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%I #37 Mar 29 2018 12:02:57
%S 2,6,30,210,510510
%N Primorial numbers of the form n^2 + n for some integer n.
%C Primorial numbers m such that 4m+1 is a square.
%C Intersection of the sequences A002110 and A002378.
%C If it exists, a(6) > A034386(10^11). - _Max Alekseyev_, Oct 23 2011
%C The form is n^2 + n = n(n + 1), and the values n + 1 constitute A215659. - _Jeppe Stig Nielsen_, Mar 27 2018
%H C. Nelson, D. E. Penney, and C. Pomerance, <a href="https://math.dartmouth.edu/~carlp/714and715.pdf">714 and 715</a>, J. Recreational Mathematics (1974) 7(2), 87-89. [Warning: As of March 2018 this site appears to have been hacked. Proceed with great caution. The original content should be retrieved from the Wayback machine and added here. - _N. J. A. Sloane_, Mar 29 2018]
%F a(n) = A034386(A215658(n)). - _Jeppe Stig Nielsen_, Mar 27 2018
%e 2 = 1*2 = 2
%e 2*3 = 2*3 = 6
%e 2*3*5 = 5*6 = 30
%e 2*3*5*7 = 14*15 = 210
%e 2*3*5*7*11*13*17 = 714*715 = 510510
%t p=1; Do[p=p*Prime[c]; f=Floor[Sqrt[p]]; If[p==f*(f+1), Print[p]],{c,1000}]
%o (PARI) N=10^8;si=30;q=vector(si,i,nextprime(i*N));a=vector(si,i,1);forprime(p=2,N,for(i=1,si,a[i]=(a[i]*p)%q[i]);v=1;for(i=1,si,if(kronecker(4*a[i]+1,q[i])==-1,v=0;break));if(v,T=1;forprime(r=2,p,T*=r);print1(T",")))
%o (PARI) pr=1;forprime(p=2,,pr*=p;s=sqrtint(pr);s*(s+1)==pr&&print1(pr,", ")) \\ _Jeppe Stig Nielsen_, Mar 27 2018
%Y Cf. A002110, A002378, A215658, A215659.
%K nonn,hard,more
%O 1,1
%A _Daniel Tisdale_, Jun 14 2009
%E Edited by _Hans Havermann_, Dec 02 2010
%E Edited by _Max Alekseyev_, Dec 03 2010
%E Edited by _Robert Gerbicz_, Dec 04 2010
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