|
Search: id:A053606
|
|
| |
|
| 0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432
(list; graph; listen)
|
|
|
OFFSET
|
0,2
|
|
|
COMMENT
|
Define a(1)=0,a(2)=8 with 5*(a(1)^2)+5*a(1)+1=j(1)^2=1^2 and 5*(a(2)^2)+5*a(2)+1=j(2)^2=19^2. Then a(n)=a(n-2)+8*sqrt(5*(a(n-1)^2)+5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2)+5*a(n)+1=j(n)^2. - Pierre CAMI (pierrecami(AT)tele2.fr), Mar 30 2005
It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Apr 03 2005
sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey (Gerald.McGarvey(AT)comcast.net), Apr 19 2005
a(n) is such that 5*a(n)^2+5*a(n)+1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI (pierrecami(AT)tele2.fr), Apr 21 2005
|
|
LINKS
|
F. Ellermann, Illustration of binomial transforms
|
|
FORMULA
|
a(n+1)=9*a(n)+2*(5*(2*a(n)+1)^2-1)^0.5+4 - Richard Choulet (richardchoulet(AT)yahoo.fr), Aug 30 2007
G.f.: f(z)=a(0)+a(1)*z+a(2)*z^2+...=(8*z)/((1-z)*(1-18*z+z^2)) - R. Choulet (richardchoulet(AT)yahoo.fr), Oct 09 2007
|
|
CROSSREFS
|
Cf. A049664, A049629.
Sequence in context: A116876 A075351 A003491 this_sequence A123770 A003400 A059510
Adjacent sequences: A053603 A053604 A053605 this_sequence A053607 A053608 A053609
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
N. J. A. Sloane (njas(AT)research.att.com), James A. Sellers, Jan 20 2000
|
|
|
Search completed in 0.002 seconds
|
