

A295265


Numbers m such that sum of its i first divisors equals the sum of its j first nondivisors for some i, j.


1



4, 8, 10, 13, 14, 16, 19, 20, 21, 22, 26, 28, 30, 32, 34, 38, 39, 40, 43, 44, 46, 50, 52, 53, 56, 58, 60, 62, 63, 64, 68, 70, 72, 74, 76, 80, 82, 86, 88, 89, 90, 92, 94, 98, 99, 100, 103, 104, 106, 110, 111, 112, 116, 117, 118, 122, 124, 128, 130, 132, 134, 135
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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 nondivisors 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



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



STATUS

approved



