|
|
A158444
|
|
a(n) = 16*n^2 + 4.
|
|
2
|
|
|
20, 68, 148, 260, 404, 580, 788, 1028, 1300, 1604, 1940, 2308, 2708, 3140, 3604, 4100, 4628, 5188, 5780, 6404, 7060, 7748, 8468, 9220, 10004, 10820, 11668, 12548, 13460, 14404, 15380, 16388, 17428, 18500, 19604, 20740, 21908, 23108, 24340, 25604, 26900, 28228
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
The identity (8*n^2 + 1)^2 - (16*n^2 + 4)*(2*n)^2 = 1 can be written as A081585(n)^2 - a(n)*A005843(n)^2 = 1.
Sequence found by reading the line from 20, in the direction 20, 68, ... in the square spiral whose vertices are the generalized decagonal numbers A074377. - Omar E. Pol, Nov 02 2012
|
|
LINKS
|
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
|
|
FORMULA
|
G.f.: 4*x*(5 + 2*x + x^2)/(1-x)^3.
Sum_{n>=1} 1/a(n) = (coth(Pi/2)*Pi/2 - 1)/8.
Sum_{n>=1} (-1)^(n+1)/a(n) = (1 - cosech(Pi/2)*Pi/2)/8. (End)
|
|
MATHEMATICA
|
a[n_] := 16*n^2 + 4; Array[a, 50] (* Amiram Eldar, Mar 05 2023 *)
|
|
PROG
|
(Magma) [16*n^2+4: n in [1..50]]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|