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%I #26 Sep 08 2022 08:45:07
%S 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,
%T 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,
%U 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
%N a(n) = floor(n/10) + 2*(n mod 10).
%C Delete the last digit from n and add twice this digit to the shortened number. - _N. J. A. Sloane_, May 25 2019
%C (n==0 modulo 19) iff (a(n)==0 modulo 19); applied recursively, this property provides a useful test for divisibility by 19.
%D Erdős, Paul, and János Surányi. Topics in the Theory of Numbers. New York: Springer, 2003. Problem 6, page 3.
%D Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 79A.
%H Reinhard Zumkeller, <a href="/A076312/b076312.txt">Table of n, a(n) for n = 0..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DivisibilityTests.html">Divisibility Tests</a>.
%H Wikipedia, <a href="http://en.wikipedia.org/wiki/Divisibility_rule">Divisibility rule</a>
%F 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]
%e 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.
%e Is 12882 divisible by 19? 12882 -> 1288+2*2=1292 -> 129+2*2=133 -> 13+2*3=19, therefore the answer is YES.
%t f[n_]:=Module[{idn=IntegerDigits[n]},FromDigits[Most[idn]]+2idn[[-1]]]; Array[ f,80,0] (* _Harvey P. Dale_, Mar 01 2020 *)
%o (Haskell)
%o a076312 n = n' + 2 * m where (n', m) = divMod n 10
%o -- _Reinhard Zumkeller_, Jun 01 2013
%o (Magma) [Floor(n/10) + 2*(n mod 10): n in [0..100]]; // _Vincenzo Librandi_, Mar 05 2020
%Y Cf. A008601, A076309, A076310, A076311.
%K nonn
%O 0,2
%A _Reinhard Zumkeller_, Oct 06 2002
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