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

1,1

Links

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

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)

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]

Prog

(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

Crossrefs

Cf. A007494, A010888, A378837.

Cf. A378560, A378564, A378835, A378836.

Sequence in context: A156767 A365025 A174921 * A010180 A109013 A343102

Adjacent sequences: A378835 A378836 A378837 * A378839 A378840 A378841

Keyword

nonn,base

Author

Stefano Spezia, Dec 09 2024

Extensions

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

Status

approved