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Number of integers of the form (x^4 + y^4) mod 3^n; a(n) = A289559(3^n).
+20
1
1, 3, 7, 19, 55, 165, 493, 1477, 4429, 13287, 39859, 119575, 358723, 1076169, 3228505, 9685513, 29056537
COMMENTS
It appears that for n > 4: a(n) = 2*3^(n-1) + a(n-4).
For n < 5: a(n) = 2*3^(n-1) + 1.
Conjecture in closed form: a(n) = 2*ceiling(3^(n+3)/80) - 1.
FORMULA
Conjecture: a(n) = 2*ceiling(3^(n+3)/80) - 1.
PROG
(PARI) a(n) = #setbinop((x, y)->Mod(x, 3^n)^4+Mod(y, 3^n)^4, [0..3^n-1]);
(Python)
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
Largest prime number p such that x^n + y^n mod p does not take all values on Z/pZ.
+10
4
7, 29, 61, 223, 127, 761, 307, 911, 617, 1741, 1171, 2927, 3181, 2593, 1667, 4519, 2927, 10781, 5167, 6491, 8419, 7177, 5501, 7307, 9829, 11117, 12703, 20011, 10789, 13249, 18217, 22271, 14771, 29629, 13691, 18773, 22543, 21601, 19927, 46411, 18749, 32957
COMMENTS
As discussed in the link below, the Hasse-Weil bound assures that x^n + y^n == k (mod p) always has a solution when p - 1 - 2*g*sqrt(p) > n, where g = (n-1)*(n-2)/2 is the genus of the Fermat curve. For example, p > n^4 is large enough to satisfy this condition, so we only need to check finitely many p below n^4.
Conjecture: a(n) == 1 (mod n). - Jason Yuen, May 18 2024
EXAMPLE
a(3) = 7 because x^3 + y^3 == 3 (mod 7) does not have a solution, but x^3 + y^3 == k (mod p) always has a solution when p is a prime greater than 7.
PROG
(Python)
import sympy
def a(n):
g = (n - 1) * (n - 2) / 2
plist = list(sympy.primerange(2, n ** 4))
plist.reverse()
for p in plist:
# equivalent to p-1-2*g*p**0.5 > n:
if (p - 1 - n) ** 2 > 4 * g * g * p:
continue
solution = [False] * p
r = sympy.primitive_root(p)
rn = r ** n % p # generator for subgroup {x^n}
d = sympy.n_order(rn, p) # size of subgroup {x^n}
nth_power_list = []
xn = 1
for k in range(d):
xn = xn * rn % p
nth_power_list.append(xn)
for yn in nth_power_list:
solution[(xn + yn) % p] = True
for yn in nth_power_list: # consider the case x=0
solution[yn] = True
solution[0] = True
if False in solution:
return p
return -1
print([a(n) for n in range(3, 18)])
Number of modulo n residues among sums of two sixth powers.
+10
2
1, 2, 3, 3, 5, 6, 3, 3, 3, 10, 11, 9, 5, 6, 15, 5, 17, 6, 10, 15, 9, 22, 23, 9, 25, 10, 7, 9, 29, 30, 16, 9, 33, 34, 15, 9, 19, 20, 15, 15, 41, 18, 29, 33, 15, 46, 47, 15, 15, 50, 51, 15, 53, 14, 55, 9, 30, 58, 59, 45, 51, 32, 9, 17, 25, 66, 56, 51, 69, 30, 71
COMMENTS
This sequence appears to be multiplicative (verified through n = 10000).
This sequence is multiplicative. In general, by the Chinese remainder theorem, the number of distinct residues modulo n among the values of any multivariate polynomial with integer coefficients will be multiplicative. - Andrew Howroyd, Aug 01 2018
EXAMPLE
a(5) = 5 because (j^6 + k^6) mod 5, where j and k are integers, can take on all 5 values 0..4; e.g.:
(1^6 + 2^6) mod 5 = ( 1 + 64) mod 5 = 65 mod 5 = 0;
(0^6 + 1^6) mod 5 = ( 0 + 1) mod 5 = 1 mod 5 = 1;
(1^6 + 1^6) mod 5 = ( 1 + 1) mod 5 = 2 mod 5 = 2;
(2^6 + 2^6) mod 5 = (64 + 64) mod 5 = 128 mod 5 = 3;
(0^6 + 2^6) mod 5 = ( 0 + 64) mod 5 = 64 mod 5 = 4.
a(7) = 3 because (j^6 + k^6) mod 7 can take on only the three values 0, 1, and 2. (This is because j^6 mod 7 = 0 for all j divisible by 7, 1 otherwise.)
PROG
(PARI) a(n) = #Set(vector(n^2, i, ((i%n)^6 + (i\n)^6) % n)); \\ Michel Marcus, Jul 10 2017
CROSSREFS
Cf. A057760, A155918 (gives number of modulo n residues among sums of two squares), A289559 (Number of modulo n residues among sums of two fourth powers).
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