OFFSET
1,1
COMMENTS
It seems that the two sums are never both a square or a cube.
Conjecture [False!]: All squares belonging to a pair are associated with a unique cube. Conversely, all cubes are associated with a unique square.
The corresponding pairs (sum of even divisors, sum of odd divisors) are (2^3, 2^2), (4^2, 2^3), (8^3, 16^2), (36^2, 6^3), (36^2, 6^3), (32^2, 8^3), 11 times the pair (24^3, 48^2), 3 times the pair (108^2, 18^3), (30^3, 30^2), (32^3, 128^2), 16 times the pair (288^2, 24^3),...
We observe several classes of numbers that generate identical pairs, for example:
{636, 748} => pair (36^2, 6^3);
{4620, 5964, 6204, 6324,... } => pair (24^3, 48^2);
{9010, 9710, 11342} => pair (108^2, 18^3);
{29820, 31020, 31620, 32844, 35420,... } => pair (288^2, 24^3);
{69576, 72168, 87752, 98552,...} => pair (56^3, 112^2);
The conjecture above is false. Consider for example the triples of numbers {69576, 938184, 7505472} or {958528, 952520, 12382760}. For the first one the (even, odd) sum of divisors pairs are (56^3, 112^2), (1568^2, 56^3), and (4704^2, 56^3). - Giovanni Resta, May 28 2016
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
EXAMPLE
434 is in the sequence because the divisors are {1, 2, 7, 14, 31, 62, 217, 434} => sum of even divisors = 2+14+62+434 = 512 = 8^3 and sum of odd divisors = 1+7+31+217 = 256 = 16^2.
636 is in the sequence because the divisors are {1, 2, 3, 4, 6, 12, 53, 106, 159, 212, 318, 636} => sum of even divisors = 2+4+6+12+106+212+318+636 = 1296 = 36^2 and sum of odd divisors = 1+3+53+159 = 216 = 6^3.
MAPLE
with(numtheory):
for n from 2 by 2 to 500000 do:
y:=divisors(n):n1:=nops(y):s0:=0:s1:=0:
for k from 1 to n1 do:
if irem(y[k], 2)=0
then
s0:=s0+ y[k]:
else
s1:=s1+ y[k]:
fi:
od:
ii:=0:
for a from 1 to 1000 while(ii=0)do:
for i from 2 to 3 do:
if s0=a^i
then
for b from 1 to 1000 while(ii=0) do:
if s1=b^(5-i)
then
ii:=1:printf(`%d, `, n):
else
fi:
od:
fi:
od:
od:
od:
MATHEMATICA
es[n_] := 2 DivisorSigma[1, n/2]; os[n_] := DivisorSigma[1, n] - es[n]; powQ[n_] := Or @@ IntegerQ /@ (n^(1/{2, 3})); Select[2 Range[10^4], powQ@ es@ # && powQ@ os@ # &] (* Giovanni Resta, May 28 2016 *)
PROG
(PARI) isA002760(n)=issquare(n) || ispower(n, 3)
is(n)=n%2==0 && isA002760(2*sigma(n/2)) && isA002760(sigma(n>>valuation(n, 2))) \\ Charles R Greathouse IV, Jun 08 2016
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, May 28 2016
STATUS
approved