A365452 Largest number whose digits, in some base, sum to n and include no zeros.

1, 3, 7, 22, 67, 213, 853, 3413, 13653, 66406, 332031, 1660156, 8734003, 52404019, 314424115, 1886544691, 12193514915, 85354604406, 597482230843, 4182375615902, 31414617936457, 251316943491657, 2010535547933257, 16084284383466057, 135502101309790873, 1219518911788117858

Offset: 1

Author

Elliott Line, Sep 04 2023

Keywords

• nonn
• base

Comments

There is a proviso with this: the first digit must be one less than the base; otherwise we could claim that a number was in an arbitrarily large base.

The maximum value will always be found by following the initial digit with a string of 1s, which is n candidate values for a(n).

Examples

			For n=5, the candidate digits are 11111_2, 2111_3, 311_4, 41_5 and 5_6. These have decimal values 31, 67, 53, 21 and 5, respectively and the largest of is 67, so a(5)=67.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..597

Crossrefs

Cf. A000225.

Programs

Maple

a:= n-> max((i+1)^(n-i)*(1/i+i)-1/i$i=1..n):
seq(a(n), n=1..25);  # Alois P. Heinz, Sep 06 2023

PARI

a(n) = {my(res=2^n-1, v = vector(n, i, 1)); for(i = 2, n, v[i] += v[i-1]; v[i-1] = 0; res = max(res, fromdigits(v, i+1))); res} \\ David A. Corneth, Sep 04 2023

Python

def A365452(n): return max(((i + 1)**(n-i)*(i**2 + 1) - 1)//i for i in range(1,n+1)) # Chai Wah Wu, Oct 01 2023

Formula

a(n) >= A000225(n). - David A. Corneth, Sep 04 2023

a(n) = max_{i=1..n} ((i+1)^(n-i)*(1/i+i)-1/i). - Alois P. Heinz, Sep 09 2023

Extensions

More terms from David A. Corneth, Sep 04 2023