A365097 Smallest k > 1 such that the total number of digits "1" required to write the numbers 1..k in base n is equal to k.

2, 4, 25, 181, 421, 3930, 8177, 102772, 199981, 3179142, 5971945, 143610511, 210826981, 4754446846, 8589934561, 222195898593, 396718580701, 13494919482970, 20479999999961, 764527028941797, 1168636602822613, 41826814261329722, 73040694872113105, 2855533828630999398

Offset: 2

Author

Andrew Pope, Aug 21 2023

Keywords

• nonn
• base

Comments

a(10) = A014778(3), being the smallest term > 1 there.

An upper bound is a(n) <= A226238(n) = u, since the digits of u show there are u 1's in numbers 1..u (in base n). - Kevin Ryde, Sep 28 2023

Examples

			For n=2, the first k=2 positive integers are 1 = 1_2 and 2 = 10_2, which have a total of two 1's, so a(2) = 2.
For n=3, the first k=4 positive integers, which are 1_3, 2_3, 10_3, and 11_3, have a total of four 1's, which is equal to k, so a(3) = 4.
For n=4, a total of 25 1's occur in the first k=25 positive integers (they occur in 1_4, 10_4, 11_4, 12_4, 13_4, 21_4, 31_4, 100_4, 101_4, 102_4, 103_4, 110_4, 111_4, 112_4, 113_4, 120_4, and 121_4 = 25), so a(4) = 25.
		

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 2..200

Crossrefs

Cf. A014778, A094798, A226238.

Programs

Mathematica

a[n_] := Module[{k = 1, sum = 1}, While[sum == 1 || sum != k, k++; sum += Count[IntegerDigits[k, n], 1]]; k]; Array[a, 6, 2] (* Amiram Eldar, Aug 29 2023 *)

Python

from itertools import count
from sympy.ntheory.factor_ import digits
def A365097(n):
    c, a, q, m = 1, 1, 0, 1
    for k in count(2):
        m += 1
        if m == n:
            m = 0
            q += 1
            a = digits(q,n).count(1)
        elif m==1:
            a += 1
        elif m==2:
            a -= 1
        c += a
        if c == k:
            return k # Chai Wah Wu, Sep 28 2023

Formula

For even n > 2, a(n) = 2*n^(n/2) - 2*n + 1. - Jon E. Schoenfield, Sep 30 2023

Extensions

a(11)-a(15) from Amiram Eldar, Aug 29 2023

a(16)-a(19) from Chai Wah Wu, Sep 29 2023

a(20)-a(25) from Jon E. Schoenfield, Sep 30 2023