A378838 a(n) is the number of n-digit nonnegative integers with the mean of the digits equal to the digital root.

10, 1, 1, 748, 1, 1, 373327, 1, 1, 900000000, 1, 1, 118641180477, 1, 1, 70265700376176, 1, 1, 473609016175792282, 1, 1, 25843609164809475416, 1, 1, 15917111940073972644247, 1, 1, 319908753084273214311674685, 1, 1, 6159986083122001233681300544, 1, 1, 3860982462614939076156553701616, 1, 1

Offset: 1

Author

Stefano Spezia, Dec 09 2024

Keywords

• nonn
• base

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..81

Crossrefs

Cf. A007494, A010888, A378837.

Cf. A378560, A378564, A378835, A378836.

Programs

Mathematica

A010888[n_]:=If[n==0,0,n - 9*Floor[(n-1)/9]]; a[n_]:=If[n==1,10,Module[{c=0}, For[k=10^(n-1), k<=10^n-1, k++, If[Mean[IntegerDigits[k]]==A010888[k], c++]]; c]]; Array[a, 6]

Python

from math import prod, factorial
from sympy.utilities.iterables import partitions
def A378838(n):
    if n==1: return 10
    if n%3!=1: return 1
    k, c, f = n%9, 0, factorial(n-1)
    a = 3*n if k==4 or k==7 else n
    for i in range(a,9*n+1,a):
        for s,p in partitions(i,m=n,k=9,size=True):
            v = list(p.values())
            p = prod((factorial(i) for i in v))*factorial(n-s)
            c += sum(f*i//p for i in v)
    return c # Chai Wah Wu, Dec 12 2024

Formula

Conjecture: a(A007494(n)) = 1.

From Chai Wah Wu, Dec 12 2024: (Start)

The above conjecture is true, i.e. if n == 0 or 2 mod 3, then a(n) = 1.

Proof: if m is a n-digit integer with mean of digits equal to its digital root k, then sum of digits of m is k*n.

Then m mod 9 = k*n mod 9. Since the digital root is k = 1 + (m-1) mod 9, this means that k = 1 + (k*n-1) mod 9. As 1<=k<=9, for n == 2, 3, 5, 6, 8, 9 mod 9 the only k that satisfies this equation is k=9. Then the only corresponding m whose digit sum is 9*n is 10^n-1. Thus a(n) = 1.

Other results:

Theorem 1: If n == 4 or 7 mod 9, then a(n) is the number of n-digit nonnegative integers with digit sum a multiple of 3*n.

Proof: Follows from the fact that the only k that satisfies k = 1 + (k*n-1) mod 9 is k = 3, 6, or 9.

Theorem 2: If n>1 and n == 1 mod 9, then a(n) is the number of n-digit nonnegative integers with digit sum a multiple of n.

Proof: Since n>1, the digital root of an n-digit integer is > 0. The result then follows from the fact that k = 1 + (k*n-1) mod 9 is satisfied for all 1<=k<=9.

(End)

Extensions

a(11)-a(36) from Chai Wah Wu, Dec 12 2024