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a(n) = floor((n+1/n)^7).
2

%I #20 Apr 17 2023 15:34:54

%S 128,610,4572,25045,102807,339119,948645,2337564,5211939,10721353,

%T 20642867,37610342,65394162,109236388,176246402,275862066,420381442,

%U 625570111,911349137,1302568701,1829872465,2530657691,3450136161,4642500937,6172204001,8115349815

%N a(n) = floor((n+1/n)^7).

%H Vincenzo Librandi, <a href="/A197708/b197708.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (8,-28,56,-70,56,-28,8,-1).

%F For n > 35, a(n) = n^7 + 7*n^5 + 21*n^3 + 35*n. - _Charles R Greathouse IV_, Dec 27 2011

%F a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8). - _Wesley Ivan Hurt_, Apr 17 2023

%t Table[Floor[(n + 1/n)^7], {n, 40}] (* _T. D. Noe_, Dec 27 2011 *)

%o (Magma) [Floor((n+1/n)^7): n in [1..40]]

%o (PARI) a(n)=floor((n+1/n)^7) \\ _Charles R Greathouse IV_, Dec 27 2011

%Y Cf. A014052, A197602, A197603, A197604.

%K nonn,easy

%O 1,1

%A _Vincenzo Librandi_, Oct 18 2011