A356660 Numbers k that can be written as the sum of 10 divisors of k (not necessarily distinct).

10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 40, 42, 44, 48, 50, 52, 54, 56, 60, 64, 66, 68, 70, 72, 76, 78, 80, 84, 88, 90, 92, 96, 98, 100, 102, 104, 108, 110, 112, 114, 116, 120, 126, 128, 130, 132, 136, 138, 140, 144, 150, 152, 154, 156, 160, 162

Offset: 1

Author

Wesley Ivan Hurt, Aug 20 2022

Keywords

• nonn

Comments

From David A. Corneth, Oct 08 2022: (Start)

All terms are even. Proof: suppose a term is odd. Then all divisors are odd. Adding 10 odd numbers gives an even number. A contradiction.

If k is a term then so is k*m for m >= 1. Proof: Multiply each divisor in this sum of 10 divisors that give k with m. Then each term is a divisor of k*m and their sum is k*m. (End)

Examples

			14 is in the sequence since 14 = 2+2+2+2+1+1+1+1+1+1, where each summand divides 14.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10000

Crossrefs

Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), A355200 (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), this sequence (j=10).

Programs

PARI

upto(n) = { my(v = vector(n,i,-1), t = 0); forstep(i = 2, n, 2, if(v[i] == -1, v[i] = is(i); if(v[i] == 1, for(j = 2, n \ i, v[i*j] = 1; ) ) ); ); select(x->x >= 1, v, 1); }
is(n, {qd = 10}) = { my(d = divisors(n), res = 0); d = d[^#d]; forvec(x = vector(qd-1, i, [1, #d]), s = sum(i = 1, qd-1, d[x[i]]); if(n - s >= d[x[qd - 1]], if(n % (n - s) == 0,  return(1); ) ) , 1 ); 0 } \\ David A. Corneth, Oct 08 2022

Python

from sympy import divisors
def t_sum_of_n_div(n, target):
    out, p = [], divisors(n)[::-1][1:]
    def dfs(t, divs,  index_s, kk):
        if len(out)!=0 or kk>target:return
        if kk == target and t == 0:
            out.append(divs)
            return
        for i in range(index_s, len(p)):
            if t >= p[i]:
                temp_divs = divs.copy()
                temp_divs.append(p[i])
                dfs(t-p[i], temp_divs, i, kk+1)
    dfs(n, [], 0, 0)
    return out
terms = [i for i in range(2, 200) if len(t_sum_of_n_div(i,10))!=0]
print(terms) # Gleb Ivanov, Sep 02 2022