2015 Countdown Problem #16: Some Brute Force & Luck Needed – Part I

Number Theory Level 4

A sequence is defined by the recurrence relation $u_1=2015$ and $u_n$ equals the sum of the squares of each digit in $u_{(n-1)}$ for $n≥2$ .

How many more square numbers than prime numbers are there between $u_1$ and $u_{2015}$ inclusive?

#GoodLuckHaveFun

This problem is part of the set 2015 Countdown Problems .

The answer is 0.

1 solution

Josh Banister
Dec 21, 2014

The sequence is 2015, 30, 9, 81, 65, 61, (37, 58, 89, 145, 42, 20, 4, 16) and loops back to 37. The first part (unlooped) of the sequence has 2 squares and 1 prime and the looped portion has 2 square and 2 primes. By repeating that loop, the 2015th number in the sequence is 37 (which is a prime). Without calculation, There is one more square than prime in the unlooped portion, no changes after the completed loops and one more prime than square in the unfinished loop leaving a total of zero.

You have assumed by finding a pattern that the loop continues till 2015. Is there any much better solid proof ?

Venkata Karthik Bandaru - 6 years, 3 months ago

2015 Countdown Problem #16: Some Brute Force & Luck Needed – Part I

The answer is 0.

1 solution

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