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.

A365771 a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n) for n >= 0.

Original entry on oeis.org

1, 2, 20, 280, 4620, 84084, 1633632, 33256080, 701149020, 15191562100, 336424047960, 7584833081280, 173575987821600, 4022766574898400, 94247674040476800, 2228957491057276320, 53150802525726081660, 1276661433215969608500, 30863850087221160009000
Offset: 0

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Author

Paul D. Hanna, Oct 10 2023

Keywords

Comments

Equals the central terms of triangle A365770.
Conjectures: given A033042 is the sums of distinct powers of 5, then
(1) a(5*A033042(n)) == 4 (mod 5) for n > 0,
(2) a(5*A033042(n) + 1) == 2 (mod 5) for n > 0,
(3) a(n) == 0 (mod 5) for n > 0 except when n or n-1 equals 5*A033042(k) for some k >= 0.

Crossrefs

Programs

  • Mathematica
    A365771[n_] := Binomial[2*n + 1, n]/(2*n + 1)*Binomial[3*n - 1, n];
    Array[A365771, 20, 0] (* Paolo Xausa, Oct 12 2024 *)
  • PARI
    {a(n) = binomial(2*n+1, n)/(2*n+1) * binomial(3*n-1, n)}
    for(n=0,30,print1(a(n),", "))
    
  • Python
    from math import comb
    def A365771(n): return comb(m:=(n<<1)+1,n)*comb(m+n-2,n)//m if n else 1 # Chai Wah Wu, Oct 11 2023

Formula

a(n) = A365770(2*n,n) for n >= 0.
a(n) = A000108(n) * A165817(n) for n >= 0.
a(n) = 2*A319578(n) = (2/3) * A007004(n) for n >= 1. - Peter Bala, Aug 25 2025

A357714 a(n) is the number of equations in the set E_{n,b} := {x+2^b*y=n^b, 2^b*x+3^b*y=n^b, ..., k^b*x+(k+1)^b*y=n^b, ..., n^b*x+(n+1)^b*y=n^b} which admit at least one nonnegative integer solution when b is sufficiently large.

Original entry on oeis.org

1, 2, 3, 4, 3, 5, 4, 6, 5, 6, 4, 8, 5, 7, 7, 8, 5, 9, 5, 9, 8, 8, 6, 12, 7, 8, 8, 10, 6, 12, 7, 11, 9, 9, 9, 14, 7, 9, 9, 13, 7, 13, 8, 12, 12, 10, 8, 16, 9, 12, 10, 12, 8, 14, 10, 14, 11, 11, 9, 19, 9, 11, 13, 14, 11, 15, 9, 13, 11, 15, 9, 19, 10, 12, 14, 14, 12, 16, 10, 18, 13
Offset: 1

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Author

Luca Onnis, Oct 10 2022

Keywords

Comments

Defining a(n,b) as the number of equations of the set E_{n,b} which admit at least one nonnegative integer solution, it's possible to prove the existence of b_0 such that for all b > b_0, a(n,b) = a(n) whose value does not depend on b anymore.
a(n) is the number of positive integers k such that k(k+1) <= n or k divides n or k+1 divides n.

Examples

			a(11) = 4 since for all b >= 29 the number of equations of the set E_{11,b} which admit at least one nonnegative integer solution is exactly equal to 4.
a(4) = 4 since for all b >= 1 the number of equations of the set E_{11,b} which admit at least one nonnegative integer solution is exactly equal to 4.
		

Crossrefs

Programs

  • Mathematica
    Table[Ceiling[Sqrt[n] - 3/2] + Length[Divisors[n]], {n, 1, 100}]

Formula

a(n) = ceiling(sqrt(n) - 3/2) + A000005(n).
a(n) ~ A356770(n)/2 as n->infinity.
a(n) <= A356770(n) for all n >= 1.

A361197 a(n) is the number of equations in the set {x^2 + 2y^2 = n, 2x^2 + 3y^2 = n, ..., k*x^2 + (k+1)*y^2 = n, ..., n*x^2 + (n+1)*y^2 = n} which admit at least one nonnegative integer solution.

Original entry on oeis.org

1, 2, 3, 3, 3, 3, 3, 4, 4, 2, 5, 5, 3, 3, 3, 5, 4, 4, 5, 5, 5, 3, 3, 6, 4, 3, 6, 5, 5, 3, 5, 6, 4, 4, 4, 8, 3, 3, 5, 4, 6, 2, 5, 8, 6, 3, 3, 7, 6, 4, 6, 6, 4, 6, 3, 7, 4, 2, 7, 5, 6, 3, 6, 8, 3, 5, 5, 6, 7, 2, 5, 8, 4, 4, 6, 8, 4, 2, 6, 7, 8, 4, 5, 9, 3, 5, 4, 5, 6, 4, 6, 5, 4, 3, 4, 9
Offset: 1

Views

Author

Luca Onnis, Mar 04 2023

Keywords

Comments

Compared to the "linear" case given by A356770, the "quadratic" case given by this sequence has a more chaotic behavior.
a(n) >= 2 for all n > 1 since (n-1)*x^2 + n*y^2 = n and n*x^2 + (n+1)*y^2 = n always admit one integer solution (respectively (0,1) and (1,0)).
Conjecture: a(n) = 2 for infinitely many n.

Examples

			a(5) = 3. Consider the equations: x^2 + 2y^2 = 5, 2x^2 + 3y^2 = 5, 3x^2 + 4y^2 = 5, 4x^2 + 5y^2 = 5, 5x^2 + 6y^2 = 5. Only three of them admit at least one nonnegative integer solution, since 3x^2 + 4y^2 = 5 and x^2 + 2y^2 = 5 have no nonnegative integer solutions.
		

Crossrefs

Cf. A356770.

Programs

  • Mathematica
    b[m_] := m;
    f[n_] := Table[Dimensions[Solve[b[k]*x^2 + b[k + 1]*y^2 == n, {x, y}, NonNegativeIntegers]][[1]], {k, 1, n}];
Showing 1-3 of 3 results.