OFFSET
1,1
COMMENTS
Using induction, it is easy to prove that a(n)==3 (mod 10).
The largest prime factors of these terms are large (they start 3, 13, 83, 673, 6203, 61613, 642683, 161411, 9221, 870136013, 751453, 4016443, 6267060337, 16318813519213,..)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 1..200
FORMULA
Recurrence: n*a(n) = 3*(5*n-7)*a(n-1) - (23*n-48)*a(n-2) + 9*(n-3)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ sqrt(13*sqrt(10)-40)*(7+2*sqrt(10))^n/(4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
From Ehren Metcalfe, Mar 02 2025: (Start)
G.f.: 1/2 + sqrt(9*x^2 - 14*x + 1)/(2*(x - 1)).
a(n) = (1/2)*Sum_{j=0..n} (-1)^(j+1)*Sum_{k=0..j} binomial(1/2,k)*binomial(1/2, j-k)*(7 + 2*sqrt(10))^k*(7 - 2*sqrt(10))^(j-k) = 1 + 2*Sum_{j=0..n} A379103(j), for n>=1. (End)
MATHEMATICA
aa=ConstantArray[0, 20]; aa[[1]]=3; Do[aa[[n]]=n+2+Sum[aa[[i]]*aa[[n-i]], {i, 1, n-1}], {n, 2, 20}]; aa (* Vaclav Kotesovec, Oct 20 2012 *)
PROG
(Python)
from sympy import series, sqrt, Symbol, Poly
x = Symbol("x")
p = Poly(series((1 + sqrt(9*x**2 - 14*x + 1)/(x - 1))/2, n=20).removeO(), x)
print([p.coeff_monomial(x**n) for n in range(1, p.degree())]) # Ehren Metcalfe, Mar 03 2025
(PARI) my(x='x+O('x^30)); Vec(1/2 + sqrt(9*x^2 - 14*x + 1)/(2*(x - 1))) \\ Michel Marcus, Mar 05 2025
CROSSREFS
KEYWORD
nonn,changed
AUTHOR
Vladimir Shevelev, Mar 05 2010
STATUS
approved