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Locate the node which contains n in binary tree A005940 and traverse towards the root (which is 1), counting separately all numbers of the form 4k+1 and of the form 4k+3 that occur on the path (including also starting n itself, if it is of the either form), until 1 is reached, which however, is never included in the count of 4k+1. a(n) is the count of 4k+3 numbers minus the count of 4k+1 numbers that were encountered.

Numbers n such that sum of divisors of n (A000203) is odd.

Also the numbers with an odd number of run sums (trapezoidal arrangements, number of ways of being written as the difference of two triangular numbers). - Ron Knott, Jan 27 2003

Pell(n)*Sum_{k|n} 1/Pell(k) is odd, where Pell(n) is A000129(n). - Paul Barry, Oct 12 2005

Number of odd divisors of n (A001227) is odd. - Vladeta Jovovic, Aug 28 2007

A071324(a(n)) is odd. - Reinhard Zumkeller, Jul 03 2008

Sigma(a(n)) = A000203(a(n)) = A152677(n). - Jaroslav Krizek, Oct 06 2009

Numbers n such that sum of odd divisors of n (A000593) is odd. - Omar E. Pol, Jul 05 2016

A187793(a(n)) is odd. - Timothy L. Tiffin, Jul 18 2016

If k is odd (k = 2m+1 for m >= 0), then 2^k = 2^(2m+1) = 2*(2^m)^2. If k is even (k = 2m for m >= 0), then 2^k = 2^(2m) = (2^m)^2. So, the powers of 2 sequence (A000079) is a subsequence of this one. - Timothy L. Tiffin, Jul 18 2016

Numbers n such that A175317(n) = Sum_{d|n} pod(d) is odd, where pod(m) = the product of divisors of m (A007955). - Jaroslav Krizek, Dec 28 2016

Positions of zeros in A292377 and A292383, positions of ones in A286357 and A292583. (See A292583 for why.) - Antti Karttunen, Sep 25 2017

Numbers of the form A000079(i)*A016754(j), i,j>=0. - R. J. Mathar, May 30 2020

Equivalently, numbers whose odd part is square. Cf. A042968. - Peter Munn, Jul 14 2020

These are the Heinz numbers of the partitions counted by A119620. - Gus Wiseman, Oct 29 2021

Numbers m whose abundance, A033880(m), is odd. - Peter Munn, May 23 2022

Numbers with an odd number of middle divisors (cf. A067742). - Omar E. Pol, Aug 02 2022

Term a(n) essentially records the run lengths of numbers of form 4k+3 encountered when starting from that node in binary tree A005940 which contains n, and by then traversing towards the root by iterating the map n -> A252463(n). The actual run lengths can be read from the exponents of primes in the prime factorization of A278222(A292383(m)), where m = min_{k=1..n} for which a(k) = a(n). In compound filter A292584 this is combined with similar information about the run lengths of the numbers of the form 4k+1 (A292585).

From Antti Karttunen, Sep 25 2017: (Start)

For all i, j: a(i) = a(j) => A053866(i) = A053866(j).

This follows from the interpretation of A053866 (A093709) as the characteristic function of squares and twice-squares. In binary tree A005940 each number is "born" by repeated applications of two functions: when we descend leftward we apply A003961, which shifts all prime factors of n one step towards larger primes. On the other hand, when we descend rightward the terms grow by doubling: n -> 2n (A005843). No square is ever of the form 4k+3, and for any square x, A003961(x) is also a square. Multiplying a square by 2 gives twice a square, and then multiplying by 2 again gives 4*square, which is also a square. In general, applying an even number of doubling steps in succession keeps a square as a square, while an odd number of doubling steps gives twice a square. Applying A003961 to any 2*square gives 3*(some square) which is always of the form 4k+3. Moreover, after any such "wrong turn" in A005940-tree no square nor twice a square can ever be encountered under any of the further descendants, because with this process it is impossible to find a pair for the lone prime factor now present. On the other hand, when turning left from any (2^2k)*s (where s is a square), one again gets a square of the form (3^2k)*A003961(s). All this implies that there are no numbers of the form 4k+3 in any trajectory leading to a square or twice a square in A005940-tree, while all trajectories to any other kind of number contain at least one number of the form 4k+3. Because each a(n) in this sequence contains enough information to count the 4k+3 numbers encountered on a A005940-trajectory to n (being 1 iff there are none), this filter matches A053866.

(End)

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+1 encountered until 1 has been reached, which is also included in the count. The count includes also n itself if it is of the form 4k+1 (A016813), thus a(1) = 1.

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+1 that occur on the path.

Because A292590(n) = a(A245611(n)), the sequence works as a "masking function" where the 1-bits in a(n) (always a subset of the 1-bits in binary expansion of n) indicate which numbers are multiples of 3 in binary tree A245612 on that trajectory which leads from the root of the tree to the node containing A245612(n). See the examples.

Starting from x=n, iterate the map x -> A332893(x) which divides even numbers by 2, and for odd n, changes every 4k+1 prime in the prime factorization to 4k+3 prime and vice versa (except 3 --> 2), like in A332819. 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 A332815 and traverse from that node towards the root, counting all numbers of the form 4k+3 that occur on the path.

From Antti Karttunen, Sep 25 2017: (Start)

Some of the sequences this filter matches to:

For all i, j: a(i) = a(j) => A053866(i) = A053866(j).

For all i, j: a(i) = a(j) => A061395(i) = A061395(j). (also A006530, etc.)

For all i, j: a(i) = a(j) => A292378(i) = A292378(j).

The latter two implications follow simply because:

A001222(A278222(A292383(n))) = A000120(A292383(n)) = A292377(n)

and, similarly, for n > 1,

A001222(A278222(A292385(n))) = A000120(A292381(n)) = A292375(n),

and the sum of A292375(n) and A292377(n) is A061395(n) [index of the largest prime dividing n], while A292378 has been defined as 1 + their difference.

The case A053866 follows because of the component A292583, see comments under that entry. (End)

Positions for the first occurrence of each n, for n >= 0, are: 1, 4, 16, 32, 144, 512, 2048, 6912, 20736, 62208, ...

For odd numbers > 1, iterate the map x -> A064989(x), which shifts every prime in the prime factorization of n one index step towards smaller primes. a(n) counts the numbers of the form 4k+3 encountered until the first number which is even has been reached. This count includes also n itself if it is of the form 4k+3 (A004767).

In other words, locate the position where n is in square array A246278 and moving up by that column, count all numbers of the form 4k+3 before an even number at the top of the column is reached.