OFFSET
0,3
COMMENTS
16*k*a(n) provides infinitely many x-values solutions (x,y) of x*(5*x + k) = y^2.
This follows from the fact that 5*16*a(n) + 1 is a perfect square: more precisely, 80*a(n) + 1 = A023039(n)^2.
This is a divisibility sequence, that is a(n) divides a(m) if n divides m. It is the case P1 = 324, P2 = 644, Q = 1 of the 3 parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Jan 19 2018
LINKS
Colin Barker, Table of n, a(n) for n = 0..300
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
Index entries for linear recurrences with constant coefficients, signature (323,-323,1).
FORMULA
G.f.: x*(1 + x)/((1 - x)*(1 - 322*x + x^2)).
a(n) = a(-n) = ((2 + sqrt(5))^(4*n) + (2 - sqrt(5))^(4*n) - 2)/320.
a(n) = A225786(n)/48. This is the case k=3 of the first comment. Example: for n = 2, 16*3*a(2) = A225786(2) = 15552 and 15552*(5*15552+3) = 34776^2.
a(n) = A049660(n)^2.
a(n)*(80*a(n) + 1) = 81*A253368(n)^2 for n>0.
a(n)*a(n-2) = (a(n-1) - 1)^2.
a(n) = 322*a(n-1) - a(n-2) + 2.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3). - Iain Fox, Jan 12 2018
MAPLE
P:=proc(n) trunc(evalf(((2+sqrt(5))^(4*n)+(2-sqrt(5))^(4*n)-2)/320, 1000));
end: seq(P(i), i=0..13); # Paolo P. Lava, Jan 18 2018
MATHEMATICA
CoefficientList[x (1 + x)/((1 - x) (1 - 322 x + x^2)) + O[x]^20, x]
PROG
(Sage)
gf = x*(1+x)/((1-x)*(1-322*x+x^2))
print(taylor(gf, x, 0, 20).list())
(Maxima) makelist(coeff(taylor(x*(1+x)/((1-x)*(1-322*x+x^2)), x, 0, n), x, n), n, 0, 20);
(PARI) first(n) = Vec(x*(1 + x)/((1 - x)*(1 - 322*x + x^2)) + O(x^n), -n) \\ Iain Fox, Jan 12 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruno Berselli, Jan 12 2018
STATUS
approved