cp's OEIS Frontend

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)

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)

If there are no more 1's in this sequence after the initial one, then there are no odd terms of A007691 (multiply perfect numbers) larger than one.

Denominator of the ratio A003973(n) / A003961(n), also denominator of the ratio (A341528(n)/A341529(n)) / (n / sigma(n)). - Antti Karttunen, Feb 16 2021

Denominator of ratio A341528(n)/A341529(n). A341526 gives the numerator, see comments there.

Like the ratios sigma(n)/n, A003973(n)/A003961(n) and A003961(n)/n, also the ratio r(n) = A341528(n)/A341529(n) is multiplicative: if gcd(x,y) = 1, r(x*y) = r(x)*r(y).