1000: cube of 10

1024: tenth power of 2

1184: Paganini’s number

1210: the number amicable to Paganini’s number

1225: square of 35

1331: cube of 11

1444: square of 38

1681: square of 41

1728: cube of 12

1729: Hardy’s taxicab number

1854: subfactorial 7

2025: square of 45

2047: first composite Mersenne number

2310: primorial of 11

2538: Trigg’s number

4096: twelfth power of 2

5040: factorial of 7

5041: square of 71

6174: Kaprekar’s constant

7560: smallest number having the number of its factors: 64

7824: Boolean Pythagorean triples limit

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# Day: July 18, 2020

# Pisano periods

Here is the link to the Wikipedia article on that topic.

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# article: Some conditional results on primes between consecutive squares

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# a result on perfect numbers on Mathematics Stack Exchange

Definition. (suggested new terminology) If n is an even perfect number, then the statement that p is the generating prime for n means that p is the integer such that n = (2^(p – 1))*(2^p – 1).

The following theorem is posted on Mathematics Stack Exchange:

Theorem. If n is an even perfect number, then n has 2p – 1 proper divisors, and their product is n^(p-1), where p is the generating prime for n.

Corollary. If n is an even perfect number, then the product of the divisors of n is a power of n.

(However, the converse is not true. For example, the product of the divisors of 8 is 64, and 64 = 8^2, but 8 is not a perfect number.)

Here is the link.

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[even perfect numbers]

[a necessary condition for an even number to be perfect]

# Cototient characterization of even perfect numbers

An even number is perfect if, and only if, its cototient is the square of the highest power of 2 contained in it.

Details:

For even n, call the highest power of 2 that divides n bouba(n), and call n/bouba(n) kiki(n). (Saying that a number is ‘totally kiki’ (or, ‘purely kiki’) is just a picturesque way of saying that it is odd.)

Theorem. If n is an even number, then n is perfect fif the cototient of n is (bouba(n))^2.

Proof:

First, we address the ‘only if’ portion.

Suppose that n is an even number. Let bouba(n) = 2^(w – 1), where w > 1.

If n is perfect, then 2^w – 1 is prime and:

cototient of n = n – phi(n) = bouba(n)*kiki(n) – phi((bouba(n)*kiki(n))

= bouba(n)*kiki(n) – phi(bouba(n))*phi(kiki(n))

= (2^(w-1))*(2^w-1) – phi(2^(w-1)*phi(2^w-1)

= (2^(w-1))*(2^w-1) – 2^(w-2)*phi(2^w-1)

= (2^(w-1))*(2^w-1) – 2^(w-2)*(2^w-2) good

= 2^(2w-1) – 2^(w-1) – 2^(2w-2) + 2^(w-1)

= 2^(2w-1) – 2^(2w-2) – 2^(w-1) + 2^(w-1)

= 2^(2w-2)

= (2^(w-1))*(2^(w-1))

= (2^(w-1))^2

= (bouba(n))^2

Now we address the ‘if’ portion.

Suppose that the cototient of n is (bouba(n))^2.

Then:

cototient of n is (bouba(n))^2

implies

bouba(n)*kiki(n) – phi(bouba(n)*kiki(n)) = (bouba(n))^2

implies

(2^(w-1))*kiki(n) – phi((2^(w-1))*kiki(n)) = (2^(w-1))^2

implies

(2^(w-1))*kiki(n) – phi(2^(w-1))*phi(kiki(n)) = (2^(w-1))^2.

implies

(2^(w-1))*kiki(n) – 2^(w-2)*phi(kiki(n)) = 2^(2w – 2)

implies

2*kiki(n) – phi(kiki(n)) = 2^w

implies

phi(kiki(n)) – 2*kiki(n) = -2^w

implies

phi(kiki(n)) = 2*kiki(n) – 2^w

implies

phi(kiki(n))/2 = kiki(n) – 2^(w-1)

implies

kiki(n) is a power of an odd prime, p (because the RHS is odd, and therefore the LHS must also be odd, but if kiki(n) has more than one prime factor, then 4 is a divisor of phi(kiki(n)), and so the LHS would be even)

implies

kiki(n) = p (because if there exists a k > 1 such that kiki(n) = p^k, then phi(kiki(n)) = phi(p^k) = (p^(k-1))(p-1), and therefore (p^(k-1))*(p-1)/2 = p^k – 2^(w-1), and therefore (p^(k-1))*(p-1) = 2*p^k – 2^w, and therefore p^k – p^(k-1) = 2*p^k – 2^w and therefore -p^(k-1) = p^k – 2^w, and therefore 2^w = p^k – p^(k-1) and therefore p^k – p^(k-1) = 2^w, and therefore p*(p^(k-1) – p^(k-2)) = 2^w and therefore p is a divisor of 2^w and therefore an odd number > 1 is a divisor of a power of 2, a contradiction)

implies

(p-1)/2 = p – 2^(w-1)

implies

p – 1 = 2p – 2^w

implies

p = 2^w – 1

implies

n = (2^(w-1)*(2^w-1), where 2^w-1 is a prime

implies

n is perfect QED

notation: ‘fif’ means ‘if, and only if,’.

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# article: The Humble Sum of Remainders Function

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# ‘has a convergent reciprocal sum’

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# harmonic divisor numbers

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# radical of a number

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[rad(n)]