cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A066443 Number of distinct paths of length 2n+1 along edges of a unit cube between two fixed adjacent vertices.

Original entry on oeis.org

1, 7, 61, 547, 4921, 44287, 398581, 3587227, 32285041, 290565367, 2615088301, 23535794707, 211822152361, 1906399371247, 17157594341221, 154418349070987, 1389765141638881, 12507886274749927, 112570976472749341
Offset: 0

Views

Author

John W. Layman, Aug 12 2002

Keywords

Comments

All members of sequence are also hex, or central hexagonal, numbers (A003215). (If n is a hex number, 9n - 2 is always a hex number; see recurrence.) - Matthew Vandermast, Mar 30 2003
The sequence 1,1,7,61,547,... with g.f. (1-9x+6x^2)/((1-x)(1-9x)) and a(n) = A054879(n)/3 + 2*0^n/3 gives the denominators in the probability that a random walk on the cube returns to its starting corner on the 2n-th step. - Paul Barry, Mar 11 2004
Equals row sums of even row terms of triangle A158303. - Gary W. Adamson, Mar 15 2009
It appears that a(n) is the n-th record value in A120437, which gives the differences of A037314 (positive integers n such that the sum of the base 3 digits of n equals the sum of the base 9 digits of n). - John W. Layman, Dec 14 2010
Numbers in base 9 are 1, 6+1, 66+1, 666+1, 6666+1, 66666+1, etc.; that is, n 6's + 1. - Yuchun Ji, Aug 15 2019
All prime factors of a(n) are 1 mod 6. In addition, if n is not 1 mod 3 (first index being n=0), then 3 is a cubic residue modulo all prime factors of a(n). This provides a simple proof that there are infinitely many primes 1 mod 6 that have 3 as a cubic residue. - William Hu, Jul 26 2024

Examples

			From _Michael B. Porter_, Aug 22 2016: (Start)
Give coordinates (a,b,c) to the vertices of the cube, where a, b, and c are either 0 or 1. For n = 1, the a(1) = 7 paths of length 2n + 1 = 3 from (0,0,0) to (0,0,1) are:
(0,0,0) -> (0,0,1) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (0,0,1) -> (0,1,1) -> (0,0,1)
(0,0,0) -> (0,0,1) -> (1,0,1) -> (0,0,1)
(0,0,0) -> (0,1,0) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (0,1,0) -> (0,1,1) -> (0,0,1)
(0,0,0) -> (1,0,0) -> (0,0,0) -> (0,0,1)
(0,0,0) -> (1,0,0) -> (1,0,1) -> (0,0,1) (End)
		

Crossrefs

Cf. A158303, A037314, A120437, A083234 (binomial transform), A083233 (inverse binomial transform), A054879 (recurrent walks), A125857 (walks ending on face diagonal), A054880 (walks ending on space diagonal).

Programs

  • Magma
    [(3^(2*n+1)+1)/4: n in [0..20]]; // Vincenzo Librandi, Jun 16 2011
    
  • Maple
    seq((3^(2*n+1) + 1)/4, n=0..18); # Zerinvary Lajos, Jun 16 2007
  • Mathematica
    NestList[9 # - 2 &, 1, 18] (* or *)
    Table[(3^(2 n + 1) + 1)/4, {n, 0, 18}] (* or *)
    CoefficientList[Series[(1 - 3 x)/((1 - x) (1 - 9 x)), {x, 0, 18}], x] (* Michael De Vlieger, Aug 22 2016 *)
  • PARI
    a(n)=3^(2*n+1)\/4 \\ Charles R Greathouse IV, Jul 02 2013
    
  • PARI
    Vec((1-3*x)/((1-x)*(1-9*x)) + O(x^50)) \\ Altug Alkan, Nov 13 2015

Formula

a(n) = (3^(2*n+1)+1)/4. - Vladeta Jovovic, Dec 22 2002
a(n) = 9*a(n-1) - 2. - Matthew Vandermast, Mar 30 2003
From Paul Barry, Apr 21 2003: (Start)
G.f.: (1-3*x)/((1-x)*(1-9*x)).
E.g.f.: (3*exp(9*x) + exp(x))/4. (End)
a(n) = (-1)^n times the (i, i)-th element of M^n (for any i), where M = ((1, 1, 1, -2), (1, 1, -2, 1), (1, -2, 1, 1), (-2, 1, 1, 1)). - Simone Severini, Nov 25 2004
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k)*4^(n-k). - Paul Barry, Jan 22 2005
a(n) = A054880(n) + 1.
a(n) = A057660(3^n). - Henry Bottomley, Nov 08 2015
a(n) = Sum_{k=0..2n} (-3)^k == 1 + Sum_{k=1..n} 2*3^(2k-1). - Bob Selcoe, Aug 21 2016
a(n) = 3^(2*n+1) * a(-1-n) for all n in Z. - Michael Somos, Jul 02 2017
a(n) = 6*A002452(n) + 1. - Yuchun Ji, Aug 15 2019

Extensions

Corrected by Vladeta Jovovic, Dec 22 2002

A122983 a(n) = (2 + (-1)^n + 3^n)/4.

Original entry on oeis.org

1, 1, 3, 7, 21, 61, 183, 547, 1641, 4921, 14763, 44287, 132861, 398581, 1195743, 3587227, 10761681, 32285041, 96855123, 290565367, 871696101, 2615088301, 7845264903, 23535794707, 70607384121, 211822152361, 635466457083
Offset: 0

Views

Author

Paul Barry, Sep 22 2006

Keywords

Comments

Old definition was: "Binomial transform of aeration of A081294".
Binomial transform is A063376.
A122983 = (1,1,3,7,1,1,3,7,...) mod 10. - M. F. Hasler, Feb 25 2008
Equals row sums of triangle A158301. - Gary W. Adamson, Mar 15 2009
a(n) = the number of ternary sequences of length n where the numbers of (0's, 1's) are both even. A015518 covers the (odd, even) and (even, odd) cases, and A081251 covers (odd, odd). - Toby Gottfried, Apr 18 2010
This sequence also describes the number of moves of the k-th disk solving (non-optimally) the [RED ; NEUTRAL ; BLUE] pre-colored Magnetic Tower of Hanoi (MToH) puzzle. The sequence A183119 is the partial sums of the sequence in question (obviously describing the total number of moves associated with the specific solution algorithm). For other MToH-related sequences, Cf. A183111 - A183125.
Let B=[1,sqrt(2),0; sqrt(2),1,sqrt(2); 0,sqrt(2),1] be a 3 X 3 matrix. Then a(n)=[B^n](1,1), n=0,1,2,.... - _L. Edson Jeffery, Dec 21 2011
Also the domination number of the n-Hanoi graph. - Eric W. Weisstein, Jun 16 2017
Also the matching number of the n-Sierpinski gasket graph. - Eric W. Weisstein, Jun 17 2017
Let M = [1,1,1,0; 1,1,0,1; 1,0,1,1; 0,1,1,1], a 4 X 4 matrix. Then a(n) is the upper left entry in M^n. - Philippe Deléham, Aug 23 2020
Also the lower matching number (=independent domination number) of the n-Hanoi graph. - Eric W. Weisstein, Aug 01 2023

Crossrefs

Cf. a(j+1) = A137822(2^j) and these are the record values of A137822.
Cf. A054879 (bisection), A066443 (bisection). Row sums of A158303.

Programs

Formula

From Paul Barry, Jun 14 2007: (Start)
G.f.: (1-2*x-x^2)/((1-x)*(1+x)*(1-3*x));
a(n) = 3^n/4+(-1)^n/4+1/2;
E.g.f.: cosh(x)^2*exp(x). (End)
a(n) = 3*a(n-1) + a(n-2) - 3*a(n-3); a(0)=1, a(1)=1, a(2)=3. - Harvey P. Dale, Sep 03 2013
E.g.f.: Q(0)/2, where Q(k) = 1 + 3^k/( 2 - 2*(-1)^k/( 3^k + (-1)^k - 2*x*3^k/( 2*x + (k+1)*(-1)^k/Q(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Dec 22 2013
a(2*n) = 3*a(2*n-1); a(2*n+1) = 3*a(2*n) - 2. - Philippe Deléham, Aug 23 2020

Extensions

Extended and corrected (existing Maple code) by M. F. Hasler, Feb 25 2008
Description changed to formula by Eric W. Weisstein, Jun 16 2017
Showing 1-2 of 2 results.