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%I #16 Mar 14 2021 15:37:47
%S 1,2,3,5,8,11,14,17,21,25,29,33,37,41,45,49,53,57,61,65,70,75,81,87,
%T 93,99,105,111,117,123,129,135,141,147,153,159,165,171,177,183,189,
%U 195,201,207,213,219,225,231,237,243,249,255,261,267,273,279,285,291,297,303,309,315,321,327
%N Lexicographically earliest sequence with a(n) odd terms among the first a(n+1) terms, for any n; a(1) = 1, a(2) = 2.
%C Without the requirement a(2) = 2 (or: increasing, or: without repeated terms), the trivial sequence (1, 1, 1, ...) = A000012 would also satisfy the definition, repeatedly stating that the first term is odd, ignoring all other terms. But a(2) - a(1) > 0 implies that the sequence will be strictly increasing throughout, and therefore make infinitely many nontrivial statements concerning all of its terms.
%C The formula comes from the fact that if we have a(n-1) odd terms among the first a(n) terms, and we need a(n) odd terms among the first a(n+1) terms, then we must use exactly a(n) - a(n-1) odd terms after the a(n)-th terms up to and including the a(n+1)-th terms, whence a(n+1) >= a(n) + a(n) - a(n-1).
%C Inspired by _Eric Angelini_'s variant based on digits, cf. LINKS.
%C First differences are: 1*2, 2*1, 3*4, 4*12, 5*2, 6*47, 7*2, 8*9, 9*2, 10*281, 11*2, 12*13, 13*2, 14*71, 15*2, 16*17, 17*2, 18*..., where a*b means that a is repeated b times (i.e., the gap of 10 appears 281 times in a row).
%H Eric Angelini, <a href="https://cinquantesignes.blogspot.com/2020/05/a-proportion-odd-digits-vs-all-digits.html">A proportion (odd digits vs all digits)</a>, personal blog "Cinquante signes" on blogspot.com, May 14 2020.
%F a(n+1) >= 2*a(n) - a(n-1) for all n > 1, with equality for n not in {3, 4, 8, 20, 22, 69, 71, 80, 82, ...}.
%e There is a(1) = 1 odd term, namely: 1, among the first a(2) = 2 terms which are 1 and 2.
%e There are 2 odd terms, 1 and 3, among the first a(3) = 3 terms, 1, 2 and 3.
%e Then there must be a(3) = 3 odd terms among the first a(4) terms. Since there are only 2 odd terms coming earlier, there must come at least 1 more odd terms. This excludes a(4) = 4, but makes the next smaller choice a(4) = 5 possible, provided it will be followed by an even term in order to satisfy the requirement that there be exactly 3 odd terms among the first 5 terms.
%e Knowing that a(5) must be even (see just above), and that a(5) >= 2 a(4) - a(3) = 7 (see formula, explained in comments), we have as smallest possibility a(5) = 8, which does not raise a contradiction.
%e Then again we can use a(6) = 2*8 - 5 = 11, least possibility according to the formula, and indeed not yet leading to a contradiction.
%e Then again a(7) = 2*11 - 8 = 14 does not raise a contradiction, nor does a(8) = 2*14 - 11 = 17, since we can see that it will be possible to have 11 odd terms among the first 14, and 14 odd terms among the first 17.
%e However, for a(9) >= 2*17 - 14 = 20 we see that 20 would not make it possible to satisfy the requirement of 8 odd terms among the first 11. Indeed, we have only 5 odd terms so far, so all of a(9), a(10) and a(11) must be odd. So the smallest possibility is a(9) = 20.
%e And so on.
%o (PARI) upto(N)={my(a=Vec([1,2],N), s=1, o=Vec([1,1],N), m=1); for(n=3,N, a[n]=a[n-1]*2-a[n-2]; while( m<n && a[m] <n, m++); until( !a[n]++, my(ns=s+a[n]%2); o[n]=ns; for(i=m,n, a[i] <= n && next(( a[i-1] != o[a[i]] )+1); o[n]+a[i]-n >= a[i-1] || next(2)); s=ns; break));a}
%Y Cf. A332070 (variant based on digits).
%K nonn
%O 1,2
%A _M. F. Hasler_, May 18 2020
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