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 A076312 a(n) = floor(n/10) + 2*(n mod 10). 7
 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. REFERENCES 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. LINKS Reinhard Zumkeller, Table of n, a(n) for n = 0..10000 Eric Weisstein's World of Mathematics, Divisibility Tests. Wikipedia, Divisibility rule FORMULA 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, Jan 24 2009] EXAMPLE 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. PROG (Haskell) a076312 n =  n' + 2 * m where (n', m) = divMod n 10 -- Reinhard Zumkeller, Jun 01 2013 CROSSREFS Cf. A008601, A076309, A076310, A076311. Sequence in context: A076309 A088133 A115299 * A061762 A136614 A245627 Adjacent sequences:  A076309 A076310 A076311 * A076313 A076314 A076315 KEYWORD nonn AUTHOR Reinhard Zumkeller, Oct 06 2002 STATUS approved

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Last modified November 13 02:59 EST 2019. Contains 329085 sequences. (Running on oeis4.)