Problems & Puzzles: Puzzles

Puzzle 419. Four SOPF questions Fred Schneider sent the following puzzle:  SOPF(N) =sum of prime factors of N ND(N) = num of divisors of N Sigma(N) = divisor sum for N  Let N=35, SOPF(N)=5+7=12 ND(N)=2*2=4 Sigma(N)=35+7+5+1 = 48 So SOPF(N)*ND(N)=Sigma(N)  Q1: 35 is the only number to have this property. Can you prove it? ============================== Let N=22446139 SOPF(N)=31+67+101*107=306 ND(N) = 2^4 = 16 Sigma(N) = 23970816 (calculated in PARI) So SOPF(N)*ND(N))^2 = Sigma(N)  Q2: This is the first such number. Are there an infinite number of these cases? ====================================================== Here's a number such that  (Sigma(N)) = (SOPF(N) * ND(N))^3 N=14,844,221,560,107,739  =  31*71*79*109*127*131*179*263 Sigma(N) = 2^5 * (2^3 * 3^2) * (2^4 * 5) * (2*5*11) * (2^7) * (2^2*3*11) * (2^2*3^2*5) * (2^3*3*11) = 2^27 * 3^6 * 5^3 * 11^3 SOPF(N) = 990 = 2*3^2*5*11 ND(N) = 2^8  Q3:  Can you find a smaller solution? ================================================== Q4: Can you a find a number N such that (Sigma(N)) = (SOPF(N) * ND(N))^P where P is an integer > 3? ==================================================

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Contributions came from Enoch Haga & Farideh Firoozbakht & J. Wroblewski.

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Enoch wrote for Q1:

1) SOPF(20) = 2+5 = 7
ND(20) = 6 (1+2+4+5+10+20) = 42
SIGMA(20) = 42
7*6=42

2) SOPF(42) = 2+3+7 = 12
ND(42) = 8 (1+2+3+6+7+14+21+42) = 96
12*8=96
 

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Farideh wrote for Q1:

It seems that 35, 42 & 68 are the only numbers N such that

SOPF(N)*ND(N)=Sigma(N).

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Jarek found many solutions related to Q3 & Q4 for P =3 to 7 & several SOPFs.

I am putting selected results in the directory http://www.math.uni.wroc.pl/~jwr/pp419/ with file names being of the form "P-SOPF.TXT". I am assuming numbers to have distinct factors, and with that assumption my search for given P and SOPF is exhaustive.

It looks like the smallest number for P=3 I have at the moment is 7803872444012429 = 47*139*149*223*239*359*419 (first line in 3-1575.TXT)

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