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.
%I A269657 #31 Aug 18 2021 16:45:58
%S A269657 1,15,79,253,621,1291,2395,4089,6553,9991,14631,20725,28549,38403,
%T A269657 50611,65521,83505,104959,130303,159981,194461,234235,279819,331753,
%U A269657 390601,456951,531415,614629,707253,809971,923491,1048545,1185889,1336303
%N A269657 Number of length-4 0..n arrays with no adjacent pair x,x+1 repeated.
%C A269657 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
%H A269657 R. H. Hardin, <a href="/A269657/b269657.txt">Table of n, a(n) for n = 0..210</a> (a(0) = 1 inserted by _M. F. Hasler_, Feb 29 2020).
%H A269657 John Elias, <a href="/A269657/a269657.png">Illustration of initial terms: chain-linked cubes</a>
%F A269657 Empirical: a(n) = n^4 + 4*n^3 + 6*n^2 + 3*n + 1.
%F A269657 Conjectures from _Colin Barker_, Jan 25 2019: (Start)
%F A269657 G.f.: (1 + 10*x + 14*x^2 - 2*x^3 + x^4) / (1 - x)^5.
%F A269657 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
%F A269657 (End)
%F A269657 a(n) = (n+1)^4 - n, cf. comment, confirming the above conjectured formulas. - _M. F. Hasler_, Feb 29 2020
%e A269657 From _M. F. Hasler_, Feb 29 2020: (Start)
%e A269657 For n=0, the only length-4 0..0 array is (0,0,0,0) and it satisfies the restriction, so a(0) = 1.
%e A269657 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.
%e A269657 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.
%e A269657 (End)
%e A269657 Some solutions for n=3 (length-4 arrays shown as columns):
%e A269657 1 1 0 2 0 2 2 3 0 3 2 1 0 3 1 1
%e A269657 1 0 0 1 3 2 0 1 2 3 2 1 2 0 0 2
%e A269657 1 1 2 2 1 0 0 2 2 0 2 0 0 0 0 1
%e A269657 3 3 0 1 0 0 1 2 2 1 3 3 2 2 0 3
%t A269657 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 *)
%o A269657 (PARI) apply( {A269657(n)=(n+1)^4-n}, [0..44]) \\ _M. F. Hasler_, Feb 29 2020
%Y A269657 Row 4 of A269656.
%K A269657 nonn
%O A269657 0,2
%A A269657 _R. H. Hardin_, Mar 02 2016
%E A269657 Extended to a(0) = 1 by _M. F. Hasler_, Feb 29 2020