A263573 Intersection of A024365 and A129912.

6, 30, 60, 180, 210, 2310, 4620, 60060, 510510, 10810800, 116396280, 200560490130, 401120980260

Offset: 1

Author

Bill McEachen, Oct 21 2015

Keywords

• nonn
• more

Comments

The two sequences involve areas of primitive Pythagorean triples and primorial products. Intersections are only considered once (no repeats). Conjecture: the sequence is infinite.

Conjecture: The next two entries are a(12) = 200560490130, a(13) = 401120980260.

From G. C. Greubel, Dec 29 2015: (Start)

6|a(n) for n>=1,

30|a(n) for n>=2,

a(n)/6 = {1, 5, 10, 30, 35, 385, 770, 10010, ...} is a subset of values found in A008706.

(End)

a(12) and a(13) confirmed. a(14) > 2*10^31, if it exists. - Giovanni Resta, Mar 31 2017

Examples

			A024365 begins {6, 30, 60, 84, 180, 210, 210, 330, 504, 546, 630, 840, 924, 990, 1224, 1320, 1386, 1560, 1710, 1716, 2310, ...}.
A129912 begins {1, 2, 6, 12, 30, 60, 180, 210, 360, 420, 1260, 2310, 2520, ...}.
So, common entries encountered are {6, 30, 60, 180, 210, 2310, ...}.
Specifically, we see that A024365(1) = A129912(3), A024365(2) = A129912(5), A024365(3) = A129912(6), A024365(5) = A129912(7).
These are then the first four entries of the sequence (6, 30, 60, 180).
		

Crossrefs

Cf. A024365, A129912.

Programs

Mathematica

s = 6 Take[Sort[(Times @@ #)/12 & /@ ({Times @@ #, (Last[#]^2 - First[#]^2)/2} & /@ Select[Subsets[Range[1, 3600, 2], {2}], GCD @@ # == 1 &])], 1800]; f[m_] := f[m] = Union[Times @@@ Subsets[FoldList[Times, 1, Prime[Range[m]]]]][[1 ;; 100]]; f[10]; f[m = 11]; While[f[m] != f[m - 1], m++]; t = f[m]; Intersection[s, t] (* Michael De Vlieger, Oct 22 2015, after Harvey P. Dale at A020885 and Jean-François Alcover at A129912 *) (* or *)
ok[n_] := Block[{a, f = Power @@@ FactorInteger[2 n]}, SelectFirst[ Subsets[f, {1, Floor[ Length[f]/2]}], (a = Times @@ #; IntegerQ@ Sqrt[a^2 + (2 n/a)^2]) &, {}] != {}]; pr[n_] := Product[ Prime[n+1-i]^i, {i, n}];  upto[mx_] := Block[{ric, j=1}, ric[n_, ip_, ex_] := If[n < mx, Block[{p = Prime[ip + 1]}, If[ex == 1 && ok[n], Sow@ n]; ric[n p^ex, ip + 1, ex]; If[ex > 1, ric[n p^(ex - 1), ip+1, ex-1]]]]; Sort@ Reap[ While[pr[j] < mx, ric[2^j, 1, j]; j++]][[2, 1]]]; upto[10^12] (* much faster, Giovanni Resta, Mar 31 2017 *)

PARI

\\note: code does not generate the sequence, just checks for a matching PPT entry
genit(area)={myMax=floor(sqrt(2*area));i5=myMax;endless=0;soln=List();
while(i5>=2,dun=0;j=2.*myVal/i5; k=floor(j); if(j>k, dun=1 );if(dun<1,
c=sqrt(i5^2 + k^2);w=floor(c);if(c>w,dun=1); if(dun<1,if(gcd(k,i5)>1,dun=1 ));
if(dun<1,listput(soln,k); listput(soln,i5);listput(soln,w);listsort(soln);
print("soln a,b,c = ", soln[1],"  ",soln[2],"  ",soln[3] );dun=2;break ));
i5--;endless++);if(i5<=2&&dun<1,print("no solution ") );if(i5>2&&dun<2,
print("max iteration limit was hit ",endless) );print (endless);}
(C++)
#include 
#include 
using namespace std;
int main(){ifstream fin1,fin2;
int myValue,myValue2,ptr,fptr,i5,j5;
unsigned long list1[9999]={0};
unsigned long list2[999]={0};
unsigned long final[31]={0};
fin1.open("A024365.txt"); fin2.open("A129912.txt");
ptr=1;
while(ptr<9999)
{fin1>> myValue;fin1.get();list1[ptr]=myValue;
    if(ptr<999)
       {fin2>> myValue2;fin2.get();list2[ptr]=myValue2;}
    ptr++;}
fin1.close();fin2.close();fptr=1;
for(i5=1;i5<9990;i5++)
{for(j5=1;j5<999;j5++){
if(list1[i5]==list2[j5] )
{
    fptr++;
    if(fptr>30){break;}
    final[fptr]=list1[i5];
    cout << final[fptr] << ",";
    break;
}}if(fptr>30){break;}}}

Extensions

a(12)-a(13) from Giovanni Resta, Mar 31 2017