%I A076312
%S A076312 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,
%T A076312 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,
%U A076312 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 A076312 floor(n/10) + 2*(n mod 10).
%C A076312 (n==0 modulo 19) iff (a(n)==0 modulo 19); applied recursivly, this property 
               provides a divisibility test for numbers given in base 10 notation.
%D A076312 Karl Menninger, Rechenkniffe, Vandenhoeck & Ruprecht in Goettingen (1961), 
               79A.
%F A076312 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). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Jan 24 2009]
%e A076312 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 A076312 is 12882 divisible by 19? 12882 -> 1288+2*2=1292 -> 129+2*2=133 -> 13+2*3=19, 
               therefore the answer is YES.
%Y A076312 Cf. A008601, A076309, A076310, A076311.
%Y A076312 Sequence in context: A076309 A088133 A115299 this_sequence A061762 A136614 
               A097586
%Y A076312 Adjacent sequences: A076309 A076310 A076311 this_sequence A076313 A076314 
               A076315
%K A076312 nonn
%O A076312 0,2
%A A076312 Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Oct 06 2002