Introduction
let σ(n) denote the sum of the positive divisors of n. It is well-known that a positive integer n is said to be abundant, perfect or deficient according as σ(n) > 2n, σ(n) = 2n or σ(n) < 2n. One can see that the set of abundant numbers as well as the set of deficient numbers are both infinite. In fact, the numbers of the form 2k.3 with k > 1 are all abundant, while every prime is deficient. But it is not known whether the set of perfect numbers is infinite or not.
Cattaneo 1 has called a positive integer n quasi perfect if σ(n) = 2n + 1. It is not known whether such numbers exist at all. Abbott, Kishore and Cohen 2, 3, 4, 5 have made significant contributions to the study of quasi perfect numbers. In 1978, Kishore 4 used the inequality 2−10−12 < σ(N)/N< 2 +10−12 and proved that there are no odd perfect numbers, no quasi perfect numbers and no odd almost perfect numbers with five distinct prime factors, proving ω(n) ≥ 6, where w(n) denotes number of divisors of n.
Cohen 6 proved, if any quasi perfect number n exists, then
(1.1.1) ω(n) 7 ⩾and n > 1035
Subsequently, Peter Hagis and Cohen 7 have used computations to improve some of the results on quasi perfect numbers. In fact, they established that
(1.1.2) for any quasi perfect number n, ω(n) ⩾15 if (15, n) = 1, ω(n) ⩾ 9 if 3 ł n and in other
case ω(n) > 7 and also n > 1035., where ω(n) denotes the number of divisors of n.
Cattaneo 1 has proved the following theorem:
1.1 Theorem: If n is a quasi perfect number then
(1.1.3) n=p12e1p22e2⋯pr2er
where pi’s are odd primes
Moreover,
(1.1.4) i) pi ≡ 1 (mod 8) ⇒ei ≡ 0 or 1 (mod 4)
ii) pi ≡3(mod 8) ⇒ ei ≡0 (mod 2)
iii) pi ≡ 5 (mod 8) ⇒ ei ≡0 or -1 (mod 4)
iv) pi ≡ 7 (mod 8) ⇒ ei ≥1
In a different direction, Cohen 8 has considered R(n), the sum of the reciprocals of the distinct primes dividing the quasi perfect number n (if exists) and obtained bounds for R(n). They are
(1.1.5)
i) 0.667450 < Rn< 0.693148 if (15, n)=1
ii) 0.603831 < Rn < 0.625140 if (15, n)=3
iii) 0.647387 < Rn < 0.670017 if (15, n)=5
iv) 0.596063 < Rn < 0.602009 if (15, n)=15
New methods were introduced by Tang and Feng 9 and they established that there are no odd deficient-perfect numbers with three distinct prime divisors. The alternate proof was given in 10 using special components, which is
(1.1.6) If n is a quasi perfect number and is of the form (1.1.3) then for at least one factor pi2ei, we have either
pi≡1mod 8 and ei≡1(mod 4)orpi≡5mod 8 and ei≡3(mod 4)
Calling pi2eias a special component if it satisfies (1.1.6)
and proved every quasi perfect number has an odd number of special components.
In 2019, Tomohiro 11 has given some lower bounds concerning quasi perfect numbers of the form N = m2 where m is square free and Prasad 12 obtained a lower bound for the product of the distinct primes dividing n in terms of ω(n).
In this paper, the upper bounds given in (1.1.5) are improved and subsequently an important result regarding divisibility with primes of n is obtained.
Methodology
In this section, we will prove some lemmas used in the sequel. We shall give detailed proof of the required lemmas in 13 which were used for bounds of odd perfect numbers.
2.1 Lemma:
For o<p≤13 we have 1+p+p2+p3+p4>ep+t1p2where t1=0.373.
Proof: For simplicity of notation we write t for t1. Consider the function
fp=ep+tp2-1+p+p2+p3+p4
= 1+p+tp21!+p+tp222!+p+tp233!+…-1+p+p2+p3+p4
= p2t-12+p3t-1+p4t22-1+13! p+tp23+…
Now the inequality of the lemma holds
⇔fp<0
⇔13!p+tp23+14!p+tp24+…<p212-t+p31-t+p41-t22
⇔13!p1+tp3+14!p21+tp4+…<12-t+p1-t+p21-t22
⇔16p1+tp31+14p1+tp+…<12-t (1)
But for 0<p≤13, we have
gp=16p1+tp31+14p1+tp+120p21+tp2+…
< 16p1+tp31+p+tp2+p+tp22+…
= 16p1+tp31-p+tp2-1,
Whenever t is chosen such that p+tp2<1. Therefore for 0<p≤13, we have
gp<16p1+tp311-p+tp2
≤1181+t33 11-13+t9
118∙3+t333∙96-t (2)
Now from (1) and (2), the inequality of the lemma holds if
118∙3+t333⋅96-t<12-t,
and this holds for t=0.373. Hence the lemma.
In similar way, we can prove the following Lemma 2.2, Lemma 2.3 and Lemma 2.4.
2.2 Lemma:
For 0<p≤17 we have 1+p+p2>ep+t2p2 where t2=0.406.
2.3 Lemma:
For 0<p≤15 we have 1+p+p2+p3+p4+p5+p6>e(p+t3p2) Where t3=0.496
2.4 Lemma:
For 0<p≤15 we have 1+p+p2+p3+p4+p5+p6+p7+p8>e(p+t4p2) Where t4=0.4998.
2.5 Lemma:
Suppose p is a quasi perfect number, Rkp is the sum of reciprocals of the kth powers of the distinct primes dividing p and t=mint1,t2,t3,t4=0.373 where t1,t2,t3,t4 are as given in Lemma 2.1 to 2.4. Then R1p+cR2p<log2.000000001.
Proof: Given that p is quasiperfect, so that in view of (1.1.3) and (1.1.6) we can write p as
(2.5.1)
p=a12e1a22e2…ar2er⋅b12f1b22f2…bs2fs
Where each of the ai2e is a special component and bj2fj is any component.
Then by (1.1.6), r is odd and also for each i, we have either
(2.5.2)
ai≡1mod 8 and ei≡1(mod 4)orai≡5mod 8 and ei≡3(mod 4)
Further by (1.1.4), for each j,
(2.5.3)
(2.5.3) bj≡1(mod8)⇒fj≡0(mod4)bj≡3(mod8)⇒fj≡0(mod2)bj≡5(mod8)⇒fj≡0(mod4)bj≡7(mod8)⇒fj≥1
Therefore by (2.5.2) and (2.5.3) we have
(2.5.4)
2+1p=σ(p)p=∏i=1r1+1ai+⋯+1ai2ei∏j=1S1+1bj+⋯+1bj2fj
˃∏ai≡1mod 81+1ai+1ai2 ∏ai≡5mod 81+1ai+1ai2+1ai3+1ai4+1ai5+1ai6
∏bj≡1mod 81+1bj+1bj2+1bj3+1bj4+1bj5+1bj6+1bj7+1bj8
∏bj≡3mod 81+1bj+1bj2+1bj3+1bj4⋅∏bj≡7mod 81+1bj+1bj2
∏bj≡5mod 81+1bj+1bj2+1bj3+1bj4+1bj5+1bj6+1bj7+1bj8
Now using Lemmas 2.1 to 2.4 on the right of (2.5.4) we get
2+1p>∏ai≡1mod 8e1ai+t1ai2⋅∏ai≡5mod 8e1ai+t3ai2⋅∏bj≡1mod 8e1bj+t4bj2⋅∏bj≡3mod 8e1bj+t2bj2
⋅∏bj≡5mod 8e1bj+t4bj2⋅∏bj≡7mod 8e1bj+t1bj2
Which on taking logarithm gives
log2+1n>∑ai≡1mod 81ai+t1ai2+∑ai≡5mod 81ai+t3ai2+∑bj≡1mod 81qj+t4qj2
+∑bj≡3mod 81qj+t2qj2+∑bj≡5mod 81qj+t4qj2+∑bj≡7mod 81qj+t1qj2
= R1p+t1∑ai≡1mod 81ai2+t3∑ai≡5mod 81ai2+t4∑bj≡1mod 81bj2
+t2∑bj≡3mod 81qj2+t4∑bj≡5mod 81qj2+t4∑bj≡7mod 81qj2
> R1p+tR2p
Since t=mint1,t2,t3,t4, proving the lemma because p>1035.
2.6 Lemma:
Suppose p is quasi perfect and α=2+1p. Then for any divisor of q of p with q<p, we have R1p<R1q+tR2q+logαqβσq-tR2p, where β = 1 if(p,15)=1 or (p,15)=(q,15)=3
or p,15=q,15=5 or p,15=q,15=15
1+13+132 ifp,15=3 and q,15=1 or p,15=15 and q,15=5
1+15+152 ifp,15=5 and q,15=1 or p,15=15 and q,15=3
1+13+1321+15+152 ifp,15=15 and q,15=1
Proof: Suppose p is quasiperfect, it is of the form p=∏i=1rai2ei, where ai’s are odd primes, then
(2.6.1)
α=2+1p=σpp=∏i=1r1+1ai+1ai2+…+1ai2ei
Suppose q=∏i=1raibi is a divisor of p with q<p. Then 0≤bi≤2ei for each i and bi<2ei for at least one i. Also
(2.6.2)
σqq=∏i=1r1+1pi+1pi2+…+1pibi
Now from (2.6.1), (2.6.2) and Lemma 2.1, we get
\geq \prod_{\substack{i=1 \\ a_i \mid q}}^r\left(1+\frac{1}{a_i}+\cdots+\frac{1}{a_i^{b_i}}\right) \prod_{\substack{i=1 \\ a_i \mid a}}^r\left(1+\frac{1}{a_i}+\cdots+\frac{1}{a_i^2}\right)
=\frac{\sigma(q)}{q} \exp \left\{\sum_{a \mid p}\left(\frac{1}{a}+t \cdot \frac{1}{a^{2}}\right)\right\} \cdot \beta
Which on taking logarithm gives
\log \alpha>\log \left(\frac{\sigma(q)}{q}\right)+\sum_{\substack{a \mid p \\ a \nmid q}}\left(\frac{1}{a}+t \cdot \frac{1}{a^{2}}\right)+\log \beta
=\log \left(\frac{\sigma(q)}{q}\right)+\sum_{a \nmid q} \frac{1}{a}+t \cdot \sum_{a \mid p} \frac{1}{a^{2}}+\log \beta
{{log}{\left(\frac{\sigma\left(q\right)}q\right)+R_1\left(p\right)-R_1\left(q\right)+t\left(R_2\left(p\right)-R_2\left(q\right)\right)+{{log}\beta}\;}}
Therefore,
R_1\left(p\right)<R_1\left(q\right)+{{log}{\left(\frac\alpha\beta\right)-{{log}{\left(\frac{\sigma\left(q\right)}q\right)}}-t\left(R_2\left(p\right)-R_2\left(q\right)\right)}}
= \;\;R_1\left(q\right)+tR_2\left(q\right)+{{log}{\left(\frac{\alpha\left(q\right)}{\beta\sigma\left(q\right)}\right)}}-tR_2\left(p\right)
Proving the lemma.
Results and Discussion
3.1 Theorem:
Suppose 3 | n and 5 ł n. Then
i) R1(n) < 0.2727439241 if 7 | n.
and
ii) R1(n) < 0.2839779561 if 7 ł n.
Proof: (i) Suppose 3 | n and 7 | n.
It was proved in 2 that if n\equiv0(mod\;3) and p\equiv1(mod\;3) is a divisor of n then p^k | n with k \geq 4. Therefore 7^4| n. Also since w(n) \geq 7 we get that m=7^4 is a divisor of n with m < n, so that by Lemma 2.6 we have
R1(n) < R_1(m)+cR_2(m)+{{log}{\left(\frac{\alpha m}{\beta\sigma(m)}\right)}}-cR_2(n)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}{\left(\frac{\alpha⋅7^4⋅6}{\beta⋅\left(7^5-1\right)}\right]}}-(0.373)\left(\frac1{3^2}+\frac1{7^2}\right)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}\alpha}+{{log}{\left(\frac{7^4⋅6}{(1.444444444)⋅\left(7^5-1\right)}\right]}}-(0.373)\left(\frac1{3^2}+\frac1{7^2}\right)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}2}+{{log}{\left(1+\frac1{2n}\right)}}+{{log}{\left(\frac{14406}{24275.33333}\right]}}-(0.373)\left(\frac1{3^2}+\frac1{7^2}\right)
= 0.272743924 + {{log}{\left(1+\frac1{2n}\right)}}
= 0.272743924 + 0.0000000001, since n > 1035. Thus R1(n) < 0.2727439241.
Suppose 3 | n and 7 ł n
It follows from (1.1.4) (ii) that 3^4| n . Also let m = 1 is a divisor of n with m < n, so that by Lemma 2.6 we have R1(n)\;<\;log\left(\frac\alpha\beta\right)-cR_2(n)
= log \alpha– log \beta– (0.373) \left(\frac1{3^2}\right)
= log 2 + {{log}{\left(1+\frac1{2n}\right)}} – log (1.444444444) – (0.373) \left(\frac1{3^2}\right)
= 0.283977956 + {{log}{\left(1+\frac1{2n}\right)}}
= 0.283977956 + 0.0000000001, since n > 1035. Thus R1(n) < 0.2839779561.
3.2 Theorem:
Suppose 3 n and 5 | n. Then
i) R1(n) < 0.4547419741 if 7 | n.
and
ii) R1(n) < 0.4631158001 if 7 ł n
Proof: (i) Suppose 5 | n and 7 | n.
It follows from (1.1.4) (iii) and (1.1.4) (iv) that 5^6| n and 7^2 | n. Also since w(n) \geq 7 we get that m=7^2 is a divisor of n with m < n, so that by Lemma 2.6 we have
R1(n) < R_1(m)+cR_2(m)+{{log}{\left(\frac{\alpha m}{\beta\sigma(m)}\right)}}-cR_2(n)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}{\left(\frac{\alpha⋅7^2⋅6}{\beta⋅\left(7^3-1\right)}\right]}}-(0.373)\left(\frac1{5^2}+\frac1{7^2}\right)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}\alpha}+{{log}{\left(\frac{7^2⋅6}{(1.24)⋅\left(342\right)}\right]}}-(0.373)\left(\frac1{5^2}+\frac1{7^2}\right)
= \frac17+0.373\left(\frac1{7^2}\right)+{{log}2}+{{log}{\left(1+\frac1{2n}\right)}}+{{log}{\left(\frac{294}{424.08}\right]}}-(0.373)\left(\frac1{5^2}+\frac1{7^2}\right)
= 0.454741974 + {{log}{\left(1+\frac1{2n}\right)}}
= 0.454741974 + 0.0000000001, since n > 1035. Thus R1(n) < 0.4547419741.
Suppose 5 | n and 7 ł n
It follows from (1.1.4) (iii) that 5^6| n. Also let m = 1 is a divisor of n with m < n, so that by Lemma 2.6 we have
R1(n) < {{log}{\left(\frac\alpha\beta\right)}}-cR_2(n)
= log \alpha– log \beta– (0.373) \left(\frac1{5^2}\right)
= log 2 + {{log}{\left(1+\frac1{2n}\right)}} – log (1.24) – (0.373) \left(\frac1{5^2}\right)
= 0.4631158 + {{log}{\left(1+\frac1{2n}\right)}}
= 0.4631158 + 0.0000000001, since n > 1035. Thus R1(n) < 0.4631158001.
3.3 Theorem:
If n is quasi perfect then (15, n) = 1 or 15. More explicitly, every quasi perfect number is divisible by both 3 and 5 or by none of them.
Proof: If n is quasi perfect and (15, n) = 3.
Then by Lemma 3.1, R1(n) < 0.2839779561 while R1(n) > 0.603831 by (1.1.5) (ii), which gives a contradiction. Hence (15, n) \neq3.
Again, if (15, n) = 5.
Then by Lemma 3.2, R1(n) < 0.4631158001 while (1.1.5) (iii) gives R1(n) > 0.647387. These two contradict each other. Hence (15, n) \neq5.
Thus (15, n) = 1 or 15, proving the theorem.
3.4 Theorem:
If n is quasi perfect then 3.5.7 cannot divide n.
Proof: Suppose 3.5.7 divide n. Then it follows from (1.1.4) (ii), (1.1.4) (iii) and (1.1.4) (iv) that 34 | n, 56 | n, 72 | n. But it was proved in 2 that if n \equiv0 (mod 3) and p \equiv1 (mod 3) is a divisor of n then pk | n with k \geq 4. Therefore 74 | n. Also since w(n) \geq7 we get that m = 34.56.74 is a divisor of n with m < n, so that by Lemma 2.6 we have
R1(n) < R_1(m)+cR_2(m)+{{log}{\left(\frac{\alpha m}{\beta\sigma(m)}\right)}}-cR_2(n)
= \frac13+\frac15+\frac17+0.373\left(\frac1{3^2}+\frac1{5^2}+\frac1{7^2}\right)+{{log}{\left(\frac{\alpha⋅3^4⋅5^6⋅7^4⋅2⋅4⋅6}{\beta⋅\left(3^5-1\right)⋅\left(5^7-1\right)⋅\left(7^5-1\right)}\right]}}-(0.373)\left(\frac1{3^2}+\frac1{5^2}+\frac1{7^2}\right)
= \frac13+\frac15+\frac17+{{log}\alpha}+{{log}{\left(\frac{3^4⋅5^6⋅7^4⋅48}{1⋅(3^5-1)⋅(5^7-1)(7^5-1)}\right]}}
= \frac13+\frac15+\frac17+{{log}2}+{{log}{\left(1+\frac1{2n}\right)}}+{{log}{\left(\frac{3^4⋅5^6⋅7^4⋅48}{(3^5-1)⋅(5^7-1)(7^5-1)}\right]}}
= 0.590774338 + {{log}{\left(1+\frac1{2n}\right)}}
= 0.590774338 + 0.0000000001, since n > 1035. Thus R1(n) < 0.5907743381.
But (1.1.5) (iv) gives R1(n) > 0.596063, which gives a contradiction.
Hence 3.5.7 cannot divide n.