^3-2x^4-2x^5-2x^6-2x^7-2x^8)/((1-x)^2(1+x)(x^4+x^3+x^2+x+1)(x^4-x^3+x^2-x+1)). a(n)=A059995(n)+2*A010879(n). [R. J. Mathar, Jan 24 2009]
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EXAMPLE
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26468 is not a multiple of 19, as 26468 -> 2646+2*8=2662 -> 266+2*2=270 -> 27+2*0=27=19*1+8, therefore the answer is NO.
Is 12882 divisible by 19? 12882 -> 1288+2*2=1292 -> 129+2*2=133 -> 13+2*3=19, therefore the answer is YES.
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MATHEMATICA
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f[n_]:=Module[{idn=IntegerDigits[n]}, FromDigits[Most[idn]]+2idn[[-1]]]; Array[ f, 80, 0] (* Harvey P. Dale, Mar 01 2020 *)
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PROG
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(Haskell)
a076312 n = n' + 2 * m where (n', m) = divMod n 10
(Magma) [Floor(n/10) + 2*(n mod 10): n in [0..100]]; // Vincenzo Librandi, Mar 05 2020
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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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