|
|
A075528
|
|
Triangular numbers that are half other triangular numbers.
|
|
23
|
|
|
0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,2
|
|
COMMENTS
|
This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012
|
|
LINKS
|
|
|
FORMULA
|
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
Lim_{n->infinity} a(n)/a(n-1) = 17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
|
|
MATHEMATICA
|
CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)
|
|
PROG
|
(PARI) concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015
|
|
CROSSREFS
|
Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|