Problems & Puzzles: Puzzles

 Puzzle 635. 34155 On March 2003, Jordi Domènech i Arnau computed the first terms of a sequence composed by those integers such that are equal to the sum of the proper divisors greater than its square root. See A182147.Claudio Meller (See his entry 914) says that Jordi speculates that 34155 is the only odd number contained in this sequence. Divisors of 34155 (square root of 34155 is 184.8): 1, 3, 5, 9, 11, 15, 23, 27, 33, 45, 55, 69, 99, 115, 135, 165, 207, 253, 297, 345, 495, 621,759, 1035, 1265,1485, 2277, 3105, 3795, 6831, 11385, 34155 34155 = Σ{207, 253, 297, 345, 495, 621,759, 1035, 1265, 1485, 2277, 3105, 3795, 6831, 11385} On April 2012 Donovan Johnson wrote that no other odd number less than 2*10^11 is in this sequence. See A182147. Q. Can you prove the Jordi's speculation or get another odd number for this sequence?

Contributions cam from Seiji Tomita

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Seiji wrote:

I could not get another odd number, but... I found the interesting property of a(n) for even number n.
Almost of even number n satisfies the form n=2^m*p*q where p,q are prime.

1) n = 2*3*p, p > 2*3.

Let n = r*p*q.
a(n) = p+p*q+p*r = n = r*p*q.
Thus, q=(1+r)/(-1+r),then r-1=1,we get r=2 and q=3.
Smallest n of this case is 2*3*7 = 42.

2) n = 2^2*7*p, p > 2^2*7.

Let n = r^2*p*q.
a(n) = p+r*p+r^2*p+q*p+r*q*p = n = r^2*q*p.
Thus, q=(1+r+r^2)/(-1-r+r^2),then -1-r+r^2=1,we get r=2 and q=7.
Smallest n of this case is 2^2*7*29 = 812.

3) n = 2^4*31*p, p > 2^4*31.

Let n = r^4*p*q.
a(n) = p+r*p+r^2*p+r^3*p+r^4*p+p*q+r*p*q+r^2*p*q+r^3*p*q = n = r^4*q*p.
Thus, q=(1+r+r^2+r^3+r^4)/(-1-r-r^2-r^3+r^4),then -1-r-r^2-r^3+r^4=1,
we get r=2 and q=31.
Smallest n of this case is 2^4*31*499 = 247504.

4) n = 2^6*127*p, p > 2^6*127.
Let n = r^6*p*q.
Similarly,we get r=2 and q=127.
Smallest n of this case is 2^6*127*8147 = 66218816.

Generally,if q=2^m-1 is mersenne prime number and p > 2^(m-1)*q then
n= 2^(m-1)*q*p satisfies the sequence.
For example,if m= 13 then smallest n = 2^12*(2^13-1)*33550337 = 1125625079263232.

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