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URL: https://oeis.org/A056021

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A056021
Numbers k such that k^4 == 1 (mod 5^2).
12
1, 7, 18, 24, 26, 32, 43, 49, 51, 57, 68, 74, 76, 82, 93, 99, 101, 107, 118, 124, 126, 132, 143, 149, 151, 157, 168, 174, 176, 182, 193, 199, 201, 207, 218, 224, 226, 232, 243, 249, 251, 257, 268, 274, 276, 282, 293, 299, 301, 307, 318, 324, 326, 332, 343, 349
OFFSET
1,2
COMMENTS
Numbers congruent to {1, 7, 18, 24} mod 25.
These terms (apart from 1) are tetration bases characterized by a constant convergence speed strictly greater than 1 (see A317905). - Marco Ripà, Jan 25 2024
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Colin Barker)
FORMULA
G.f.: x*(x^2+3*x+1)^2 / ((1+x)*(x^2+1)*(x-1)^2). - R. J. Mathar, Oct 25 2011
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
From Mike Sheppard, Feb 18 2025: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5).
a(n) ~ (5^2/4)*n. (End)
MATHEMATICA
Select[ Range[ 400 ], PowerMod[ #, 4, 25 ]==1& ]
(* or *)
LinearRecurrence[{1, 0, 0, 1, -1}, {1, 7, 18, 24, 26}, 100] (* Mike Sheppard, Feb 18 2025 *)
PROG
(PARI) a(n) = (-25 - (-1)^n + (9-9*I)*(-I)^n + (9+9*I)*I^n + 50*n) / 8 \\ Colin Barker, Oct 16 2015
(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
(PARI) for(n=0, 1e3, if(n^4 % 5^2 == 1, print1(n", "))) \\ Altug Alkan, Oct 16 2015
(PARI) isok(k) = Mod(k, 25)^4 == 1; \\ Michel Marcus, Jun 30 2021
CROSSREFS
Cf. A381319 (general case mod n^2)
Sequence in context: A391309 A070609 A007236 * A090098 A101865 A138391
KEYWORD
nonn,easy
AUTHOR
Robert G. Wilson v, Jun 08 2000
STATUS
approved