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 A081215 #26 Dec 12 2020 15:52:36
%S A081215 1,0,1,5,41,434,5713,90075,1657009,34867844,826446281,21794641505,
%T A081215 633095889817,20088655029078,691413758034721,25657845139503479,
%U A081215 1021273028302258913,43404581642184336392,1961870762757168078553
%N A081215 a(n) = (n^(n+1)+(-1)^n)/(n+1)^2.
%C A081215 From _Mathew Englander_, Oct 19 2020: (Start)
%C A081215 The sum of two adjacent terms of the sequence cannot be prime.
%C A081215 In base n, a(n) has n-1 digits, which are (beginning from the left): n-2, 2, n-4, 4, and so on, except that if n is even the rightmost digit is 1 instead of 0. In that case, the other digits form a palindrome with every even digit from 2 to n-2 appearing twice. For example, a(14) in base 14 is c2a486684a2c1. If n is odd, then all digits from 1 to n-1 occur exactly once. For example, a(15) in base 15 is d2b496785a3c1e.
%C A081215 For any positive integer k, any prime p, and any positive integer h such that h*p > 2, a(h*p^k - 2) == (-1)^h * (1 - 2^(h-1)) (mod p). For example, a(7*p^k - 2) == 63 (mod p); a(10*p^k - 2) == -511 (mod p).
%C A081215 Suppose k and m are positive integers. If k is even, then a(k*m) == 1, a(k*m+1) == 0, and a(k*m-1) == -1 (all mod m). If k is odd, then a(k*m) == (-1)^m and a(k*m+1) == ceiling(m/2) (both mod m), while a(k*m-1) == m/2 - 1 for m even, and a(k*m-1) == 1 for m odd (mod m).
%C A081215 For proofs of the above, see the Englander link. (End)
%H A081215 Robert Israel, <a href="/A081215/b081215.txt">Table of n, a(n) for n = 0..387</a>
%H A081215 Mathew Englander, <a href="/A081215/a081215.pdf">Comments on OEIS A081215</a>
%F A081215 a(n) = (-1)^n + Sum_{k=1..n} (-1)^(k+1)*(n+1)^(n-k)*C(n+1,n+2-k). - _Gionata Neri_, May 19 2016
%F A081215 E.g.f.: (Ei(1,x) - Ei(1,-LambertW(-x)))/x. - _Robert Israel_, May 19 2016
%F A081215 For n > 1, a(n) = Sum_{k=1..floor(n/2)} (n^(n-2*k) * (2*k/n + n - 2*k)). - _Mathew Englander_, Oct 19 2020
%p A081215 seq((j^(j+1)+(-1)^j)/(j+1)^2, j=0..50); # _Robert Israel_, May 19 2016
%t A081215 Array[(#^(# + 1) + (-1)^#)/(# + 1)^2 &, 19, 0] (* _Michael De Vlieger_, Nov 13 2020 *)
%o A081215 (PARI) a(n) = (n^(n+1)+(-1)^n)/(n+1)^2; \\ _Michel Marcus_, Oct 20 2020
%Y A081215 Cf. A081209, A110567, A076951, A273319, A193746, and A060073.
%K A081215 easy,nonn
%O A081215 0,4
%A A081215 _Vladeta Jovovic_, Apr 17 2003