cp's OEIS Frontend

A321225 Break the infinite word 1232112321... into chunks so that the sum of the digits in the n-th chunk is n.

%I A321225 #37 Apr 18 2024 11:00:44
%S A321225 1,2,3,211,23,2112,3211,2321,12321,123211,232112,321123,21123211,
%T A321225 2321123,211232112,321123211,232112321,1232112321,12321123211,
%U A321225 23211232112,32112321123,2112321123211,232112321123,21123211232112,32112321123211,23211232112321,123211232112321
%N A321225 Break the infinite word 1232112321... into chunks so that the sum of the digits in the n-th chunk is n.
%C A321225 For every n=3*k, a(n) must be divisible by 3 and is therefore a palindrome. - _Ivan N. Ianakiev_, Nov 01 2018
%C A321225 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
%H A321225 Seiichi Manyama, <a href="/A321225/b321225.txt">Table of n, a(n) for n = 1..1800</a>
%F A321225 Conjectures from _Chai Wah Wu_, Apr 18 2024: (Start)
%F A321225 a(n) = 100001*a(n-9) - 100000*a(n-18) for n > 18.
%F A321225 G.f.: x*(10000*x^16 + 20000*x^15 + 30000*x^14 + 21100*x^13 + 23000*x^12 + 21120*x^11 + 32110*x^10 + 23210*x^9 + 12321*x^8 + 2321*x^7 + 3211*x^6 + 2112*x^5 + 23*x^4 + 211*x^3 + 3*x^2 + 2*x + 1)/(100000*x^18 - 100001*x^9 + 1). (End)
%e A321225 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, ... .
%o A321225 (PARI) getd(n) = {my(m = n % 5); if (!m, m = 1); [1, 2, 3, 2, 1][m];}
%o A321225 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
%Y A321225 Cf. A028355, A028359, A321232.
%K A321225 nonn,base
%O A321225 1,2
%A A321225 _Seiichi Manyama_, Oct 31 2018