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a(n) = floor((n+3)^(n+2)/((n+3)^2-1)).
1

%I #8 Sep 08 2022 08:45:12

%S 1,4,26,222,2451,33288,538084,10101010,216145205,5195862732,

%T 138679078110,4070332170534,130325562613351,4521260802379792,

%U 168962471790509960,6767528048726614650,289242639716420115369

%N a(n) = floor((n+3)^(n+2)/((n+3)^2-1)).

%H G. C. Greubel, <a href="/A089816/b089816.txt">Table of n, a(n) for n = 0..385</a>

%t Table[Floor[(n + 3)^(n + 2)/((n + 3)^2 - 1)], {n, 0, 50}] (* _G. C. Greubel_, Oct 11 2017 *)

%o (PARI) for(n=0, 25, print1(floor((n+3)^(n+2)/((n+3)^2-1)), ", ")) \\ _G. C. Greubel_, Oct 11 2017

%o (Magma) [Floor((n+3)^(n+2)/((n+3)^2-1)): n in [0..25]]; // _G. C. Greubel_, Oct 11 2017

%Y Cf. A000975, A033113-A033119, A059848, A062864, A056830.

%K nonn

%O 0,2

%A _Paul Barry_, Nov 12 2003