A341529 - OEIS

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

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OFFSET

1,2

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).

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

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 *)

PROG

(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));

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.

Sequence in context: A345727 A332372 A344818 * A013573 A146388 A230833

Adjacent sequences: A341526 A341527 A341528 * A341530 A341531 A341532

KEYWORD

nonn,mult

AUTHOR

Antti Karttunen, Feb 16 2021

STATUS

approved