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a(n) = (n+1)^4 written in base n.
0

%I #13 May 28 2024 18:28:58

%S 1111111111111111,1010001,100111,21301,20141,15041,14641,14641,14641,

%T 14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,

%U 14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641,14641

%N a(n) = (n+1)^4 written in base n.

%C a(n) = 14641 for n >= 7.

%D GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See pages 108 and 300.

%H <a href="/index/Rec#order_01">Index entries for linear recurrences with constant coefficients</a>, signature (1).

%F G.f.: x*(400*x^6 + 5100*x^5 + 1160*x^4 + 78810*x^3 + 909890*x^2 + 1111111110101110*x - 1111111111111111)/(x - 1). - _Chai Wah Wu_, Apr 23 2023

%e For n=2, 3^4 = 81 = 1010001 in base 2.

%t PadRight[{1111111111111111, 1010001, 100111, 21301, 20141, 15041}, 50, 14641] (* _Paolo Xausa_, May 28 2024 *)

%K nonn,base

%O 1,1

%A _N. J. A. Sloane_, Apr 22 2023