cp's OEIS Frontend

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

Numbers k whose only divisor in A246282 is k itself, i.e., A003961(k) > 2k, but for none of the proper divisors d|k, dA003961(d) > 2d.

Question: Do the odd terms in A326134 all occur here? Answer is yes, if the following conjecture holds: This is a subsequence of A263837, nonabundant numbers. In other words, we claim that any abundant number k (A005101) has A337345(k) > 1 and thus is a term of A341610. (The conjecture indeed holds. See the proof below).

From Antti Karttunen, Dec 06 2024: (Start)

Observation 1: The thirteen initial terms (4, 6, 9, ..., 69, 91) are only semiprimes in A246282, all other semiprimes being in A246281 (but none in A341610), and there seems to be only 678 terms m with A001222(m) = 3, from a(14) = 125 to the last one of them, a(2691) = 519963. There are more than 150000 terms m with A001222(m) = 4. In general, there should be only a finite number of terms m for any given k = A001222(m). Compare for example with A287728.

Observation 2: The intersection with A005101 (and thus also with A091191) is empty, which then implies the claims made in the sequences A378662, A378664, from which further follows that there are no 1's present in any of these sequences: A378658, A378736, A378740.

(End)

Proof of the latter observation by Jianing Song, Dec 11 2024: (Start)

Let's write p' for the next prime after the prime p. Also, write Q(n) = A003961(n)/sigma(n) which is multiplicative.

Proposition: For n > 1 not being a prime nor twice a prime, n has a factor p such that Q(n) > p'/p.

This implies that if n is abundant [including any primitively abundant n in A091191], then n has a factor p such that A003961(n/p)/(n/p) = (A003961(n)/n)/(p'/p) > sigma(n)/n [which is > 2 because n is abundant], so n/p is in A246282, meaning that n cannot be in this sequence.

Proof. We see that 1 <= Q(p) <= Q(p^2) <= ..., which implies that if n verifies the proposition, then every multiple of n also verifies it. Since n = p^2 > 4 and n = 8 verify the proposition, it suffices to consider the case where n = pq is the product of two distinct odd primes. Suppose WLOG that p < q, so q >= p', then using q/(q+1) >= p'/(p'+1) we have

Q(n) = p'q'/((p+1)(q+1)) >= p'^2*q'/(q(p+1)(p'+1)) > (p'^2-1)*q'/(q(p+1)(p'+1)) = (p'-1)/(p+1) * q'/q >= q'/q.

(End)

These are prime-factorization representations of single-variable polynomials where the coefficient of term x^(k-1) (encoded as the exponent of prime(k) in the factorization of n) is equal to the number of times a nonzero digit k occurs in the factorial base representation of n. See the examples.

Terms a(n) larger than 1 and equal to A252748(n) occur at n = 6, 28, 69, 91, 496, ..., see A326134. See also A349753.

Records 1, 3, 43, 45, 2005, 79243, ... occur at n = 1, 6, 28, 360, 496, 8128, ...

Numbers k such that A348990(k) [= k/gcd(k, A003961(k))] is equal to A348992(k), which is the odd part of A349162(k), thus all terms must be odd, as A348990 preserves the parity of its argument.

Equally, numbers k for which gcd(A064987(k), A191002(k)) is equal to A000265(gcd(A064987(k), A341529(k))).

Also odd numbers k for which A348993(k) = A319627(k).

Odd terms of A336702 are given by the intersection of this sequence and A349174.

Conjectures:

(1) After 1, all terms are multiples of 3. (Why?)

(2) After 1, all terms are in A104210, in other words, for all n > 1, gcd(a(n), A003961(a(n))) > 1. Note that if we encountered a term k with gcd(k, A003961(k)) = 1, then we would have discovered an odd multiperfect number.

(3) Apart from 1, 15, 105, 3003, 13923, 264537, all other terms are abundant. [These apparently are also the only terms that are not Zumkeller, A083207. Note added Dec 05 2024]

(4) After 1, all terms are in A248150. (Cf. also A386430).

(5) After 1, all terms are in A348748.

(6) Apart from 1, there are no common terms with A349753.

Note: If any of the last four conjectures could be proved, it would refute the existence of odd perfect numbers at once. Note that it seems that gcd(sigma(k), A003961(k)) < k, for all k except these four: 1, 2, 20, 160.

Questions:

(1) For any term x here, can 2*x be in A349745? (Partial answer: at least x should be in A191218 and should not be a multiple of 3). Would this then imply that x is an odd perfect number? (Which could explain the points (1) and (4) in above, assuming the nonexistence of opn's).