Problems & Puzzles: Problems

Problem 21.- Divisors (III). Certain questions about s(N), the sum of divisors of a number N.

Let's define s(N) as the sum of the positive divisors of N.

Example: s(14) = 1+2+7+14 = 24

a) Can you calculate s(2097727632)? Can you explain your algorithm?

b) What non-trivial can you say about the existence of solutions for s(N) = K, for any positive number K?

c) Can you calculate or - at least - bound the minimal N for a given K, supposed sigma(N) = K has solutions.

d) There are many couples of consecutive numbers N and N+1 such that s(N) = s(N+1). Some of these are 14 & 15, 206 & 207, 957 & 958, etc. (See B13, p. 68 of Richard K. Guy's book "Unsolved Problems in Number Theory". Also the Sloane's sequence A002961). But as far as I know, it hasn't been discovered any triplet of consecutive numbers such that s(N) = s(N+1) = s(N+2). Can you find the first of such triplets? (Hint: Jud McCranie - who has done exhaustive search in this subject also, informs at 22/07/99: "I found no solutions for s(n)=s(n+1)=s(n+2) for n < 4,250,000,000").

Solution

a) T.W.A Baumann & Enoch Haga independently sent the following solution at (25/07/99):
"If N = p1^a1*p2^a2*p3^a3*... | p(i) are the prime factors of N
then
s(N) = (p1^(a1+1)-1)/(p1-1)*(p2^(a2+1)-1/(p2-1)*...

N = 2097727632 = 2^4 * 3^5 * 7^3 * 11^2 * 13
s(2097727632) = (2^5-1)/1*(3^6-1)/2*(7^4-1)/6*(11^3-1)/10*
(13^2-1)/12 = 8404323200
"

Enoch sent the following reference for this: Albert Beiler, "Recreations in Theory of Numbers", p. 20.


Dominique Toublanc wrote (April, 2006):

I don't know if this has been done since the last update of problem 21 but i have checked s(N)=s(N+1)=s(N+2) up to 15,000,000,000 and found no solutions...

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