A230764 Length of chunks of even digits, delimited by an odd digit, in the concatenation of all terms; smallest such sequence without repeated terms.

1, 2, 3, 4, 6, 5, 8, 20, 7, 22, 24, 21, 26, 28, 9, 40, 42, 44, 23, 46, 48, 60, 62, 64, 66, 68, 80, 82, 25, 84, 86, 88, 10, 200, 202, 204, 206, 208, 220, 201, 222, 224, 226, 228, 240, 242, 244, 203, 246, 248, 260, 262, 264, 266, 205, 268, 280, 282, 284, 286

Offset: 1

Author

Eric Angelini and M. F. Hasler, Oct 29 2013

Keywords

• nonn
• base

Comments

Cut the sequence of digits of the concatenation of all terms after each odd digit: 1|23|465|8207|222421|26289|40424423|46..., then the length of the segments is given by the sequence itself. This defines the sequence, with the additional requirements that no number can appear more than once, and that the smallest possible term must always be chosen.

There cannot be two odd digits in a row, which excludes many numbers such as 11, 13, 15, .... Moreover, after the initial terms, no "small" numbers with more than one odd digit can be present. Therefore this cannot be a permutation of the integers, and numbers with odd digit(s) (and, in particular, primes) become increasingly sparse.

See A230765 for the "dual" version where chunks of odd digits are delimited by an even digit.

See A230763 for a version where, instead of digits, the English spelling of the terms is considered.

Links

• E. Angelini, Cut the sequence after it's odd/even digits, SeqFan List, Oct 29 2013

Programs

PARI

A230764_vec(n, a=[], s=[],u=0)={my(check(a, s)=my(L=0, j=1); !for(i=1, #s, ( i>L+a[j] || (( i==L+a[j] )!= bittest(s[i],0))) && return; bittest(s[i],0) || next; L=i; j++)); check(a, s)||return; n-- || return(a); my(t); for(k=1, 9e9, bittest(u,k)&&next; (t=A230764_vec(n, concat(a, k), concat(s, digits(k)), u+2^k))&& return(t))}