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A173998
For n>=1, a(n) = n + 2 + Sum_{i=1..n-1} a(i)*a(n-i).
1
3, 13, 83, 673, 6203, 61613, 642683, 6940673, 76930803, 870136013, 10002590883, 116521027873, 1372486213803, 16318813519213, 195599588228683, 2360929398934273, 28671940652447203, 350089944825571213, 4295280755452388083, 52926654021145267873
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
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
Sequence in context: A219906 A000904 A201304 * A135743 A123114 A104032
KEYWORD
nonn,changed
AUTHOR
Vladimir Shevelev, Mar 05 2010
STATUS
approved