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 A321225 Break the infinite word 1232112321... into chunks so that the sum of the digits in the n-th chunk is n. 2
 1, 2, 3, 211, 23, 2112, 3211, 2321, 12321, 123211, 232112, 321123, 21123211, 2321123, 211232112, 321123211, 232112321, 1232112321, 12321123211, 23211232112, 32112321123, 2112321123211, 232112321123, 21123211232112, 32112321123211, 23211232112321, 123211232112321 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS For every n=3*k, a(n) must be divisible by 3 and is therefore a palindrome. - Ivan N. Ianakiev, Nov 01 2018 For every n=3*k, the number of digits of a(n) equals he number of digits of a(n-9)+5 and the starting/ending digits of a(n) and a(n-9) are the same. For any possible natural number m, there are five possible candidate numbers for a(3*k) that are of length m, of which only one, the palindrome, is divisible by 3. - Ivan N. Ianakiev, Nov 02 2018 LINKS Seiichi Manyama, Table of n, a(n) for n = 1..1800 EXAMPLE 1, 2, 3, 2+1+1=4, 2+3=5, 2+1+1+2=6, 3+2+1+1=7, 2+3+2+1=8, 1+2+3+2+1=9, ... . PROG (PARI) getd(n) = {my(m = n % 5); if (!m, m = 1); [1, 2, 3, 2, 1][m]; } lista(nn) = {my(k = 1); for (n=1, nn, my (s = 0, list = List(), d); while (s != n, d = getd(k); listput(list, d); s += d; k++; ); print1(fromdigits(Vec(list)), ", "); ); } \\ Michel Marcus, Nov 11 2018 CROSSREFS Cf. A028355, A028359, A321232. Sequence in context: A195264 A316941 A037274 * A037275 A142960 A117324 Adjacent sequences:  A321222 A321223 A321224 * A321226 A321227 A321228 KEYWORD nonn,base AUTHOR Seiichi Manyama, Oct 31 2018 STATUS approved

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Last modified October 14 14:31 EDT 2019. Contains 328019 sequences. (Running on oeis4.)