cp's OEIS Frontend

A257925 a(n) = (n^2 - n + 1)*(n^2 + n - 1).

%I A257925 #30 Mar 03 2020 07:26:23
%S A257925 1,15,77,247,609,1271,2365,4047,6497,9919,14541,20615,28417,38247,
%T A257925 50429,65311,83265,104687,129997,159639,194081,233815,279357,331247,
%U A257925 390049,456351,530765,613927,706497,809159,922621
%N A257925 a(n) = (n^2 - n + 1)*(n^2 + n - 1).
%C A257925 Subsequence of a(m,n)=(m^2 + n).(n^2 + m)/(m - n)^3 with m=n-1. Q N4 of the 2012 International Mathematical Olympiad paper poses the problem of proving more than 500 solutions exist below 2012 for the equation: a(m,n).(m - n)^3=(m^2 + n).(n^2 + m). Such solutions a(m,n) were called 'Friendly'. If m=2k-1 and n=k-1, solutions of the form a=4k-3 for some integer k, satisfy this requirement although others do exist for other (m,n) pairs e.g. if (m,n)=(1,2), a(m,n)=15.
%C A257925 If m=n-2, a(n)=(n^2 - 3*n + 4)*(n^2 + n - 2)/8. This is the sequence A176145 [t*(t-3)*(t^2-7*t+14)/8] with t=n+2.
%C A257925 Satisfies a linear recurrence having signature (5, -10, 10, -5, 1). - _Harvey P. Dale_, Apr 18 2019
%H A257925 International Mathematical Olympiad 2012, <a href="https://www.imo-official.org/problems/IMO2012SL.pdf">Number Theory Question 4</a>
%F A257925 a(n) = (n^2 - n + 1)*(n^2 + n - 1).
%F A257925 a(n) = A002061(n)*A028387(n-1). - _Michel Marcus_, Apr 17 2016
%e A257925 For n=1, a(1) = 1;
%e A257925 For n=2, a(2) = 15;
%e A257925 For n=3, a(3) = 77.
%t A257925 Table[(n^2-n+1)(n^2+n-1),{n,40}] (* _Harvey P. Dale_, Apr 18 2019 *)
%o A257925 (PARI) a(n) = (n^2 - n + 1)*(n^2 + n - 1); \\ _Michel Marcus_, Apr 17 2016
%Y A257925 Cf. A002061, A028387, A176145.
%K A257925 nonn,easy
%O A257925 1,2
%A A257925 _Matthew Ryan_, Apr 17 2016