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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
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
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)
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
KEYWORD
nonn,base
AUTHOR
Seiichi Manyama, Oct 31 2018
STATUS
approved