A322489 - OEIS

%I #51 Feb 27 2023 04:02:34

%S 2,18,22,38,42,58,62,78,82,98,102,118,122,138,142,158,162,178,182,198,

%T 202,218,222,238,242,258,262,278,282,298,302,318,322,338,342,358,362,

%U 378,382,398,402,418,422,438,442,458,462,478,482,498,502,518,522,538,542,558

%N Numbers k such that k^k ends with 4.

%C Also numbers k == 2 (mod 4) such that 2^k and k^2 end with the same digit.

%C Numbers congruent to {2, 18} mod 20. - _Amiram Eldar_, Feb 27 2023

%H Colin Barker, <a href="/A322489/b322489.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1,1,-1).

%F O.g.f.: 2*x*(1 + 8*x + x^2)/((1 + x)*(1 - x)^2).

%F E.g.f.: 2 + 3*exp(-x) + 5*(2*x - 1)*exp(x).

%F a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).

%F a(n) = 10*n + 3*(-1)^n - 5. Therefore:

%F a(n) = 10*n - 8 for odd n;

%F a(n) = 10*n - 2 for even n.

%F a(n+2*k) = a(n) + 20*k.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = tan(2*Pi/5)*Pi/20 = sqrt(5+2*sqrt(5))*Pi/20. - _Amiram Eldar_, Feb 27 2023

%p select(n->n^n mod 10=4,[$1..558]); # _Paolo P. Lava_, Dec 18 2018

%t Table[10 n + 3 (-1)^n - 5, {n, 1, 60}]

%o (Sage) [10*n+3*(-1)^n-5 for n in (1..70)]

%o (Maxima) makelist(10*n+3*(-1)^n-5, n, 1, 70);

%o (GAP) List([1..70], n -> 10*n+3*(-1)^n-5);

%o (Magma) [10*n+3*(-1)^n-5: n in [1..70]];

%o (Python) [10*n+3*(-1)**n-5 for n in range(1, 70)]

%o (Julia) [10*n+3*(-1)^n-5 for n in 1:70] |> println

%o (PARI) apply(A322489(n)=10*n+3*(-1)^n-5, [1..70]) \\ _M. F. Hasler_, Dec 14 2018

%o (PARI) Vec(2*x*(1 + 8*x + x^2) / ((1 - x)^2*(1 + x)) + O(x^70)) \\ _Colin Barker_, Dec 13 2018

%Y Cf. A004526, A056849.

%Y Subsequence of A139544, A235700.

%Y Numbers k such that k^k ends with d: A008592 (d=0), A017281 (d=1), A067870 (d=3), this sequence (d=4), A017329 (d=5), A271346 (d=6), A322490 (d=7), A017377 (d=9).

%K nonn,base,easy

%O 1,1

%A _Bruno Berselli_, Dec 12 2018