A304978 Numbers that can be expressed in more than one way as 6xy + x + y with x >= y > 0.

106, 155, 197, 204, 253, 288, 302, 351, 379, 400, 421, 449, 470, 498, 504, 535, 547, 554, 561, 596, 645, 652, 687, 694, 704, 729, 743, 779, 782, 792, 820, 834, 841, 873, 890, 904, 925, 939, 953, 988, 1016, 1029, 1037, 1042, 1054, 1079, 1086, 1107, 1121, 1135, 1184, 1198, 1204, 1211, 1219, 1233, 1254, 1276, 1282, 1289, 1329

Offset: 1

Author

Pedro Caceres, May 22 2018

Keywords

• nonn

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

Examples

			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.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Subsequence of A067611. A279060.

Programs

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 *)

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

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