|
Search: id:A105099
|
|
|
| A105099 |
|
Nonnegative numbers n such that 23*n^2 + 23*n + 1 = j^2 = a square. |
|
+0
1
|
|
| 0, 335, 815, 772320, 1877280, 1777881455, 4321498895, 4092682338240, 9948088580160, 9421352964748175, 22900495590030575, 21687950432167961760, 52716930900161804640, 49925652473497683224495, 121354352031676884251855
(list; graph; listen)
|
|
|
OFFSET
|
1,2
|
|
|
COMMENT
|
a(5)=2649601*(2*a(1)+1)-1-a(4), a(6)=2649601*(2*a(2)+1)-1-a(3), a(7)=2649601*(2*a(3)+1)-1-a(2), a(8)=2649601*(2*a(4)+1)-1-a(1), a(9)=2649601*(2*a(5)+1)-1-a(1), a(10)=2649601*(2*a(6)+1)-1-a(2), a(11)=2649601*(2*a(7)+1)-1-a(3), a(12)=2649601*(2*a(8)+1)-1-a(4), a(13)=2649601*(2*a(9)+1)-1-a(1), a(14)=2649601*(2*a(10)+1)-1-a(1). This is a strange recurrence - does it continue ? Remark : 2649601 = 23*24*25*192+1
In terms of indices of triangular numbers: A000217(n) = 4*A000217[(j-1)/2]/23. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
|
|
FORMULA
|
Union of two sequences defined by the recurrence a(n+1)=2302*a(n)-a(n-1)+1150 a(0)=0, a(1)=335, a(2)=772320, ... a(0)=0, a(1)=815, a(2)=1877280, ... - Max Alekseyev (maxale(AT)gmail.com), Apr 09 2005
O.g.f.: -5*(67*x^2+96*x+67)*x^2/((x^2+48*x+1)*(x^2-48*x+1)*(-1+x)). - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
|
|
CROSSREFS
|
Sequence in context: A020367 A119760 A046015 this_sequence A038648 A137522 A090487
Adjacent sequences: A105096 A105097 A105098 this_sequence A105100 A105101 A105102
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Pierre CAMI (pierrecami(AT)tele2.fr), Apr 07 2005
|
|
EXTENSIONS
|
More terms from Max Alekseyev (maxale(AT)gmail.com), Apr 09 2005
More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Dec 05 2007
|
|
|
Search completed in 0.002 seconds
|
