cp's OEIS Frontend

Complement of A014601. - Reinhard Zumkeller, Oct 04 2004

Let S(x) = (1, 2, 2, 2, ...). Then A042963 = ((S(x))^2 + S(x^2))/2 = ((1, 4, 8, 12, 16, 20, ...) + (1, 0, 2, 0, 2, 0, 2, ...))/2 = (1, 2, 5, 6, 9, 10, ...). - Gary W. Adamson, Jan 03 2011

(a(n)*(a(n) + 1 + 4*k))/2 is odd, for k >= 0. - Gionata Neri, Jul 19 2015

Equivalent to the following variation on Fermat's Diophantine m-tuple: 1 + the product of any two distinct terms is not a square; this sequence, which we'll call sequence S, is produced by the following algorithm. At the start, S is initially empty. At stage n, starting at n = 1, the algorithm checks whether there exists a number m already in the sequence, such that mn+1 is a perfect square. If such a number m is found, then n is not added to the sequence; otherwise, n is added. Then n is incremented to n + 1, and we repeat the procedure. Proof by Clark R. Lyons: We prove by strong induction that n is in the sequence S if and only if n == 1 (mod 4) or n == 2 (mod 4). Suppose now that this holds for all k < n. In case 1, either n == 1 (mod 4) or n == 2 (mod 4), and we wish to show that n does indeed enter the sequence S. That is, we wish to show that there does not exist m < n, already in the sequence at this point such that mn+1 is a square. By the inductive hypothesis m == 1 (mod 4) or m == 2 (mod 4). This means that both m and n are one of 1, 2, 5, or 6 mod 8. Using a multiplication table mod 8, we see that this implies mn+1 is congruent to one of 2, 3, 5, 6, or 7 mod 8. But we also see that mod 8, a perfect square is congruent to 0, 1, or 4. Thus mn+1 is not a perfect square, so n is added to the sequence. In case 2, n == 0 (mod 4) or n == 3 (mod 4), and we wish to show that n is not added to the sequence. That is, we wish to show that there exists m < n already in the sequence such that mn+1 is a perfect square. For this we let m = n - 2, which is positive since n >= 3. By the inductive hypothesis, since m == 1 (mod 4) or m == 2 (mod 4) and m < n, m is already in the sequence. And we have m*n + 1 = (n - 2)*n + 1 = n^2 - 2*n + 1 = (n - 1)^2, so mn+1 is indeed a perfect square, and so n is not added to the sequence. Thus n is added to the sequence if and only if n == 1 (mod 4) or n == 2 (mod 4). This completes the proof. - Robert C. Lyons, Jun 30 2016

Also the number of maximal cliques in the (n + 1) X (n + 1) black bishop graph. - Eric W. Weisstein, Dec 01 2017

Lexicographically earliest sequence of distinct positive integers such that the average of any two or more consecutive terms is never an integer. (For opposite property see A005408.) - Ivan Neretin, Dec 21 2017

Numbers whose binary reflected Gray code (A014550) ends with 1. - Amiram Eldar, May 17 2021

Also: append its negated last bit to n-1. - M. F. Hasler, Oct 17 2022

Like A129594 this sequence utilizes the fact that compositions (i.e., ordered partitions) can be bijectively mapped to (unordered) partitions by taking the partial sums of the list of composants after one has been subtracted from each except the first one. Compositions in turn are mapped to nonnegative integers via the runlength encoding, where the lengths of maximum runs of 0's or 1's in binary representation of n give the composants. See the OEIS Wiki page and the example below.

Each n occurs A000041(n) times in total and occurs for the first time at A227368(n) and for the last time at position A000225(n). See further comments and conjectures at A227368 and A227370.

a(even) = number of trailing binary zeros;

a(odd) = number of trailing binary ones.

For n>0, power of 2 associated with n^2 + n, e.g. n=4 gives 20, so a(4)=2. - Jon Perry, Sep 12 2014

Discarding the trailing zero terms, on each row n there is a unique partition of integer A227183(n). All possible partitions of finite natural numbers eventually occur. The first partition that sums to n occurs at row A227368(n).

Irregular table A227739 lists only the nonzero terms.

A(n,k) is set to zero if there are less than k+1 runs.

The irregular table A101211 gives the nonzero terms of each row in reverse order. The terms on row n sum to A029837(n+1). The product of nonzero terms on row n>0 is A167489(n). Number of nonzero terms on each row: A005811.

a(n) is also the map n -> A065423(n+1) applied A007814(n+1) times. - Federico Provvedi, Dec 14 2021

A(n,k) is set to zero if there are fewer runs in n than k+1.

Equally, when A005811(n) > 1, A(n,k) gives the zero-based bit-index where the (k+2)-th run in the binary expansion of n starts, counted from the least significant end.

Each row gives the partial sums of the terms on the corresponding row in A227186, up to the first zero.

After n=3, only composites may obtain value 0. (But not all of them do; see A227762.) The first nine n for which a(n)=2 are 7, 13, 23, 33, 47, 61, 79, 97, 119, of which all are primes except 33 and 119. Conjecture: these values are given by A227786.

Are there any terms larger than 2?