Let S1= 1,4,9,16,…,n^2 and S2=1,5,14,30,…,n(n+1)(2n+1)/6
Clearly the first sequence represents the squares of positive integers while the second their interim sums.
For what smallest value of n (excluding the trivial n=1) will the average of the first n squares be a square number itself?
As I interpret the problem;
First n squares(n^2):
1,4,9,16,25,36,49,...
Sum of first n squares n(n+1)(2n+1)/6
1, 5, 14, 30, 55, 91, 140, 204,...
Sum of sum of first n squares n^2*(n^2-1)/12
1, 6, 20, 50, 105, 196, 336, 540,...
Average of these is (n^2*(n^2-1)/12)/n, but that has a leading zero, so use ((n+1)^2*((n+1)^2-1)/12)/n
{{1, 1}, {2, 3}, {3, 20/3}, {4, 25/2}, {5, 21}, {6, 98/3}, {7, 48}, {8, 135/2}, {9, 275/3}, {10, 121}}
If n=10, then 1 + 5 + 14 + 30 + 55 + 91 + 140 + 204 + 285 + 385 = 1210, divided by 10 = 121 = 11^2
Small values of ((n+1)^2*((n+1)^2-1)/12)/n = m^2 for n:
1,10,25,46,73,106,145,190...
If we ignore the 'interim sums' then the average becomes simply
(n+1)(2n+1)/6= 1,5/2,14/3,15/2,11,91/6,...
a Pellian 2n^2-6m^2+3n+1=0, with small values:
n m
-1 0
1 1
-25 -14
337 195
-4705 -2716
65521 37829
-912601 -526890
12710881 7338631
where n/m converges to sqrt(3) and n(k)/-n(k-1) to 7+4sqrt(3)
for which the smallest qualifying value of n is 337.
Edited on November 16, 2023, 12:39 am
|
|
Posted by broll
on 2023-11-16 00:04:35 |
Possible Solution
