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Ordered by increasing m with k < m, a(n) is the n-th record value of gcd(k!+1, m!+1).
1

%I #31 Jan 07 2022 19:32:38

%S 2,7,71,661,733,2371,3529,13499,46549,98101,163517,197933,1924217,

%T 3322441,5370731

%N Ordered by increasing m with k < m, a(n) is the n-th record value of gcd(k!+1, m!+1).

%C The pairs (m,k) generating records are (1,0), (6,3), (9,7), (17,8), (89,51), (174,144), (349,228), (422,81), (650,406), (1415,1718), (1697,161), (1622,773), (1884,1219), (7003,2031) and (17057,660).

%C Heuristics in concert with a database of 'small' (less than, say, 10^12) prime factors of numbers of this kind would generate faster accurate results with near certainty, while any truly proving program is doomed to be relatively slow by comparison (and see following on a(15)).

%C Note: An auxiliary program employed a limit of 10^8--in lieu of a database--to generate the likely-but-not-certain value of a(15) shown last.

%e a(1)=2, corresponding to m=1 and k=0. 7 is the first value other than 1 to be the greatest common divisor of two different numbers k!+1 and m!+1, where m is increasing and k is allowed to increase to m-1 for a given m (for m=6 and k=3, m!+1=7*103 and k!+1=7); so that a(2)=7.

%o (PARI)

%o {

%o \\ The constant L here is arbitrary.\\

%o \\ This does not generate a(1).\\

%o rec=2;L=10000;F=vector(L);n=2;

%o for(k=1,L,n--;n*=k;n++;F[k]=n);

%o for(m=2,L,

%o for(k=1,m-1,

%o a=gcd(F[m],F[k]);if(a>rec,

%o rec=a;print1(a": "m","k"\n"))))

%o }

%Y Cf. A002583, A002981, A038507, A051301.

%K nonn,more

%O 1,1

%A _James G. Merickel_, Oct 19 2013