cp's OEIS Frontend

There is a proviso with this: the first digit must be one less than the base; otherwise we could claim that a number was in an arbitrarily large base.

The maximum value will always be found by following the initial digit with a string of 1s, which is n candidate values for a(n).

The sequence is constructed so that after the initial two 0's the next three pairs form a self-contained block beginning with 3, the subsequent nine pairs form a self-contained block beginning with 9, the following twenty-seven pairs form a block beginning with 27, etc. (powers of 3: A000244).

There are numerous sequences that satisfy the given criteria, so to fully define the continuation of this sequence I will add the following extra constraints. After the 0th block 0,0 the n-th block is found as follows: The block can be split into two halves such that one occurrence of each number appears in each half. The first half of the n-th block begins with 3^n then increases by consecutive integers until the maximum for that block: (3^(n+1) - 3)/2, before abruptly dropping to (3^n - 1)/2 and increasing by consecutive integers until (3^n - 1) is reached. The second half of the n-th block is then defined by the original constraints.

For a given n, for all k, k^n mod a(n) will always be either 0, 1 or a(n)-1. This will not be true for numbers larger than a(n).

It appears that a(m) = 4 for m in A045979. - Michel Marcus, Sep 04 2020

If the sum of the two numbers above in the triangular array is not a Fibonacci number (A000045), then a 1 is put in its place.

A307069(k) is the row number of the first instance of the k-th Fibonacci number.

Consider a version of Pascal's Triangle: a triangular array with a single 1 on row 0, with numbers below equal to the sum of the two numbers above it if and only if that sum appears in the Fibonacci sequence A000045. If the sum does not appear in A000045, a 1 is put in its place.

So the first few rows would be as follows:

row 0: 1

row 1: 1 1

row 2: 1 2 1

row 3: 1 3 3 1

row 4: 1 1 1 1 1

row 5: 1 2 2 2 2 1

row 6: 1 3 1 1 1 3 1

row 7: 1 1 1 2 2 1 1 1

row 8: 1 2 2 3 1 3 2 2 1

row 9: 1 3 1 5 1 1 5 1 3 1

...

a(n) is the row number in which the n-th Fibonacci number first appears in this triangular array.

a(16) > 2.2*10^5. - David A. Corneth, Mar 25 2019

a(16) > 3.2*10^6. - Daniel Suteu, Mar 26 2019

a(16) > 1.5*10^7. - Bert Dobbelaere, Apr 02 2019

Consider a function f(n), which looks at all of the numbers from 1 to n, counts up how many contain consecutive repeated digits, and how many do not, and subtracts the former count from the latter count. For all numbers less than 11824560, f(n) is positive; for all numbers greater than 33082200, f(n) is negative. In between, the function crosses the zero line several times, with the 35 numbers in this sequence being the only values for which f(n)=0.

The set of b values (see formula), and therefore also a(n), depends only on the prime signature of n. So, for example, a(24) will be identical to a(n) of any other n which is also of the form p_1^3*p_2, (e.g., 40, 54, 56).

The value of b_1 will always be 1. When n is prime, the only nonzero b will be b_1, so therefore a(n) will be 1.

In any arrangement, both dice will have a 0, and one will have a 1 (here called the lead die). To determine any one of the actual arrangements to numbers on the dice, select one of the permutations of divisors (for the lead die), then select another permutation that is either the same length as that of the lead die, or one less. For example, if n = 12, we might select 2*3*2 for the lead die, and 3*4 for the second die. These numbers effectively tell you when to "switch track" when numbering the dice, and will uniquely result in the numbering: (0,1,6,7,12,13,72,73,78,79,84,85; 0,2,4,18,20,22,36,38,40,54,56,58).

a(n) is the number of (unordered) pairs of polynomials c(x) = x^c_1 + x^c_2 + ... + x^c_n, d(x) = x^d_1 + x^d_2 + ... + x^d_n with nonnegative integer exponents such that c(x)*d(x) = (x^(n^2)-1)/(x-1). - Alois P. Heinz, May 13 2016

a(n) is also the number of principal reversible squares of order n. - S. Harry White, May 19 2018

From Gus Wiseman, Oct 29 2021: (Start)

Also the number of ordered factorizations of n^2 with alternating product 1. This follows from the author's formula. Taking n instead of n^2 gives a(sqrt(n)) if n is a perfect square, otherwise 0. Here, an ordered factorization of n is a sequence of positive integers > 1 with product n, and the alternating product of a sequence (y_1,...,y_k) is Product_i y_i^((-1)^(i-1)). For example, the a(1) = 1 through a(9) = 3 factorizations are:

() (22) (33) (44) (55) (66) (77) (88) (99)

(242) (263) (284) (393)

(2222) (362) (482) (3333)

(2233) (2244)

(2332) (2442)

(3223) (4224)

(3322) (4422)

(22242)

(24222)

(222222)

The even-length case is A347464.

(End)

Given n players there are a(n) different ways of arranging those players in a 3 against 3 contest.

Number of ways to select two disjoint subsets of size 3 from a set of n elements. - Joerg Arndt, Mar 29 2016

For any number that does not appear on this list, there exists an arrangement of that number of unit-diameter circles that can be enclosed in a rectangle with area of less than 1 square unit per circle.

Any number of unit-diameter circles greater than or equal to 14 can be arranged in two rows, where the upper row is offset by 1/2 horizontally and (sqrt(3/4)-1) vertically, thereby reducing the minimum size of the enclosing rectangle to less than n square units. However, this isn't necessarily the overall minimum.

In addition, 11 unit-diameter circles placed in 3 rows can be enclosed in an area less than 11 square units.

This type of lock is quite common in the real world. The lock has typically 13 'number' buttons (actually 0 1 2 3 4 5 6 7 8 9 X Y Z), plus a C (for clear) button, and a knob to turn to 'try' the combination. The way it functions is that the unlocking code is an n-digit binary number. By pressing one of the number buttons, you change the corresponding digit from 0 to 1. Pressing C reverts all digits to 0.