|
| |
|
|
A106789
|
|
Sum of two consecutive squares of Lucas 3-step numbers (A001644).
|
|
0
| |
|
|
10, 10, 58, 170, 562, 1962, 6562, 22202, 75242, 254330, 860474, 2911226, 9848050, 33316090, 112707970, 381286954, 1289885834, 4363653034, 14762129274, 49939929610, 168945571442, 571538767370, 1933501811618, 6540989771354
(list; graph; refs; listen; history; internal format)
|
|
|
|
OFFSET
| 0,1
|
|
|
COMMENTS
| A106729 is sum of two consecutive squares of Lucas numbers (A001254), for which L(n)^2 + L(n+1)^2 = 5*{F(n)^2 + F(n+1)^2} = 5*A001519(n). Sum of two consecutive squares of Lucas 3-step numbers can be expressed in terms of tribonacci numbers, but not quite as neatly, as derived from the identity A001644(n) = T(n) + 2*T(n-1) + 3*T(n-2) = 3*T(n+1) - 2*T(n) - T(n-1) where the tribonacci numbers T(n) = A000073(n).
|
|
|
FORMULA
| a(n) = A001644(n)^2 + A001644(n+1)^2.
G.f.: (-10*x^5-2*x^4-36*x^3+8*x^2-10*x+10)/[(1+x+x^2-x^3)*(1-3*x-x^2+x^3)].
|
|
|
EXAMPLE
| a(0) = A001644(0)^2 + A001644(1)^2 = 3^2 + 1^2 = 9 + 1 = 10.
a(1) = A001644(1)^2 + A001644(2)^2 = 1^2 + 3^2 = 1 + 9 = 10.
a(2) = A001644(2)^2 + A001644(3)^2 = 3^2 + 7^2 = 9 + 49 = 58.
a(3) = A001644(3)^2 + A001644(4)^2 = 7^2 + 11^2 = 49 + 121 = 170 = 13^2 + 1.
|
|
|
CROSSREFS
| Cf. A001644, A106729.
Sequence in context: A201027 A003875 A111220 * A056483 A056473 A165831
Adjacent sequences: A106786 A106787 A106788 * A106790 A106791 A106792
|
|
|
KEYWORD
| easy,nonn
|
|
|
AUTHOR
| Jonathan Vos Post (jvospost3(AT)gmail.com), May 16 2005
|
| |
|
|
