A016131 Expansion of 1/((1-2*x)*(1-8*x)).

1, 10, 84, 680, 5456, 43680, 349504, 2796160, 22369536, 178956800, 1431655424, 11453245440, 91625967616, 733007749120, 5864062009344, 46912496107520, 375299968925696, 3002399751536640, 24019198012555264, 192153584100966400

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

"Numbral" powers of 10 (see A048888 for definition). - John W. Layman, Dec 18 2001

For n > 1, a(n-1) is the (2^n-3)rd coefficient in the expansion of th(0)=y, th(n+1) = th(n)*(th(n) + 1).

If 2^(n+1) is the length of the even leg of a primitive Pythagorean triangle (PPT) then it constrains the odd leg to have a length of 4^n-1 and the hypotenuse to have a length of 4^n+1. The resulting triangle has a semiperimeter of 4^n+2^n, an area of 8^n-2^n and an inradius of 2^n-1. Now consider the term 8^n-2^n: it must at least be divisible by 6 because it is the area of a PPT. a(n) is 1/6 the area of such triangles. - Frank M Jackson, Dec 28 2017

Links

• Iain Fox, Table of n, a(n) for n = 0..1107
• Index entries for linear recurrences with constant coefficients, signature (10,-16).

Crossrefs

Cf. A000079, A002450, A016203, A048888, A059155, A120689, A248217.

Programs

Magma

[Binomial(2^(n+1)+1,3): n in [0..40]]; // G. C. Greubel, Dec 26 2024

Maple

seq(binomial(2^n,2)*(2^n + 1)/3,n=1..20); # Zerinvary Lajos, Jan 07 2008

Mathematica

CoefficientList[Series[1/((1 - 2x)(1 - 8x)), {x, 0, 100}], x] (* Stefan Steinerberger, Apr 21 2006 *)
a[n_] := (8^n-2^n)/6; Array[a, 20] (* Frank M Jackson, Dec 28 2017 *)

PARI

Vec(1/((1-2*x)*(1-8*x))+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

Python

def A016131(n): return binomial(pow(2,n+1) +1, 3)
print([A016131(n) for n in range(41)]) # G. C. Greubel, Dec 26 2024

Sage

[lucas_number1(n,10,16) for n in range(1, 21)] # Zerinvary Lajos, Apr 26 2009

Sage

[(8^n - 2^n)/6 for n in range(1,21)] # Zerinvary Lajos, Jun 05 2009

Formula

a(0) = 1, a(n) = (2^(3n+2) - 2^n)/3 = A059155(n)/12 = A000079(n)*A002450(n+1) = A016203(n+1) - A016203(n). - Ralf Stephan, Aug 14 2003

a(n) = binomial(2^n,2)*(2^n + 1)/3, n >= 1. - Zerinvary Lajos, Jan 07 2008

a(n) = (8^(n+1) - 2^(n+1))/6. - Al Hakanson (hawkuu(AT)gmail.com), Jan 07 2009

a(n) = Sum_{i=1...(2^n -1)} i*(i+1)/2. - Ctibor O. Zizka, Mar 03 2009

a(0) = 1, a(n) = 8*a(n-1) + 2^n. - Vincenzo Librandi, Feb 09 2011

a(n) = 10*a(n-1) - 16*a(n-2), n > 1. - Vincenzo Librandi, Feb 09 2011

a(n) = A248217(n+1)/6. - Frank M Jackson, Dec 28 2017

E.g.f.: e^(2*x) * (4*e^(6*x) - 1)/3. - Iain Fox, Dec 28 2017