%I #30 Sep 08 2022 08:46:16

%S 4,6,8,12,14,16,24,26,28,32,34,36,44,46,48,52,54,56,64,66,68,72,74,76,

%T 84,86,88,92,94,96,104,106,108,112,114,116,124,126,128,132,134,136,

%U 144,146,148,152,154,156,164,166,168,172,174,176,184,186,188,192,194

%N Numbers k such that the final digit of k^k is 6.

%C The values of n^n (A000312) end in every digit except for 2 and 8. The sequence of final digits of n^n (A056849) is periodic with period 20; for n=1,2,... the last digits are [1, 4, 7, 6, 5, 6, 3, 6, 9, 0, 1, 6, 3, 6, 5, 6, 7, 4, 9, 0]. Thus, 6 is the most common final digit of n^n. Since 6 does not occur at any odd index in the list above, all terms of a(n) are even. Also, from the distribution of 6's in the list, we can see that the difference between any two consecutive values of a(n) will be 2, 4 or 8.

%H Felix Fröhlich, <a href="/A271346/b271346.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,1,-1).

%F a(n) = a(n-1) + a(n-6) - a(n-7) for n > 7. - _Wesley Ivan Hurt_, Oct 08 2017

%F G.f.: 2*x*(1 + x^2)*(2 + x - x^2 + x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)). - _Colin Barker_, Dec 13 2018

%p A271346:=n->`if`(n^n mod 10 = 6, n, NULL): seq(A271346(n), n=1..500);

%t LinearRecurrence[{1, 0, 0, 0, 0, 1, -1}, {4, 6, 8, 12, 14, 16, 24},59] (* _Ray Chandler_, Mar 08 2017 *)

%o (PARI) is(n) = Mod(n, 10)^n==6 \\ _Felix Fröhlich_, Apr 07 2016

%o (Magma) I:=[4,6,8,12,14,16,24]; [n le 7 select I[n] else Self(n-1)+Self(n-6)-Self(n-7): n in [1..60]]; // _Vincenzo Librandi_, Oct 09 2017

%o (PARI) Vec(2*x*(1 + x^2)*(2 + x - x^2 + x^3 + 2*x^4) / ((1 - x)^2*(1 + x)*(1 - x + x^2)*(1 + x + x^2)) + O(x^59)) \\ _Colin Barker_, Dec 13 2018

%Y Cf. A000312 (n^n), A056849 (final digit of n^n).

%K nonn,base,easy

%O 1,1

%A _Wesley Ivan Hurt_, Apr 04 2016