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Every number starts and ends with 2 and contains only twos and ones.

Removing the last digit produces sequence A304272 of the largest even integers in base 3/2.

The value of this sequence in base 10 is A304025.

When adding 1 to the value of this sequence we get A070885.

The largest integer with a given number of digits in base 3/2 can be produced directly from the smallest number, sequence A304023, by replacing 21 at the beginning and 0 at the end with 2, and by shifting the rest up by 1, see sequence A304023.

A070885 is the smallest integer that can be written with n digits in base 3/2.

This sequence represented in base 3/2 is A304024.

a(n) is a prefix of a(n+1).

The largest, not necessarily even, integer in base 3/2 with n digits is a(n-1) with 2 added at the end.

Is this the same as A072493 with its first 8 terms removed? See also the similar conjecture concerning A005428 and A073941.

From Petros Hadjicostas, Jul 20 2020: (Start)

We describe the counting-off game of Burde (1987) using language from Schuh (1968). Suppose m people are labeled with the numbers 1 through m (say clockwise). (Burde uses the numbers 0 through m-1 probably because he relates this problem to the representation of m in the fractional base k/(k-1) = 4/3. He actually modifies the (4/3)-representation of m to include negative coefficients. See the coefficients f(n;k) below.)

Suppose we start the counting at the person labeled 1, and we remove every 4th person. This sequence gives those numbers m for which the last survivor is one of the first three people.

When m = 5, 9, 12, 16, 218, 517, ... the last survivor is the first person.

When m = 7, 29, 69, 92, 291, 388, ... the last survivor is the second person.

When m = 22, 39, 52, 123, 164, 690, ... the last survivor is the third person.

If we know m = a(n) and the number, say i(n), of the last survivor (when there are a(n) people on the circle), we may find a(n+1) and the number i(n+1) of the new last survivor (when there are a(n+1) people on the circle) in the following way:

(a) If 0 = a(n) mod 3, then a(n+1) = (4/3)*a(n), and i(n+1) = i(n).

(b) If 1 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.

(c) If 1 = a(n) mod 3 and i(n) = 2, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1.

(d) If 1 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 2.

(e) If 2 = a(n) mod 3 and i(n) = 1, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 2.

(f) If 2 = a(n) mod 3 and i(n) = 2, then a(n+1) = ceiling((4/3)*a(n)) and i(n+1) = 3.

(g) If 2 = a(n) mod 3 and i(n) = 3, then a(n+1) = floor((4/3)*a(n)) and i(n+1) = 1. (End)

From Petros Hadjicostas, Jul 22 2020: (Start)

In general, for k >= 2, it seems that when m people are placed on a circle, labeled 1 through m, and every k-th person is removed (starting the counting at person 1), we may determine those m for which the last survivor is in {1, 2, ..., k-1} in the following way.

Define the sequence (T(n;k): n >= 1) by T(n;k) = ceiling(Sum_{s=1..n-1} T(s;k)/(k-1)) for n >= 2 starting with T(1; k) = 1. Then the list of those m's for which the last survivor is in {1, 2, ..., k-1} consists of all the numbers T(n;k) >= k (thus, we exclude the cases m = 1, ..., k-1 that may be repeated more than once in the sequence (T(n;k): n >= 1)).

I do not have a general proof of this conjecture though I strongly believe that Schuh's (1968) way of solving the case k = 3 (see pp. 373-375 and 377-379, where he provides two methods of solution) may provide clues for proving the conjecture.

We have T(n; k=2) = A011782(n+1), T(n; k=3) = A073941(n), T(n; k=4) = A072493(n), T(n; k=5) = A120160(n), T(n; k=6) = A120170(n), T(n; k=7) = A120178(n), T(n; k=8) = A120186(n), T(n; k=9) = A120194(n), and T(n; k=10) = A120202(n).

We also have T(n+1;k) = floor((k/(k-1))*T(n;k)) or ceiling((k/(k-1)*T(n;k)).

To identify the last survivor that results when we place T(n; k) people on the circle (with T(n;k) >= k) in the above Josephus problem, we use a modification of Burde's algorithm due to Thériault (2000).

We use the following recursions but we start at T(k;k) (rather than at the smallest n for which T(n;k) >= k). Define the sequence (S(n;k): n >= 1) by S(n;k) = T(n+k-1; k) for n >= 1. (It is easy to prove that S(1;k) = T(k;k) = 1.)

Define also the sequences (j(n;k): n >= 1) and (f(n;k): n >= 1) by j(1;k) = 1, f(1;k) = 0, f(n+1;k) = ((j(n;k) - S(n;k) - 1) mod (k-1)) + 1 - j(n;k) and j(n+1;k) = j(n;k) + f(n+1;k) for n >= 2.

Then for all n s.t. S(n;k) >= k, j(n;k) is the number of the last survivor of the Josephus problem where every k-th person is removed (provided we start the counting at number 1). It will always be the case that j(n;k) is in {1,2,...,k-1}.

We actually have S(n+1; k) = (k*S(n;k) + f(n+1;k))/(k-1) for n >= 1.

Notice that the Burde-Thériault algorithm is a generalization of Schuh's method. (End)

This sequence exists since the smallest even integers (see A303500) are prefixes of each other.

Apparently a variant of A205083. - R. J. Mathar, Jun 09 2018

Excluding 0, every term starts with 2 and has exactly one 2.

The last digit is always zero.

Removing the last digit produces the sequence A303500 of the smallest even integers in base 3/2.

The value of this sequence in base 10 is A070885.

When subtracting 1 from the value of this sequence we get A304025.

The largest integer with a given number of digits in base 3/2 can be produced directly from this sequence by replacing 21 at the beginning and 0 at the end with 2, and by shifting the rest up by 1, see sequence A304024.

This sequence is possible due to the fact that the largest even integers are prefixes of each other.

A304272(n) is the largest even integer with n digits.