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

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

Offset: 1

Author

Seiichi Manyama, Oct 31 2018

Keywords

• nonn
• base

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

Examples

			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, ... .
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..1800

Crossrefs

Cf. A028355, A028359, A321232.

Programs

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

Formula

Conjectures from Chai Wah Wu, Apr 18 2024: (Start)

a(n) = 100001*a(n-9) - 100000*a(n-18) for n > 18.

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)