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

1, 15, 77, 247, 609, 1271, 2365, 4047, 6497, 9919, 14541, 20615, 28417, 38247, 50429, 65311, 83265, 104687, 129997, 159639, 194081, 233815, 279357, 331247, 390049, 456351, 530765, 613927, 706497, 809159, 922621

Offset

1,2

Comments

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.

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.

Satisfies a linear recurrence having signature (5, -10, 10, -5, 1). - Harvey P. Dale, Apr 18 2019

Links

Table of n, a(n) for n=1..31.

International Mathematical Olympiad 2012, Number Theory Question 4

Formula

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

a(n) = A002061(n)*A028387(n-1). - Michel Marcus, Apr 17 2016

Example

For n=1, a(1) = 1;
For n=2, a(2) = 15;
For n=3, a(3) = 77.

Mathematica

Table[(n^2-n+1)(n^2+n-1), {n, 40}] (* Harvey P. Dale, Apr 18 2019 *)

Prog

(PARI) a(n) = (n^2 - n + 1)*(n^2 + n - 1); \\ Michel Marcus, Apr 17 2016

Crossrefs

Cf. A002061, A028387, A176145.

Sequence in context: A205440 A051423 A205609 * A205433 A303097 A374219

Adjacent sequences: A257922 A257923 A257924 * A257926 A257927 A257928

Keyword

nonn,easy

Author

Matthew Ryan, Apr 17 2016

Status

approved