A367484 Number of integers of the form (x^4 + y^4) mod 3^n; a(n) = A289559(3^n).

1, 3, 7, 19, 55, 165, 493, 1477, 4429, 13287, 39859, 119575, 358723, 1076169, 3228505, 9685513, 29056537, 87169611, 261508831, 784526491, 2353579471, 7060738413, 21182215237, 63546645709, 190639937125, 571919811375, 1715759434123, 5147278302367, 15441834907099, 46325504721297

Offset

0,2

Links

Paolo Xausa, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,0,0,1,-3).

Formula

a(n) = A289559(3^n). - Thomas Scheuerle, Nov 20 2023

For n < 5: a(n) = 2*3^(n-1) + 1.

a(n) = 2*ceiling(3^(n+3)/80) - 1; a(n) = 2*3^(n-1) + a(n-4). - Conjectured by Albert Mukovskiy, Nov 19 2023; proved by Max Alekseyev, Mar 14 2026

a(n) = 3*a(n-1) + a(n-4) - 3*a(n-5). O.g.f.: (1 - 2*x^2 - 2*x^3 - 3*x^4)/(1 - 3*x - x^4 + 3*x^5). - Max Alekseyev, Mar 14 2026

Mathematica

A367484[n_] := 2*Ceiling[3^(n + 3)/80] - 1; Array[A367484, 30, 0] (* or *)
LinearRecurrence[{3, 0, 0, 1, -3}, {1, 3, 7, 19, 55}, 30] (* Paolo Xausa, Apr 02 2026 *)

Prog

(PARI) a(n) = #setbinop((x, y)->Mod(x, 3^n)^4+Mod(y, 3^n)^4, [0..3^n-1]);
(Python)
def A367484(n):
    m = 3**n
    return len({(pow(x, 4, m)+pow(y, 4, m))%m for x in range(m) for y in range(x+1)}) # Chai Wah Wu, Jan 23 2024

Crossrefs

Subsequence of A289559.

Cf. A000244, A039300, A046631, A155918.

Sequence in context: A183115 A183120 A100702 * A384725 A224031 A147586

Adjacent sequences: A367481 A367482 A367483 * A367485 A367486 A367487

Keyword

nonn,easy

Author

Albert Mukovskiy, Nov 19 2023

Extensions

Terms a(17) onward from Max Alekseyev, Mar 14 2026

Status

approved