OFFSET
1,1
COMMENTS
Or numbers m such that Sum_{k=1..i} d(k) = Sum_{k=1..j} nd(k) for some i, j where d(k) are the i first divisors and nd(k) the j non-divisors of m.
The corresponding sums are 3, 3, 3, 14, 3, 3, 20, 3, 11, 3, 3, (3 or 14), 11, 3, 3, 3, 17, 3, 44, 3, 3, 3, 3, 54, 3, 3, 15, 3, 11, 3, 3, 3, 33, 3, 3, 3, ... containing the set of primes {3, 11, 17, 23, 29, 37, 41, 43, 53, 59, 61, 71, 79, ...}.
The equality Sum_{k=1..i} d(k) = Sum_{k=1..j} nd(k) is not always unique, for instance for a(12) = 28, we find 1 + 2 = 3 and 1 + 2 + 4 + 7 = 3 + 5 + 6 = 14.
The primes of the sequence are 13, 19, 43, 53, 89, 103, 151, 229, 251, 349, 433, ... (primes of the form k(k+1)/2 - 2; see A124199).
+-----+-----+-----+------+-----------------------------------------+
| n | i | j | a(n) | Sum_{k=1..i} d(k) = Sum_{k=1..j} nd(k) |
+-----+-----+-----+------+-----------------------------------------+
| 1 | 2 | 1 | 4 | 1 + 2 = 3 |
| 2 | 2 | 1 | 8 | 1 + 2 = 3 |
| 3 | 2 | 1 | 10 | 1 + 2 = 3 |
| 4 | 2 | 4 | 13 | 1 + 13 = 2 + 3 + 4 + 5 = 14 |
| 5 | 2 | 1 | 14 | 1 + 2 = 3 |
| 6 | 2 | 1 | 16 | 1 + 2 = 3 |
| 7 | 2 | 5 | 19 | 1 + 19 = 2 + 3 + 4 + 5 + 6 = 20 |
| 8 | 2 | 1 | 20 | 1 + 2 = 3 |
| 9 | 3 | 3 | 21 | 1 + 3 + 7 = 2 + 4 + 5 = 11 |
| 10 | 2 | 1 | 22 | 1 + 2 = 3 |
| 11 | 2 | 1 | 26 | 1 + 2 = 3 |
| 12 | 2 | 1 | 28 | 1 + 2 = 3 |
| | 4 | 3 | 28 | 1 + 2 + 4 + 7 = 3 + 5 + 6 = 14 |
| 13 | 4 | 2 | 30 | 1 + 2 + 3 + 5 = 4 + 7 = 11 |
| 14 | 2 | 1 | 32 | 1 + 2 = 3 | (End)
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..10000
EXAMPLE
30 is in the sequence because d(1) + d(2) + d(3) + d(4) = 1 + 2 + 3 + 5 = 11 and nd(1) + nd(2) = 4 + 7 = 11.
MAPLE
with(numtheory):nn:=300:
for n from 1 to nn do:
d:=divisors(n):n0:=nops(d):lst:={}:ii:=0:
for i from 1 to n do:
lst:=lst union {i}:
od:
lst:=lst minus d:n1:=nops(lst):
for m from 1 to n0 while(ii=0) do:
s1:=sum(‘d[i]’, ‘i’=1..m):
for j from 1 to n1 while(ii=0) do:
s2:=sum(‘lst[i]’, ‘i’=1..j):
if s1=s2
then
ii:=1:printf(`%d, `, n):
else
fi:
od:
od:
od:
MATHEMATICA
fQ[n_] := Block[{d = Divisors@ n}, nd = nd = Complement[Range@ n, d]; Intersection[Accumulate@ d, Accumulate@ nd] != {}]; Select[ Range@135, fQ] (* Robert G. Wilson v, Mar 06 2018 *)
PROG
(PARI) isok(n) = {d = divisors(n); psd = vector(#d, k, sum(j=1, k, d[j])); nd = setminus([1..n], d); psnd = vector(#nd, k, sum(j=1, k, nd[j])); #setintersect(psd, psnd) != 0; } \\ Michel Marcus, May 05 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Feb 22 2018
STATUS
approved