cp's OEIS Frontend

Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).

Similar to A274528, but the triangle here is a right triangle.

The same rule applied to an equilateral triangle gives A274528.

By analogy, the offset for both rows and columns is the same as the offset of A274528.

We construct the triangle by reading from left to right in each row, starting with T(0,0) = 0.

Presumably every diagonal and every column is also a permutation of the nonnegative integers, but the proof does not seem so straightforward. Of course neither the rows nor the antidiagonals are permutations of the nonnegative integers, since they are finite in length.

Omar E. Pol's conjecture that every column and every diagonal of the triangle is a permutation of the nonnegative integers is true: see the link. - N. J. A. Sloane, Jun 07 2017

It appears that the numbers generally appear for the first time in or near the right border of the triangle.

Theorem 1: the middle diagonal gives A000004 (the zero sequence).

Proof: T(0,0) = 0 by definition. For the next rows we have that in row 1 there are no zeros because the first term belongs to a column that contains a zero and the second term belongs to a diagonal that contains a zero. In row 2 the unique zero is T(2,1) = 0 because the preceding term belongs to a column that contains a zero and the following term belongs to a diagonal that contains a zero. Then we have two recurrences for all rows of the triangle:

a) If T(n,k) = 0 then row n+1 does not contain a zero because every term belongs to a column that contains a zero or it belongs to a diagonal that contains a zero.

b) If T(n,k) = 0 the next zero is T(n+2,k+1) because every preceding term in row n+2 is a positive integer because it belongs to a column that contains a zero and, on the other hand, the column, the diagonal and the antidiagonal of T(n+2,k+1) do not contain zeros.

Finally, since both T(n,k) = 0 and T(n+2,k+1) = 0 are located in the middle diagonal, the other terms of the middle diagonal are zeros, or in other words: the middle diagonal gives A000004 (the zero sequence). QED

Theorem 2: all zeros are in the middle diagonal.

Proof: consider the first n rows of the triangle. Every element located above or at the right-hand side of the middle diagonal must be positive because it belongs to a diagonal that contains one of the zeros of the middle diagonal. On the other hand every element located below the middle diagonal must be positive because it belongs to a column that contains one of the zeros of the middle diagonal, hence there are no zeros outside of the middle diagonal, or in other words: all zeros are in the middle diagonal. QED

From Hartmut F. W. Hoft, Jun 12 2017: (Start)

T(2k,k) = 0, for all k >= 0, and T(n,{(n-1)/2,(n+2)/2,(n-2)/2,(n+1)/2}) = 1, for all n >= 0 with n mod 8 = {1,2,4,5} respectively, and no 0's or 1's occur in other positions. The triangle of positions of 0's and 1's for this sequence is the triangle in the Comment section of A274651 with row and column indices and values shifted down by one.

The sequence of rows containing 1's is A047613 (n mod 8 = {1,2,4,5}), those containing only 1's is A016813 (n mod 8 = {1,5}), those containing both 0's and 1's is A047463 (n mod 8 = {2,4}), those containing only 0's is A047451 (n mod 8 = {0,6}), and those containing neither 0's nor 2's is A004767 (n mod 8 = {3,7}).

(End)

Complement of A003714. It appears that n is in the sequence if and only if C(3n,n) is even. - Benoit Cloitre, Mar 09 2003

Since the binary representation of these numbers contains two adjacent 1's, so for these values of n, we will have (n XOR 2n XOR 3n) != 0, and thus a two player Nim game with three heaps of (n, 2n, 3n) stones will be a winning configuration for the first player. - V. Raman, Sep 17 2012

A048728(a(n)) > 0. - Reinhard Zumkeller, May 13 2014

The set of numbers x such that Or(x,3*x) <> 3*x. - Gary Detlefs, Jun 04 2024

Numbers k for which A337194(k) = 1+A161942(k) is a multiple of A000265(1+k).

Conjecture: After 1, all terms are of the form 4u+3 (in A004767). If this could be proved, then it would refute at once the existence of both the odd perfect numbers and the quasiperfect numbers. Concentrating on the latter is probably easier, as it is known that all quasiperfect numbers must be odd squares, thus k is of the form 4u+1, in which case the condition given in A336701 that A000265(1+A000265(sigma(k))) must be equal to A000265(1+k) reduces to a simpler form, A000265(1+sigma(k)) = (1+k)/2, and as k = s^2, with s odd, so (s^2 + 1)/2 should divide 1+sigma(s^2). Does that condition allow any other solutions than s=1 ? See A337339.

The table showing the possible modulo 3 combinations for p, q, r and the sum ((p*q) + (p*r) + (q*r)):

| p | q | r | sum ((p*q) + (p*r) + (q*r)) (mod 3)

--+------+------+------+----------------------------------------

| 0 | 0 | 0 | 0, p=q=r=3, sum is 27.

--+------+------+------+----------------------------------------

| 0 | 0 | +/-1 | 0, p=q=3, r > 3.

--+------+------+------+----------------------------------------

| 0 | +1 | +1 | +1

--+------+------+------+----------------------------------------

| 0 | -1 | -1 | +1

--+------+------+------+----------------------------------------

| 0 | -1 | +1 | -1

--+------+------+------+----------------------------------------

| 0 | +1 | -1 | -1

--+------+------+------+----------------------------------------

| +1 | +1 | +1 | 0

--+------+------+------+----------------------------------------

| -1 | -1 | -1 | 0

--+------+------+------+----------------------------------------

| -1 | +1 | +1 | -1, regardless of the order, thus x3.

--+------+------+------+----------------------------------------

| +1 | -1 | -1 | -1, regardless of the order, thus x3.

--+------+------+------+----------------------------------------

Notably a(n) is a multiple of 3 only when A046316(n) is either a multiple of 9, or all primes p, q and r are either == +1 (mod 3) or all are == -1 (mod 3), and the case a(n) == +1 (mod 3) is only possible when A046316(n) is a multiple of 3, but not of 9, and furthermore, it is required that r == q (mod 3). See how these combinations affects sequences like A369241, A369245, A369450, A369451, A369452.

For n=1..9 the number of terms of the form 3k, 3k+1 and 3k+2 in range [1..10^n-1] are:

6, 2, 1,

39, 22, 38,

291, 209, 499,

2527, 1884, 5588,

23527, 17020, 59452,

227297, 156240, 616462,

2232681, 1453030, 6314288,

22119496, 13661893, 64218610,

220098425, 129624002, 650277572.

It seems that 3k+2 terms are slowly gaining at the expense of 3k+1 terms when n grows, while the density of the multiples of 3 might converge towards a limit.

If Y is a fixed 2-subset of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - Milan Janjic, Oct 21 2007

A bisection of A016754. Sequence arises from reading the line from 9, in the direction 9, 49, ... in the square spiral whose vertices are the squares A000290. - Omar E. Pol, May 24 2008

Using (n,n+1) to generate a Pythagorean triangle with sides of lengths xJ. M. Bergot, Jul 17 2013

If we define f(n) to be the number of primes (counted with multiplicity) of the form 4k + 3 that divide n, then with this sequence f(a(n)) is always odd. For example, 95 is divisible by 17 and 99 is divisible by 3 (twice) and 11. - Alonso del Arte, Jan 13 2016

Complement of A002145 with respect to A004767. - Michel Marcus, Jan 17 2016

With the Jan 05 2004 Jovovic comment on A078703: The number of 1 and -1 (mod 4) divisors of a(n) are identical. Proof: each number 3 (mod 4) is trivially not a sum of two squares. The number of solutions of n as a sum of two squares is r_2(n) = 4*(d_1(n) - d_3(n)), where d_k(n) is the number of k (mod 4) divisors of n. See e.g., Grosswald, pp. 15-16 for the proof of Jacobi. - Wolfdieter Lang, Jul 29 2016

For numbers > 1, iterate the map x -> A252463(x) which divides even numbers by 2 and shifts every prime in the prime factorization of odd n one index step towards smaller primes. a(n) counts the numbers of the form 4k+3 encountered until 1 has been reached. The count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the node which contains n in binary tree A005940 and traverse from that node towards the root, counting all numbers of the form 4k+3 that occur on the path.

Terms that occur on the first two rows of array A257852. Odd numbers that are not of the form 4k+1, where k is an odd number. - Antti Karttunen, Jun 06 2024