A396689 Numbers k such that binomial(x + 4, 5) - binomial(x, 5) = k^2 for some integer x >= 5.

21, 119, 2059, 11661, 201761, 1142659, 19770519, 111968921, 1937309101, 10971811599, 189836521379, 1075125567781, 18602041786041, 105351333830939, 1822810258510639, 10323355589864241, 178616803292256581, 1011583496472864679, 17502623912382634299, 99124859298750874301

Offset

1,1

Comments

binomial(x + 4, 5) - binomial(x, 5) counts 5-multisets from an x-set containing at least one repeated element. The sequence lists the values k for which this count equals k^2.

Since binomial(x + 4, 5) - binomial(x, 5) = x^2*(x^2 + 5)/6, the condition binomial(x + 4, 5) - binomial(x, 5) = k^2 is equivalent to x^2 - 6*z^2 = -5 with k = x*z. Thus the terms are the products x*z taken over the positive integer solutions (x, z) of x^2 - 6*z^2 = -5.

The sequence is the increasing merge of two subsequences arising from the two branches of the Pell equation x^2 - 6*z^2 = -5. Each subsequence satisfies s(n) = 98*s(n - 1) - s(n - 2). Since the two subsequences interlace, a(n) = 98*a(n - 2) - a(n - 4) for n >= 5.

The restriction x >= 5 excludes the trivial cases binomial(x + 1, 2) - binomial(x, 2) = x and binomial(x + 2, 3) - binomial(x, 3) = x^2, which imply the ubiquitous representations k^2 = binomial(k^2 + 1, 2) - binomial(k^2, 2) = binomial(k + 2, 3) - binomial(k, 3).

In the more general equation binomial(x + y - 1, y) - binomial(x, y) = k^2, the additional solution 3333^2 = binomial(39, 7) - binomial(33, 7) was found for y >= 5. No other exceptions were found up to k <= 10^14.

Links

Felix Huber, Table of n, a(n) for n = 1..999

Index entries for linear recurrences with constant coefficients, signature (0,98,0,-1).

Formula

a(n) = 98*a(n - 2) - a(n - 4) for n >= 5, with a(1) = 21, a(2) = 119, a(3) = 2059, a(4) = 11661.

a(n) = A216073(n + 1)*A080806(n + 1).

G.f.: x*(21 + 119*x + x^2 - x^3)/(1 - 98*x^2 + x^4).

Example

a(1) = 21 because binomial(5 + 4, 5) - binomial(5, 5) = binomial(9, 5) - 1 = 126 - 1 = 125 = 21^2.

Maple

A396689 := proc(n) option remember;
    `if`(n = 1, 21, `if`(n = 2, 119, `if`(n = 3, 2059, `if`(n = 4, 11661, 98*A396689(n - 2) - A396689(n - 4)))));
end proc:
seq(A396689(n), n = 1 .. 20);

Mathematica

LinearRecurrence[{0, 98, 0, -1}, {21, 119, 2059, 11661}, 20] (* Amiram Eldar, Jun 23 2026 *)

Crossrefs

Cf. A080806, A216073.

Sequence in context: A079417 A267877 A183318 * A204214 A074088 A245031

Adjacent sequences: A396686 A396687 A396688 * A396691 A396692 A396694

Keyword

nonn,easy,changed

Author

Felix Huber, Jun 23 2026

Status

approved