

A258083


Smallest multiple of 3 not appearing earlier that ends with n.


6



21, 12, 3, 24, 15, 6, 27, 18, 9, 210, 111, 312, 213, 114, 315, 216, 117, 318, 219, 120, 321, 222, 123, 324, 225, 126, 327, 228, 129, 30, 231, 132, 33, 234, 135, 36, 237, 138, 39, 240, 141, 42, 243, 144, 45, 246, 147, 48, 249, 150, 51, 252, 153, 54, 255, 156
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OFFSET

1,1


COMMENTS

a(10*n) = 10*a(n).
The sequence is a permutation of the positive multiples of 3.  Vladimir Shevelev, May 24 2015
24 ends with 24 but as 24 is the least element divisible by 3 that ends with 4, it's already used and a(24) = 324. Generally, a(n) = k * 10^d + n with k in {0,1,2,3} and d the number of digits of n.  David A. Corneth, May 24 2015


LINKS



MATHEMATICA

a[n_] := a[n] = For[k=3, True, k=k+3, If[Divisible[kn, 10^IntegerLength[n] ] && FreeQ[Array[a, n1], k], Return[k]]]; Array[a, 56] (* JeanFrançois Alcover, Feb 06 2018 *)


PROG

(Haskell)
import Data.List (isPrefixOf, delete)
a258083 n = a258083_list !! (n1)
a258083_list = f 1 $ tail $ zip
a008585_list $ map (reverse . show) a008585_list where
f x ws = g ws where
g ((u, vs) : uvs) = if isPrefixOf xs vs
then u : f (x + 1) (delete (u, vs) ws) else g uvs
xs = reverse $ show x
(PARI) a(n) = {my(d = digits(n), s = vecsum(d), k); if(s%3 > 0, k = (3  s%3)%3, i=1; while(i < #d && d[i] == 3, i++); if(i<#d && d[i+1] >= 1 && d[i]1 == bitand(d[i]1, 1), k = 3)); k*10^#d + n} \\ David A. Corneth, May 24 2015


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



