cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Original entry on oeis.org

1, 3, 7, 19, 55, 165, 493, 1477, 4429, 13287, 39859, 119575, 358723, 1076169, 3228505, 9685513, 29056537
Offset: 0

Views

Author

Albert Mukovskiy, Nov 19 2023

Keywords

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.

Crossrefs

Subsequence of A289559.
Cf. A000244, A039300, A046631, A155918.

Programs

  • 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

Formula

Conjecture: a(n) = 2*ceiling(3^(n+3)/80) - 1.
a(n) = A289559(3^n). - Thomas Scheuerle, Nov 20 2023

A355920 Largest prime number p such that x^n + y^n mod p does not take all values on Z/pZ.

Original entry on oeis.org

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
Offset: 3

Views

Author

Seiichi Azuma, Jul 21 2022

Keywords

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

Examples

			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.

Links

Crossrefs

Cf. A289559.

Programs

  • 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)])

Extensions

a(13)-a(25) from Jinyuan Wang, Jul 22 2022
a(26)-a(27) from Chai Wah Wu, Sep 13 2022
a(28)-a(33) from Chai Wah Wu, Sep 14 2022
a(34)-a(40) from Robin Visser, Dec 09 2023
a(41)-a(49) from Robin Visser, Mar 18 2024

A289630 Number of modulo n residues among sums of two sixth powers.

Original entry on oeis.org

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
Offset: 1

Views

Author

Jon E. Schoenfield, Jul 08 2017

Keywords

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

Examples

			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.)

Links

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).

Programs

  • PARI
    a(n) = #Set(vector(n^2, i, ((i%n)^6 + (i\n)^6) % n)); \\ Michel Marcus, Jul 10 2017
Showing 1-3 of 3 results.

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