A379103 -id:A379103 - cp's OEIS frontend

A110147 10^((n^2-n)/2).

1, 1, 10, 1000, 1000000, 10000000000, 1000000000000000, 1000000000000000000000, 10000000000000000000000000000, 1000000000000000000000000000000000000

Offset: 0

Philippe Deléham, Sep 04 2005

Sequence given by the Hankel transform (see A001906 for definition) of A082148 = {1, 1, 11, 131, 1661, 22101, 305151, 4335711, ...}; example: det([1, 1, 11, 131; 1, 11, 131, 1661; 11, 131, 1661, 22101; 131, 1661, 22101, 305151]) = 10^6 = 1000000.

Also the Hankel transform of A379103. - Nathaniel Johnston, Dec 16 2024

Cf. A006125, A047656, A053763, A053764, A109345, A109354, A109493, A109966.

a(n)=10^binomial(n,2) \\ Charles R Greathouse IV, Jan 17 2012

a(n+1) is the determinant of n X n matrix M_(i, j) = binomial(10i, j).

a(n)=10a(n-1)^2/a(n-2), a(0)=a(1)=1. - Michael Somos, Sep 12 2005

A173998 For n>=1, a(n) = n + 2 + Sum_{i=1..n-1} a(i)*a(n-i).

Original entry on oeis.org

3, 13, 83, 673, 6203, 61613, 642683, 6940673, 76930803, 870136013, 10002590883, 116521027873, 1372486213803, 16318813519213, 195599588228683, 2360929398934273, 28671940652447203, 350089944825571213, 4295280755452388083, 52926654021145267873

Offset: 1

Vladimir Shevelev, Mar 05 2010

nonn

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,..)

Vincenzo Librandi, Table of n, a(n) for n = 1..200

Cf. A030431, A379103.

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 *)

my(x='x+O('x^30)); Vec(1/2 + sqrt(9*x^2 - 14*x + 1)/(2*(x - 1))) \\ Michel Marcus, Mar 05 2025

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

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)

cp's OEIS Frontend

A110147 10^((n^2-n)/2).

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