OFFSET
1,2
COMMENTS
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.
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).
Inspired by Eric Angelini's variant based on digits, cf. LINKS.
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).
LINKS
Eric Angelini, A proportion (odd digits vs all digits), personal blog "Cinquante signes" on blogspot.com, May 14 2020.
FORMULA
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, ...}.
EXAMPLE
There is a(1) = 1 odd term, namely: 1, among the first a(2) = 2 terms which are 1 and 2.
There are 2 odd terms, 1 and 3, among the first a(3) = 3 terms, 1, 2 and 3.
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.
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.
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.
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.
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.
And so on.
PROG
(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}
CROSSREFS
KEYWORD
nonn
AUTHOR
M. F. Hasler, May 18 2020
STATUS
approved