|
Search: id:A038202
|
|
|
| A038202 |
|
Least k such that n!+k^2 is a square. |
|
+0
1
|
|
| 1, 1, 3, 1, 9, 27, 15, 18, 288, 288, 420, 464, 1856, 10080, 46848, 210240, 400320, 652848, 3991680, 27528402, 32659200, 163296000, 1143463200, 1305467240, 6840489600, 9453465438
(list; graph; listen)
|
|
|
OFFSET
|
4,3
|
|
|
COMMENT
|
Let f=n!/4 and let x be the largest divisor of f such that x < sqrt(f). Then a(n) = f/x - x. The greatest k such that n!+k^2 is a square is f-1. The number of k for which n!+k^2 is a square is A038548(f). - T. D. Noe (noe(AT)sspectra.com), Nov 02 2004
|
|
LINKS
|
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
|
|
MATHEMATICA
|
Table[f=n!/4; x=Max[Select[Divisors[f], #<=Sqrt[f]&]]; f/x-x, {n, 4, 20}] (T. D. Noe)
|
|
CROSSREFS
|
Cf. A038548 (number of divisors of n that are at most sqrt(n)).
Sequence in context: A121489 A118793 A105951 this_sequence A128415 A090479 A010289
Adjacent sequences: A038199 A038200 A038201 this_sequence A038203 A038204 A038205
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
David W. Wilson (davidwwilson(AT)comcast.net)
|
|
|
Search completed in 0.002 seconds
|
