cp's OEIS Frontend

A324180 differs from this one in that it uses XOR instead of OR, and uses only the proper divisors of n.

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

After a(1) = 0, the other zeros occur for k >= 1, at A005940(1+A000396(k)), which, provided no odd perfect numbers exist, is equal to A324201(k) = A062457(A000043(k)): 9, 125, 161051, 410338673, ..., etc.

There are 2321 negative terms among the first 10000 terms.

From Gus Wiseman, Apr 06 2019: (Start)

Also the number of squares in the Young diagram of the integer partition with Heinz number n that are graph-distance 1 from the lower-right boundary, where the Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). For example, the partition (6,5,5,3) with Heinz number 7865 has diagram

o o o o o o

o o o o o

o o o o o

o o o

with inner rim

o

o

o o

o o o

of size 7, so a(7867) = 7.

(End)

The first four zeros after a(1) occur at n = 192, 288, 3645, 6075.

There are 1562 negative terms among the first 10000 terms.

Applying this function to the divisors of the first four terms of A324201 reveals the following pattern:

----------------------------------------------------------------------------------

A324201 divisors a(n) applied to each: Sum

9: [1, 3, 9] -> [0, 3, 9] 12 = 2*6

125: [1, 5, 25, 125] -> [0, 7, 21, 28] 56 = 2*28

161051: [1, 11, 121, 1331, 14641, 161051] -> [0, 31, 93, 124, 496, 248] 992 = 2*496

410338673: [1, 17, 289, 4913, 83521, 1419857, 24137569, 410338673]

-> [0, 127, 381, 508, 2032, 1016, 9144, 3048] 16256 = 2*8128

The second term (the first nonzero) of the latter list = A000668(n), and the sum is always twice the corresponding perfect number, which forces either it or at least many of its divisors to be present. For example, in the fourth case, although 8128 = A000396(4) itself is not present, we still have 127, 508, 1016 and 2032 in the list. See also A329644.

The first eleven zeros occur at n = 1, 15, 16, 40, 96, 119, 120, 160, 893, 2464, 6731. There are 3091 negative terms among the first 10000 terms.

Applying this function to the divisors of the first four terms of A324201 reveals the following pattern:

------------------------------------------------------------------------------------

A324201(n) divisors a(n) applied Sum of positive

to each: terms, A329610

9: [1, 3, 9] -> [0, 1, -1] 1

125: [1, 5, 25, 125] -> [0, 1, -5, 4] 5

161051: [1, 11, 121, 1331, 14641, 161051] -> [0, 1, -29, 4, -240, 264] 269

410338673: [1, 17, 289, 4913, 83521, 1419857, 24137569, 410338673]

-> [0, 1, -125, 4, -1008, 1032, -5048, 5144] 6181

The positive and negative terms seem to alternate, and the fourth term (from case n=125 onward) is always 4. See also array A329637.

Apart from primes, the sequence contains duplicate values at points p*q and p^3, where p*q are the product of two successive primes, with p < q (sequences A006094, A030078). Question: are there any other cases where a(x) = a(y), with x < y ?

The reason why this is not equal to A297169: Even though A297112 contains only powers of two after the initial zero, as A297112(n) = 2^A033265(A156552(d)) for n > 1, and A297168(n) is computed as Sum_{d|n, dA297112(d), still a single 1-bit in binary expansion of A297168(n) might be formed as a sum of several terms of A297112(d), i.e., could be born of carries.

From Antti Karttunen, Feb 28 2019: (Start)

A297168(n) = Sum_{d|n, dA297112(d) will not produce any carries (in base-2) if and only if n is a power of prime. Only in that case the number of summands (A000005(n)-1) is equal to the number of prime factors counted with multiplicity, A001222(n) = A000120(A156552(n)). (A notable subset of such numbers is A324201, numbers that are mapped to even perfect numbers by A156552). Precisely because there are so few points with duplicate values (apart from primes), this sequence is not particularly good for filtering other sequences, because the number of false positives is high. Any of the related sequences like A324203, A324196, A324197 or A324181 might work better in that respect. In any case, the following implications hold (see formula section of A324193 for the latter): (End)

For all i, j:

a(i) = a(j) => A297168(i) = A297168(j). (The same holds for A297169).

a(i) = a(j) => A324181(i) = A324181(j) => A324120(i) = A324120(j).

Unique sequence satisfying SumXOR_{d divides n} a(d) = A156552(n) for all n > 0, where SumXOR is the analog of summation under the binary XOR operation. See A295901 for a list of some of the properties of the Xor-Moebius transform.

The ordinary Möbius transform of A156552 is given in A297112.

It seems that A091629 gives the fixed points of this sequence.