A304978 - OEIS

%I #47 Sep 25 2023 14:32:05

%S 106,155,197,204,253,288,302,351,379,400,421,449,470,498,504,535,547,

%T 554,561,596,645,652,687,694,704,729,743,779,782,792,820,834,841,873,

%U 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

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

%C Is it possible to find a closed form formula for this sequence?

%C Numbers k such that 6*k+1 has at least 5 divisors == 1 (mod 6). - _Robert Israel_, Jan 20 2019

%H Robert Israel, <a href="/A304978/b304978.txt">Table of n, a(n) for n = 1..10000</a>

%e 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.

%p filter:= proc(n) nops(select(t -> t mod 6 =1, numtheory:-divisors(6*n+1)))>= 5 end proc:

%p select(filter, [$1..2000]); # _Robert Israel_, Jan 20 2019

%t Select[Range[1329], 2 == Length@ FindInstance[ 6*x*y+x+y == # && x >= y > 0, {x, y}, Integers, 2] &] (* _Giovanni Resta_, May 29 2018 *)

%o (Python)

%o # Finding duplicates in matrix A where {a_ij}=6*i*j+i+j

%o def cuenta(a):

%o     repa=[]

%o     dupo=[]

%o     k=0

%o     sumo=0

%o     while k<len(a)-1:

%o         j=k+1

%o         dupa = 1

%o         while j<len(a) and a[j]==a[k]:

%o             dupa +=1

%o             j+=1

%o         if dupa>1:

%o             repa.append(a[k])

%o             dupo.append(dupa)

%o             sumo +=(dupa-1)

%o         k +=dupa

%o     return repa, dupo,sumo

%o kuno = []

%o for k in range(1, a + 1):

%o     for j in range(1, k + 1):

%o         m=6 * k * j + k + j

%o         if m <= NN // 6: kuno.append(m)

%o kuno=sorted(kuno)

%o repkuno, dupkuno, suno= cuenta(kuno)

%o print(repkuno)

%o #

%o #END

%o (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

%Y Subsequence of A067611.

%K nonn

%O 1,1

%A _Pedro Caceres_, May 22 2018