Problems & Puzzles:
Puzzles
Puzzle 214.
Trotter's Curio
Terry Trotter
shows the following
property of the prime 11:
11: Begin
with 11, and continually [i.e. recursively] add the first
five powers of 2, but in reverse order (32,
16, …, 2). All sums are primes (43, 59, 67, 71, and 73).
We may generalize this property and to
ask for the least prime p for a given n>0 value, such all the following
numbers are primes:
p
p+2^{n}
p+2^{n}+2^{n1}
p+2^{n}+2^{n1}+2^{n2}
...
p+2^{n}+2^{n1}+2^{n2}+...2^{2}
+2
Here are the results of my own little
search for these pairs (n, p):
n 
p 
1 
3 
2 
7 
3 
5 
4 
13 
5 
11 (Trotter) 
6 
337 
7 
1889 
8 
25793653 
9 
? 
10 
? 
11 
? 
12 
? 
Question: Can you
complete the above Table of results?
Solution:
Luke Pebody (N=9 &10) and J. K. Andersen
(N=9 to 14) sent contributions to this puzzle.
Here is the Andersen's email
Minimal values of p
n= 9: 13,573,476,641
n=10: 232,317,865,657
n=11: 36,756,785,514,929
n=12: 36,756,785,510,833
n=13: 439,787,787,117,311
n=14: 191,128,877,173,556,587
The n=12 solution is the n=11 solution extended with a smaller prime.
All solutions were found with a modified version of the C program written
for puzzle 206. n=14 took a day on a 1333 MHz Athlon.
***
