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%I #41 Feb 27 2023 04:02:43
%S 3,17,23,37,43,57,63,77,83,97,103,117,123,137,143,157,163,177,183,197,
%T 203,217,223,237,243,257,263,277,283,297,303,317,323,337,343,357,363,
%U 377,383,397,403,417,423,437,443,457,463,477,483,497,503,517,523,537,543,557,563
%N Numbers k such that k^k ends with 7.
%C Equivalently, numbers k such that k and (7^h)^k end with the same digit, where h == 1 (mod 4).
%C Also, numbers k such that k and (3^h)^k end with the same digit, where h == 3 (mod 4).
%C Numbers congruent to {3, 17} mod 20. - _Amiram Eldar_, Feb 27 2023
%H Colin Barker, <a href="/A322490/b322490.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.: x*(3 + 14*x + 3*x^2)/((1 + x)*(1 - x)^2).
%F E.g.f.: 3 + 2*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 + 2*(-1)^n - 5. Therefore:
%F a(n) = 10*n - 7 for odd n;
%F a(n) = 10*n - 3 for even n.
%F a(n+2*k) = a(n) + 20*k.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = tan(7*Pi/20)*Pi/20. - _Amiram Eldar_, Feb 27 2023
%p select(n->n^n mod 10=7,[$1..563]); # _Paolo P. Lava_, Dec 18 2018
%t Table[10 n + 2 (-1)^n - 5, {n, 1, 60}]
%t LinearRecurrence[{1,1,-1},{3,17,23},80] (* _Harvey P. Dale_, Sep 15 2019 *)
%o (Sage) [10*n+2*(-1)^n-5 for n in (1..70)]
%o (Maxima) makelist(10*n+2*(-1)^n-5, n, 1, 70);
%o (GAP) List([1..70], n -> 10*n+2*(-1)^n-5);
%o (Magma) [10*n+2*(-1)^n-5: n in [1..70]];
%o (Python) [10*n+2*(-1)**n-5 for n in range(1, 70)]
%o (Julia) [10*n+2*(-1)^n-5 for n in 1:70] |> println
%o (PARI) apply(A322490(n)=10*n+2*(-1)^n-5, [1..70])
%o (PARI) Vec(x*(3 + 14*x + 3*x^2) / ((1 + x)*(1 - x)^2) + O(x^55)) \\ _Colin Barker_, Dec 13 2018
%Y Cf. A004526, A056849.
%Y Subsequence of A063226, A295009.
%Y Similar sequences are listed in A322489.
%K nonn,base,easy
%O 1,1
%A _Bruno Berselli_, Dec 12 2018
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