OFFSET
1,1
COMMENTS
Is it possible to find a closed form formula for this sequence?
Numbers k such that 6*k+1 has at least 5 divisors == 1 (mod 6). - Robert Israel, Jan 20 2019
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
EXAMPLE
106 is in this sequence because 106 can be expressed in two different ways as 6xy + x + y: 6*8*2 + 8 + 2 and 6*15*1 + 15 + 1.
MAPLE
filter:= proc(n) nops(select(t -> t mod 6 =1, numtheory:-divisors(6*n+1)))>= 5 end proc:
select(filter, [$1..2000]); # Robert Israel, Jan 20 2019
MATHEMATICA
Select[Range[1329], 2 == Length@ FindInstance[ 6*x*y+x+y == # && x >= y > 0, {x, y}, Integers, 2] &] (* Giovanni Resta, May 29 2018 *)
PROG
(Python)
from sympy import divisors
def ok(n): return sum(d%6 == 1 for d in divisors(6*n+1)) > 4
print([n for n in range(1330) if ok(n)]) # David Radcliffe, Jun 19 2025
(PARI) is(n) = my(i=0); for(x=1, n, for(y=1, x, if(n==6*x*y+x+y, i++; if(i==2, return(1))))); 0 \\ Felix Fröhlich, May 29 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Pedro Caceres, May 22 2018
STATUS
approved
