|
Search: id:A031149
|
|
|
| A031149 |
|
Numbers n such that n^2 with last digit deleted is still a perfect square. |
|
+0
2
|
|
| 4, 7, 13, 16, 19, 38, 57, 136, 253, 487, 604, 721, 1442, 2163, 5164, 9607, 18493, 22936, 27379, 54758, 82137, 196096, 364813, 702247, 870964, 1039681, 2079362, 3119043, 7446484, 13853287, 26666893, 33073696, 39480499
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
Square root of 'Squares from A023110'.
|
|
FORMULA
|
Appears to satisfy: a(n)=38a(n-7)-a(n-14) which would require a(-k) to look like 3, 2, 1, 4, 7, 13, 16, 57, 38, 19, 136, ... for k>0. - Henry Bottomley (se16(AT)btinternet.com), May 08 2001
|
|
EXAMPLE
|
364813^2 = 133088524969, 115364^2 = 13308852496
|
|
MAPLE
|
for i from 1 to 150000 do if (floor(sqrt(10 * i^2 + 9)) > floor(sqrt(10 * i^2))) then print(floor(sqrt(10 * i^2 + 9))) end if end do;
|
|
CROSSREFS
|
Cf. A023110, A030686, A030687, A031150.
Sequence in context: A084089 A137827 A045090 this_sequence A074273 A125758 A048297
Adjacent sequences: A031146 A031147 A031148 this_sequence A031150 A031151 A031152
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Patrick De Geest (pdg(AT)worldofnumbers.com)
|
|
|
Search completed in 0.002 seconds
|
