cp's OEIS Frontend

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021

Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021

a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001

a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014

From Antti Karttunen, Nov 01 2019: (Start)

More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).

Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.

(End)

Records for the sequence of the absolute values are in A075728 and the indices of these records in A074918. - R. J. Mathar, Mar 02 2007

a(n) = 1 iff n is a power of 2. a(n) = n - 1 iff n is prime. - Omar E. Pol, Jan 30 2014

If a(n) = 1 then n is called a least deficient number or an almost perfect number. All the powers of 2 are least deficient numbers but it is not known if there exists a least deficient number that is not a power of 2. See A000079. - Jianing Song, Oct 13 2019

It is not known whether there are any -1's in this sequence. See comment in A033880. - Antti Karttunen, Feb 02 2020

The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.

This array can be obtained by taking every second column from array A242378, starting from its column 2.

Permutation of natural numbers larger than 1.

The terms on row n are all divisible by n-th prime, A000040(n).

Each column is strictly growing, and the terms in the same column have the same prime signature.

A055396(n) gives the row number of row where n occurs,

and A246277(n) gives its column number, both starting from 1.

From Antti Karttunen, Jan 03 2015: (Start)

A252759(n) gives their sum minus one, i.e. the Manhattan distance of n from the top left corner.

If we assume here that a(1) = 1 (but which is not explicitly included because outside of the array), then A252752 gives the inverse permutation. See also A246276.

(End)

For any other number n than those in A326182 we have a(n) < A003961(n).

Fixed points k (for which a(k) = k) satisfy A003973(k) = 2^e * A003961(k) for some exponent e >= 0. Applying A003961 to such numbers gives the odd terms in A336702, of which there are likely to be just a single instance, its initial 1. (Clarified Nov 07 2021).

Conjecture: There are no other fixed points than a(1) = 1. If true, then there are no odd perfect numbers. This condition is equivalent to the condition that if A161942 has no fixed points larger than one, then there are no odd perfect numbers. This follows as whenever k is a fixed point, that is, a(k) = k, then we should also have A003961(a(k)) = A003961(A064989(sigma(A003961(k)))) = A161942(A003961(k)) = A003961(k). Note that A003961 is an injective and surjective mapping from natural numbers to odd numbers, A064989 is its (left) inverse, and composition A003961(A064989(n)) is equivalent to A000265(n).

From Antti Karttunen, Aug 05 2020: (Start)

For any hypothetical odd perfect number x, we would have A003973(k) = 2 * A003961(k), with k = A064989(x) and x = A003961(k). Thus we would have a(k) = A064989(sigma(A003961(k))) = A064989(sigma(x)) = A064989(2*x) = A064989(x) = k. On the other hand, A003973(k) = sigma(A003961(k)) < A003961(A003961(k)) [see A286385 for the reason why], so a necessary condition for this is that x should be one of the terms of A246282. (Clarified Dec 01 2020).

(End)

Apart from missing 2, this sequence gives all numbers k such that the binary expansion of A156552(k) is a prefix of that of A156552(sigma(k)), that is, for k > 1, numbers k for which sigma(k) is a descendant of k in A005940-tree. This follows because of the two transitions x -> A005843(x) (doubling) and x -> A003961(x) (prime shift) used to generate descendants in A005940-tree, using A003961 at any step of the process will ruin the chances of encountering sigma(k) anywhere further down that subtree.

Proof: Any left child in A005940 (i.e., A003961(k) for k) is larger than sigma(k), for any k > 2 [see A286385 for a proof], and A003961(n) > n for all n > 1. Thus, apart from A003961(2) = 3 = sigma(2), A003961^t(k) > sigma(k), where A003961^t means t-fold application of prime shift, here with t >= 1. On the other hand, sigma(2n) > sigma(n) for all n, thus taking first some doubling steps before a run of one or more prime shift steps will not rescue us, as neither will taking further doubling steps after a bout of prime shifts.

The first terms of A325637 not included in this sequence are 154345556085770649600 and 9186050031556349952000, as they have abundancy index 6.

From Antti Karttunen, Nov 29 2021: (Start)

Odd terms of this sequence are given by the intersection of A349169 and A349174.

A064989 applied to the odd terms of this sequence gives the fixed points of A326042, i.e., the positions of zeros in A348736, and a subset of the positions of ones in A348941.

Odd terms of this sequence form a subsequence of A348943, but should occur neither in A348748 nor in A348749.

(End)

Numbers n such that A003961(n) > 2*n.

Numbers n such that A048673(n) > n.

The sequence grows as:

a(10) = 18

a(100) = 192

a(1000) = 1830

a(10000) = 18636

a(100000) = 187350

a(1000000) = 1865226

a(10000000) = 18654333

and the powers of 10 occur at:

a(5) = 10

a(53) = 100

a(536) = 1000

a(5423) = 10000

a(53290) = 100000

a(535797) = 1000000

a(5361886) = 10000000

suggesting that the ratio a(n)/n is converging to an constant and an arbitrary natural number is slightly more likely to be in this sequence than in the complement A246281. See also comments at A246351 and compare to quite a different ratio present in the "inverse" case A246362.

From Antti Karttunen, Aug 27 2020: (Start)

Any perfect number, including all odd perfect numbers (if such numbers exist), must occur in this sequence. See A286385 and A326042 for the reason why.

Like abundancy index (ratio A000203(n)/n), also ratio A003961(n)/n is multiplicative and always > 1 for all n > 1. Thus if the number has a proper divisor that is in this sequence, then the number itself also is. See A337372 for terms included here, but with no proper divisor in this sequence. (End)

For k >= 2, if m * A130789(k) is a term then m * A130789(k-1) is a term. - Peter Munn, Sep 01 2025

Could be called "primeshift-abundant numbers", in analogy with A005101. - Antti Karttunen, Sep 01 2025

Question: Does the maximum value of ratio A341529(n)/A341528(n) stay below 2?

From Amiram Eldar and Antti Karttunen, Jan 28 2023: (Start)

Answer to the above question is yes: Sup_{n>=1} A341529(n)/A341528(n) = 2.

Proof:

f(n) = A341529(n)/A341528(n) is a multiplicative function with f(p^e) = (1 + 1/p + ... + 1/p^e)/(1 + 1/q + ... + 1/q^e), where q = nextprime(p).

First we prove a lemma which states that f(p^(1+e)) / f(p^e) > 1, for any prime p, and exponent e.

We note that (sigma(p^(1+e))/(p^(1+e))) / (sigma(p^e)/(p^e)) = (sigma(p^(1+e))/(p*sigma(p^e))) = sigma(p^(1+e)) / (sigma(p^(1+e)) - 1), so setting q = nextprime(p), we can write the ratio f(p^(1+e)) / f(p^e) as (sigma(p^(1+e))/(sigma(p^(1+e))-1)) / (sigma(q^(1+e))/(sigma(q^(1+e))-1)), and to prove this to be > 1, we just note that the denominator is less than the numerator, because sigma(p^e) is monotonically growing with respect to the increasing prime p.

Since q > p, we have f(p^e) > 1 for all p and all e>=1, and together with the above lemma this shows that f(n) <= f(n*m) for all m>=1.

Suppose n = Product_i p_i^e_i, and let pmax = max(p_i), emax = max(e_i), so n is a divisor of m = (pmax#)^emax, and f(n) < f(m), where p# = 2 * 3 * ... * p is the primorial of p, A034386(p).

Then f(m) = f(2^emax) * f(3^emax) * ... * f(pmax^emax) = (1 + 1/2 + ... + 1/2^emax)/(1 + 1/3 + ... + 1/3^emax)) * (1 + 1/3 + ... + 1/3^emax)/(1 + 1/5 + ... + 1/5^emax)) * ... * (1 + 1/p + ... + 1/p^emax)/(1 + 1/q + ... + 1/q^emax))[telescoping product] = (1 + 1/2 + ... + 1/2^emax)/(1 + 1/qmax + ... + 1/qmax^emax) <= (1 + 1/2 + ... + 1/2^emax) < 2, where qmax = nextprime(pmax).

So we have f(n) < 2 for all n.

To prove that 2 is the supremum, we have lim_{e,k -> oo) f(prime(k)#^e) = 2.

(End)

Numbers n such that A003961(n) < 2*n.

Numbers n such that A048673(n) <= n.

All primes (A000040) are members. (Cf. Bertrand's postulate).

All terms are deficient (in A005100). See A286385. - Antti Karttunen, Aug 27 2020