A269657 Number of length-4 0..n arrays with no adjacent pair x,x+1 repeated.

1, 15, 79, 253, 621, 1291, 2395, 4089, 6553, 9991, 14631, 20725, 28549, 38403, 50611, 65521, 83505, 104959, 130303, 159981, 194461, 234235, 279819, 331753, 390601, 456951, 531415, 614629, 707253, 809971, 923491, 1048545, 1185889, 1336303

Offset: 0

Author

R. H. Hardin, Mar 02 2016

Keywords

• nonn

Comments

I.e., a(n) = # {x in {0..n}^4 | x[1] != x[0]+1 or x[2] != x[0] or x[3] != x[1]}. The only possibility to have an adjacent x,x+1 pair repeated in a length-4 array is to have the array (x,x+1,x,x+1), with 0 <= x <= n-1 given the restriction on the domain of coefficients. This implies a(n) = (n+1)^4 - n and previously conjectured formulas. - M. F. Hasler, Feb 29 2020

Examples

			From _M. F. Hasler_, Feb 29 2020: (Start)
For n=0, the only length-4 0..0 array is (0,0,0,0) and it satisfies the restriction, so a(0) = 1.
For n=1, there is only one 4-tuple with coefficients in 0..1 which has a repeated pair (x,x+1), namely (0,1,0,1). Thus, a(1) = 2^4 - 1 = 15.
For n=2, there are two 4-tuples with coefficients in 0..2 which have a repeated pair (x,x+1), namely (0,1,0,1) and (1,2,1,2). Thus, a(1) = 3^4 - 2 = 79.
(End)
Some solutions for n=3 (length-4 arrays shown as columns):
  1  1  0  2  0  2  2  3  0  3  2  1  0  3  1  1
  1  0  0  1  3  2  0  1  2  3  2  1  2  0  0  2
  1  1  2  2  1  0  0  2  2  0  2  0  0  0  0  1
  3  3  0  1  0  0  1  2  2  1  3  3  2  2  0  3
		

Links

• R. H. Hardin, Table of n, a(n) for n = 0..210 (a(0) = 1 inserted by _M. F. Hasler_, Feb 29 2020).
• John Elias, Illustration of initial terms: chain-linked cubes

Crossrefs

Row 4 of A269656.

Programs

Mathematica

Denominator/@Flatten[Table[x/.Solve[m-Sqrt[-1/(1/(1/(1-x)-(m-1))-(m+1))]==0],{m,2,34}]] (* Ed Pegg Jr, Jan 14 2020 *)

PARI

apply( {A269657(n)=(n+1)^4-n}, [0..44]) \\ M. F. Hasler, Feb 29 2020

Formula

Empirical: a(n) = n^4 + 4*n^3 + 6*n^2 + 3*n + 1.

Conjectures from Colin Barker, Jan 25 2019: (Start)

G.f.: (1 + 10*x + 14*x^2 - 2*x^3 + x^4) / (1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.

(End)

a(n) = (n+1)^4 - n, cf. comment, confirming the above conjectured formulas. - M. F. Hasler, Feb 29 2020

Extensions

Extended to a(0) = 1 by M. F. Hasler, Feb 29 2020