A217541 Smallest numbers n such that s! + n^2 and (s+1)! + n^2 are squares for some s.

1, 108, 108, 1140, 288, 35280, 1068480, 88361280, 4409475840, 2094434496000, 868006971127296000

Offset

1,2

Comments

The values of s are: 4, 8, 9, 10, 12, 14, 16, 18, 22, 24, 32.

It can be seen that n is, on average, an increasing function. (It is constant at s = 8 and s = 9 and decreases at s = 12). If proved this would show there is no repetition of a value of n for which simultaneously s! + n^2 = b^2 and (s+k)! + n^2 = c^2 for general and large values of k (not only for k = 1) and would solve Brocard´s Problem: Exactly, the only 3 solutions to s! + 1 = b^2 are (4,5); (5,11) and (7,71).

Note that n^2 was chosen a square, but this is not necessary.

More terms of the sequence are hard to get if the program based on a simple algorithm, needing 10^9 bytes memory, is not improved in the sense of reducing the number of divisors used. This could probably be done.

Links

Table of n, a(n) for n=1..11.

Example

4! + 1 = 5^2 and 5! +1 = 11^2.
8! + 108^2 = 228^2 and 9! + 108^2 = 612^2.
9! + 108^2 = 612^2 and 10! + 108^2 = 1908^2.
10! + 1140^2 = 2220^2 and 11! + 1140^2 = 6420^2.

Prog

(PARI) for(n=4, 34, a=n!; b=n*a; s=sqrtint(a)+1+sqrtint((n+1)*a)+1; c=divisors(b); for(i=2, #c-1, if(s<=c[i], s=c[i]; r=b\s; if(r%2==1, s=c[i+1]); r=b/s; d=(s-r)/2; t=d^2-a; if(issquare(t), print1(sqrtint(t), ",  "); next(2)))))

Crossrefs

Cf. A217277, A217541, A217550, A217551, A217553.

Sequence in context: A036196 A304565 A327338 * A070797 A173748 A274118

Adjacent sequences: A217538 A217539 A217540 * A217542 A217543 A217544

Keyword

nonn

Author

Robin Garcia, Oct 06 2012

Status

approved