|
| |
|
|
A105076
|
|
Sequence a(n) such that 60*(a(n)^2)+60*a(n)+1 = j^2 = a square.
|
|
0
| |
|
|
0, 1, 2, 11, 19, 90, 153, 712, 1208, 5609, 9514, 44163, 74907, 347698, 589745, 2737424, 4643056, 21551697, 36554706, 169676155, 287794595, 1335857546, 2265802057, 10517184216, 17838621864, 82801616185, 140443172858, 651895745267
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 1,3
|
|
|
COMMENTS
| Proof: define a(1)=0, a(2)=1, a(3)=2, a(4)=11 the 4 first values such that 60*(a(n)^2)+60*a(n)+1=j^2= a square then a(n)=8*a(n-2)+3-a(n-4) 60*(a(n)^2)+60*a(n)+1=j^2 ? need to found positive integer solution a(n) for a(n)^2+a(n)-(j^2-1)/60=0 so (2*a(n)+1)^2=(4*(j^2)+56)/60=(j^2+14)/15 so (j^2+14)/15 needs to be = k^2 or j^2=15*(k^2)-14 put k=2*a(n)+1, 15*((2*a(n)+1)^2)-14 = 60*(a(n)^2)+60*a(n)+1. QED
|
|
|
FORMULA
| a(n) = a(n-1) +8*a(n-2) -8*a(n-3) -a(n-4) +a(n-5). G.f.: x^2*(1+x+x^2)/((1-x)*(x^4-8*x^2+1)). [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 13 2009]
|
|
|
CROSSREFS
| Cf. A103200, A001090.
Sequence in context: A186267 A067660 A103200 * A067670 A159879 A017185
Adjacent sequences: A105073 A105074 A105075 * A105077 A105078 A105079
|
|
|
KEYWORD
| nonn
|
|
|
AUTHOR
| Pierre CAMI (pierre-cami(AT)bbox.fr), Apr 06 2005
|
| |
|
|
