A382850 a(n) = least k such that binomial(n, k) > binomial(n - 1, h) for 0 <= h <= n - 1.

1, 1, 1, 2, 2, 2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 9, 9, 10, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31

Offset: 2

Author

Clark Kimberling, Apr 07 2025

Keywords

• nonn

Comments

All runs appear to have length 2 or 3, with no two consecutive runs of length 3, except for the first two (1, 1, 1; 2, 2, 2). The length of the k-th run of consecutive length-2 runs seems to be close to 0.722*k, but varies irregularly: 1, 2, 2, 4, 3, 5, 5, 7, 6, 8, 8, 9, 9, ... . - Pontus von Brömssen, Apr 15 2025

(Runlength sequence of the runlength sequence) = (2,1,1,2,1,2,1,4,1,3,1,5,...) appears to have 1 in positions given by A004280(n) for n >= 2. - Clark Kimberling, Apr 15 2025

Examples

			The least k such that binomial(5,k) > binomial(4,h) for 0<=h<=4 is 3, since 10 > max{1,4,6}.
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A001405, A007318, A382851.

Programs

Mathematica

z = 40; c[n_, k_] := Binomial[n, k];
t[n_] := Table[c[n, k], {k, 0, n}];
a[n_] := Select[Range[z], c[n, #] > c[n - 1, Floor[(n - 1)/2]] &, 1];
Flatten[Table[a[n], {n, 1, 3 z}]]  (* A382850 *)
Flatten[Table[c[n, a[n]], {n, 1, z}]]  (* A382851 *)

PARI

row(n) = vector(n+1, k, binomial(n,k-1));
a(n) = my(val = vecmax(row(n-1)), w = row(n)); for (i=1, #w, if (w[i] > val, return(i-1));); \\ Michel Marcus, Apr 14 2025

Python

from itertools import count
def A382850_generator():
    k = C0 = C = 1
    for n in count(2):
        C0 = 2*C0 if n%2 == 1 else (n-1)*C0//(n//2) # C(n-1,floor((n-1)/2))
        C = C*n//(n-k) # C(n,k)
        if C <= C0:
            C = C*(n-k)//(k+1)
            k += 1
        yield k # Pontus von Brömssen, Apr 15 2025

Formula

a(n) - a(n-1) is either 0 or 1. - Pontus von Brömssen, Apr 15 2025

Apparently, the lower and upper limits of f(n) = a(n) - n/2 + sqrt(n*log(2)/2) are 0 and 1, respectively, with f(n) < 1 for all n, but with the minimum -0.0144167... of f attained at n = 184. - Pontus von Brömssen, Apr 16 2025