A341529 a(n) = sigma(n) * A003961(n), where A003961 shifts the prime factorization of n one step towards larger primes, and sigma is the sum of the divisors of n.

1, 9, 20, 63, 42, 180, 88, 405, 325, 378, 156, 1260, 238, 792, 840, 2511, 342, 2925, 460, 2646, 1760, 1404, 696, 8100, 1519, 2142, 5000, 5544, 930, 7560, 1184, 15309, 3120, 3078, 3696, 20475, 1558, 4140, 4760, 17010, 1806, 15840, 2068, 9828, 13650, 6264, 2544, 50220, 6897, 13671, 6840, 14994, 3186, 45000, 6552, 35640

Offset: 1

Author

Antti Karttunen, Feb 16 2021

Keywords

• nonn
• mult

Comments

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)

Links

• Antti Karttunen, Table of n, a(n) for n = 1..8191
• Ratio A341529(n)/A341528(n) plotted with OEIS Plot tool
• Index entries for sequences computed from indices in prime factorization.
• Index entries for sequences related to sigma(n).

Crossrefs

Cf. A000203, A003961, A003973, A028982 (positions of odd terms), A072861, A151800, A286385, A341512, A341526, A341527, A341528, A341530, A342661, A342667, A342671, A342672.

Cf. also arrays A341605/A341606.

Programs

Mathematica

Array[DivisorSigma[1, #]*Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &, 56] (* Michael De Vlieger, Feb 22 2021 *)

PARI

A003961(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
A341529(n) = (sigma(n)*A003961(n));

Formula

Multiplicative with a(p^e) = q^e * (p^(e+1)-1)/(p-1), where q = nextPrime(p).

a(n) = A000203(n) * A003961(n).

For all n > 1, a(n) > A341528(n).

For all n >= 1, A072861(n) <= a(n) <= A003961(n)^2. [See A286385].

a(n) = A341528(n) + A341512(n) = A342671(n) * A342672(n) = A342661(A003961(n)). - Antti Karttunen, Mar 22 2021

Sum_{k=1..n} a(k) ~ c * n^3, where c = (1/3) * Product_{p prime} p^4*(p-1)/((p^3-nextprime(p))*(p^2-nextprime(p))) = 3.0664809..., where nextprime is A151800. - Amiram Eldar, Dec 08 2022