|
Search: id:A106232
|
|
|
| A106232 |
|
Least k > 0 such that (4*n^2 + 2)*(k^2) + (4*n^2 + 2)*k + 1 = j^2. |
|
+0
2
|
|
| 4, 4, 12, 24, 40, 60, 84, 112, 144, 180, 220, 264, 312, 364, 420
(list; graph; listen)
|
|
|
OFFSET
|
1,1
|
|
|
COMMENT
|
For k there is alway a recurrence for n=1 k(1,1)=0, k(2,1)=4 then k(i,1)=10*k(i-1,1)+4-k(i-3,n) for n>1 k(1,n)=0, k(2,n)=2*n^2-2*n, k(3,n)=2*n^2+2*n k(4,n)=(8*n^2+2)*k(2,n)+4*n^2 then k(i,n)=(8*n^2+2)*k(i-2,n)+4*n^2-k(i-4,n)
|
|
FORMULA
|
for n=1 k=4, for n > 1 k(n) = 2*n^2 - 2*n j sequence = A106231
a(n)=A046092(n-1), n>1. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Aug 28 2008]
|
|
CROSSREFS
|
Cf. A106231.
Sequence in context: A120033 A097073 A019085 this_sequence A038804 A088838 A127403
Adjacent sequences: A106229 A106230 A106231 this_sequence A106233 A106234 A106235
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Pierre CAMI (pierrecami(AT)tele2.fr), Apr 26 2005
|
|
|
Search completed in 0.002 seconds
|
