A383922 a(n) = A002104(n) + A002104(n+1) - 1.

0, 3, 10, 31, 112, 503, 2786, 18443, 141744, 1237755, 12088266, 130457479, 1541023936, 19769882767, 273671845058, 4065274481939, 64493941507232, 1088226653465139, 19458541429154250, 367527663494842671, 7311506648705326672, 152804399672163086695, 3347034732868985727202, 76675452816691696778843

Offset: 0

Author

Jianing Song, May 15 2025

Keywords

• nonn

Comments

Let m_0(x) = -x for x < 0 and (1/2)*m_0(x - m_0(x-1)) for x >= 0, then:

m_0(1 - 2^(-n)) = 2^(-(n+1)) for n >= -1;

m_0(2 - 2^(-n)) = 2^(-(2*n+3)) for n >= -2;

m_0(3 - 2^(-n)) = 2^(-a(n+2)) for n >= -2 (see my link for a proof).

Let F_0 = {x + m_0(x) : x in R}, then the intersection of F_0 and (-oo,n) is well-ordered with order type omega^^n. As a result, F_0 is well-ordered with order type epsilon_0 = omega^^omega. F_0 is a proper subset of F, the set of fusible numbers.

It had been believed that x + m_0(x) was the least fusible greater than x. Junyan Xu points out that this is false. Indeed, let x_n = 17/8 - 1/2^(n+1), c_n = 5/4 - 1/2^(n+1), and d_n = 2 - 1/2^(n-2) + 1/2^(2*n-1) for n >= 0, then

c_n = (1/2 + (1-1/2^n) + 1)/2,

d_n = ((1-1/2^(2*n-2)) + d_{n-1} + 1)/2, n >= 1

are both fusible numbers, hence so is (c_n + d_{n+3} + 1)/2 = 17/8 - 1/2^(n+1) + 1/2^(2*n+6); in other words, the least fusible number greater than x_n is at most x_n + 1/2^(2*n+6). But we have m_0(x_n) = 1/2^(n+8) > 1/2^(2*n+6) for n >= 3.

Junyan Xu gives a conjecture on the recursive formula of m(x), where x + m(x) is the least fusible greater than x. (A188545(n) is -log_2 m(n)). We have m(x) = m_0(x) for x < 33/16 (i.e., F_0 and F coincide on the interval (-oo,33/16]), but they differ for x >= 33/16, even if the intersection of F and (-oo,n) still has order type omega^^n if the conjecture is true. In fact, we have -log_2 m_0(3) = 1541023937, while -log_2 m(3) > 2^^^^^^^^^16 in Knuth's up-arrow notation.

Links

• Jianing Song, Notes on A383922
• Junyan Xu, Survey on Fusible Numbers, arXiv:1202.5614 [math.CO], 2012.

Crossrefs

Cf. A002104, A188545.

Programs

PARI

a(n) = A002104(n) + A002104(n+1) - 1 \\ See A002104

Formula

E.g.f.: exp(x) * (-2*log(1-x) + x/(1-x)).

E.g.f. satisfies (1-x)^2 * (A'(x) - A(x)) = (-2*x+3)*exp(x).

Recurrence: n*a(n) = (n^2+n-1)*a(n-1) - (n^2-1)*a(n-2) + 2*n + 1, a(0) = 0, a(1) = 3.

Recurrence: a(n) = n*a(n-1) - a(n-2) - (n-2)*a(n-3) + 4, a(0) = 0, a(1) = 3, a(2) = 10.

a(n) ~ exp(1)*n!.