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Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.
2

%I #52 Feb 21 2020 20:07:41

%S 6,8,20,24,42,48,72,80,110,120,156,168,210,224,272,288,342,360,420,

%T 440,506,528,600,624,702,728,812,840,930,960,1056,1088,1190,1224,1332,

%U 1368,1482,1520,1640,1680,1806,1848,1980,2024,2162,2208,2352,2400,2550,2600

%N Maximum remainder when (k + 1)^n + (k - 1)^n is divided by k^2 for variable n and k > 2.

%H Iain Fox, <a href="/A159469/b159469.txt">Table of n, a(n) for n = 3..10000</a>

%H Project Euler, <a href="http://projecteuler.net/index.php?section=problems&amp;id=120">Problem 120: Square remainders</a>

%H Iain Fox, <a href="/A159469/a159469.txt">Proof for recursive definition and generating function</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F maxr(n) = n*n - 2*n if n is even, and n*n - n if n is odd.

%F G.f.: x^3*(-6-2*x)/((x+1)^2*(x-1)^3). - Maksym Voznyy (voznyy(AT)mail.ru), Jul 26 2009 (proved by _Iain Fox_, Nov 26 2017)

%F a(n) = 2*A050187(n). - _R. J. Mathar_, Aug 08 2009 (proved by _Iain Fox_, Nov 27 2017)

%F a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 7. - _Colin Barker_, Oct 29 2017 (proved by _Iain Fox_, Nov 26 2017)

%F a(n) = n^2 - n*(3 + (-1)^n)/2. - _Iain Fox_, Nov 26 2017

%F From _Iain Fox_, Nov 27 2017: (Start)

%F a(n) = A000290(n) - A022998(n).

%F a(n) = 2*A093005(n-2) + A168273(n-1).

%F a(n) = (4*(A152749(n-2)) + A091574(n-1) - A010719(n-1))/3.

%F E.g.f.: x*(exp(x)*x - sinh(x)).

%F (End)

%e For n = 3, maxr => 3*3 - 3 = 6 since 3 is odd.

%e For n = 4, maxr => 4*4 - 2*4 = 8 since 4 is even.

%t LinearRecurrence[{1,2,-2,-1,1},{6,8,20,24,42},50] (* _Harvey P. Dale_, Apr 18 2018 *)

%o (PARI) a(n) = if (n % 2, n^2 - n, n^2 - 2*n); \\ _Michel Marcus_, Aug 26 2013

%o (PARI) first(n) = Vec(x^3*(-6-2*x)/((x+1)^2*(x-1)^3) + O(x^(n+3))) \\ _Iain Fox_, Nov 26 2017

%Y Cf. A050187.

%K nonn,easy

%O 3,1

%A _Gaurav Kumar_, Apr 13 2009