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a(n) = n^n mod 100.
0

%I #32 Aug 13 2023 13:23:48

%S 1,1,4,27,56,25,56,43,16,89,0,11,56,53,16,75,16,77,24,79,0,21,84,67,

%T 76,25,76,3,36,69,0,31,76,13,36,75,36,17,4,59,0,41,64,7,96,25,96,63,

%U 56,49,0,51,96,73,56,75,56,57,84,39,0,61,44,47,16,25,16,23

%N a(n) = n^n mod 100.

%H <a href="/index/Rec#order_100">Index entries for linear recurrences with constant coefficients</a>, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

%F For n >= 101, a(n) = a(n-100), i.e., cyclic with period A174824(100) = 100, disregarding a(0). - _Michael S. Branicky_, Sep 26 2021

%t Table[PowerMod[n,n,100],{n,0,70}] (* _Harvey P. Dale_, Aug 13 2023 *)

%o (Python)

%o def a(n): return pow(n, n, 100)

%o print([a(n) for n in range(101)]) # _Michael S. Branicky_, Sep 26 2021

%Y Cf. A000312 (n^n), A056849 (mod 10), A174824.

%K nonn,base,easy

%O 0,3

%A _John Bibby_, Sep 10 2021