A034494 a(n) = (7^n+1)/2.

1, 4, 25, 172, 1201, 8404, 58825, 411772, 2882401, 20176804, 141237625, 988663372, 6920643601, 48444505204, 339111536425, 2373780754972, 16616465284801, 116315256993604, 814206798955225, 5699447592686572, 39896133148806001, 279272932041642004

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Binomial transform of A081341. Inverse binomial transform of A081342. - R. J. Mathar, Oct 23 2008

Number of compositions of even natural numbers into n parts <=6. - Adi Dani, May 28 2011

From Charlie Marion, Jun 24 2011: (Start)

a(n)+(a(n)+1)+...+(a(n+1)-7^n-1)=(a(n+1)-7^n)+...+(a(n+1)-1). Let S(2n) and S(2n+1) be the sets of addends on the left- and right-hand sides, respectively, of the preceding equations. Then, since the intersection of any 2 different S(i) is null and the union of all of them is the positive integers, {S(i)} forms a partition of the positive integers. See also A034659.

In general, for k>0, let b(n)=((4k+3)^n+1)/2. Then b(n)+(b(n)+1)+ ... +(b(n+1)-(4k+3)^n-1)=k*((b(n+1)-(4k+3)^n)+ ... +(b(n+1)-1)). Then, for each k, the set of addends on the two sides of these equations also forms a partition of the positive integers. Also, with b(0)=1, b(n)=(4k+3)*b(n-1)-(2k+1).

For k>0, let c(0)=1 and, for n>0, c(n)=(2*(2k+1))^n/2. Then the sequence b(0),b(1),... is the binomial transform of the sequence c(0),c(1),....

For k>0, let d(2n)=(2k+1)^(2n) and d(2n+1)=0. Then the sequence b(0),b(1),... is the (2k+2)nd binomial transform of the sequence d(0),d(1),.... (End)

Examples

			From _Adi Dani_, May 28 2011: (Start)
a(2)=25: there are 25 compositions of even numbers into 2 parts <=6:
  (0,0)
  (0,2),(2,0),(1,1)
  (0,4),(4,0),(1,3),(3,1),(2,2)
  (0,6),(6,0),(1,5),(5,1),(2,4),(4,2),(3,3)
  (2,6),(6,2),(3,5),(5,3),(4,4)
  (4,6),(6,4),(5,5)
  (6,6)
(End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (8,-7).

Programs

Magma

[(7^n+1)/2: n in [0..30]]; // Vincenzo Librandi, Jun 16 2011

Maple

A034494:=n->(7^n+1)/2: seq(A034494(n), n=0..30); # Wesley Ivan Hurt, Apr 09 2017

Mathematica

(7^Range[0, 25] + 1)/2 (* or *)
LinearRecurrence[{8, -7}, {1, 4}, 26] (* Paolo Xausa, Aug 14 2025 *)

PARI

a(n)=(7^n+1)/2 \\ Charles R Greathouse IV, Jul 02 2013

PARI

Vec((1-4*x)/((1-x)*(1-7*x)) + O(x^100)) \\ Altug Alkan, Nov 01 2015

Formula

E.g.f.: exp(4*x)*cosh(3*x). - Paul Barry, Apr 20 2003

a(n) = 7*a(n-1) - 3, a(0) = 1.

G.f.: (1-4*x)/((1-x)*(1-7*x)). - Philippe Deléham, Jul 11 2005

a(n) = 8*a(n-1)-7*a(n-2), a(0)=1, a(1)=4. - Philippe Deléham, Nov 15 2008

a(n) = ((4+sqrt(9))^n+(4-sqrt(9))^n)/2. - Al Hakanson (hawkuu(AT)gmail.com), Dec 08 2008