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Numbers k such that k^4 == 1 (mod 5^2).
11

%I #59 Feb 11 2024 04:47:32

%S 1,7,18,24,26,32,43,49,51,57,68,74,76,82,93,99,101,107,118,124,126,

%T 132,143,149,151,157,168,174,176,182,193,199,201,207,218,224,226,232,

%U 243,249,251,257,268,274,276,282,293,299,301,307,318,324,326,332,343,349

%N Numbers k such that k^4 == 1 (mod 5^2).

%C Numbers congruent to {1, 7, 18, 24} mod 25.

%C These terms (apart from 1) are tetration bases characterized by a constant convergence speed strictly greater than 1 (see A317905). - _Marco RipĂ _, Jan 25 2024

%H Amiram Eldar, <a href="/A056021/b056021.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..1000 from Colin Barker)

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).

%F G.f.: x*(x^2+3*x+1)^2 / ((1+x)*(x^2+1)*(x-1)^2). - _R. J. Mathar_, Oct 25 2011

%F a(n) = (-25 - (-1)^n + (9-9*i)*(-i)^n + (9+9*i)*i^n + 50*n) / 8, where i = sqrt(-1). - _Colin Barker_, Oct 16 2015

%t Select[ Range[ 400 ], PowerMod[ #, 4, 25 ]==1& ]

%o (PARI) a(n) = (-25 - (-1)^n + (9-9*I)*(-I)^n + (9+9*I)*I^n + 50*n) / 8 \\ _Colin Barker_, Oct 16 2015

%o (PARI) Vec(x*(x^2+3*x+1)^2/((1+x)*(x^2+1)*(x-1)^2) + O(x^100)) \\ _Colin Barker_, Oct 16 2015

%o (PARI) for(n=0, 1e3, if(n^4 % 5^2 == 1, print1(n", "))) \\ _Altug Alkan_, Oct 16 2015

%o (PARI) isok(k) = Mod(k, 25)^4 == 1; \\ _Michel Marcus_, Jun 30 2021

%Y Cf. A056022, A056024, A056025, A056026, A056027, A056028, A056031, A056034, A056035, A317905.

%K nonn,easy

%O 1,2

%A _Robert G. Wilson v_, Jun 08 2000