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A105099
Nonnegative numbers n such that 23*n^2 + 23*n + 1 = j^2 = a square.
1
0, 335, 815, 772320, 1877280, 1777881455, 4321498895, 4092682338240, 9948088580160, 9421352964748175, 22900495590030575, 21687950432167961760, 52716930900161804640, 49925652473497683224495, 121354352031676884251855
OFFSET
1,2
COMMENTS
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, 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, 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, Dec 05 2007
MATHEMATICA
LinearRecurrence[{1, 2302, -2302, -1, 1}, {0, 335, 815, 772320, 1877280}, 20] (* Harvey P. Dale, May 20 2021 *)
CROSSREFS
Sequence in context: A046015 A184076 A253227 * A038648 A224533 A253346
KEYWORD
nonn,easy
AUTHOR
Pierre CAMI, Apr 07 2005
EXTENSIONS
More terms from Max Alekseyev, Apr 09 2005
More terms from R. J. Mathar, Dec 05 2007
STATUS
approved