OFFSET
1,1
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)
LINKS
David A. Corneth, Table of n, a(n) for n = 1..10000
EXAMPLE
14 is in the sequence since 14 = 2+2+2+2+1+1+1+1+1+1, where each summand divides 14.
PROG
(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
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Wesley Ivan Hurt, Aug 20 2022
STATUS
approved