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%I #43 Apr 11 2017 04:36:55
%S 7,3,18,8,117,43,239,5,378,132,843,377,2207,987,2943,73,4443,53,4662,
%T 1568,6072,5118,8307,743,8708,2112,9872,2738,31561,4929,103682,46368,
%U 271443,121393,853932,76378,1021693,91383,3539232,41218,3699356,473654
%N Pairs of integers (x, y), such that x^2 + 1 and y^2 + 1, 1 < y < x, have the same distinct prime factors.
%C The sequence appears to thin out quite abruptly; however, by solving the Diophantine equation x^2 + 1 = p (y^2 + 1) for a suitable prime p and selecting the solutions (x, y) for which p divides y^2 + 1, it is easy to generate larger pairs, such as (423222288438379883442890018716361, 66096216900526495715353522199871). - _Giovanni Resta_, Mar 27 2017
%C A very interesting property: the sequence contains a subsequence of pairs (Lucas numbers L(i), Fibonacci numbers F(i)) for i = 4, 6, 14, 16, 24, 36, ... These pairs are (L(4), F(4)), (L(6), F(6)), (L(14), F(14)), (L(16), F(16)), (L(24), F(24)), (L(26), F(26)), (L(34), F(34)), ... = (7, 3), (18, 8), (843, 377), (2207, 987), (103682, 46368), (271443, 121393), (12752043, 5702887), ... It seems that {i} = A090773(n) (numbers that are congruent to {4, 6} mod 10. - _Michel Lagneau_, Mar 28 2017
%C This is because L(i)^2+1 = 5*(F(i)^2+1) for even i, and 5 | F(i)^2 + 1 for i== 3,4,6,7 (mod 10). In fact (L(i), F(i)) for i in A090773 are the solutions of the generalized Pell equation x^2 + 1 = 5 (y^2 + 1) for which 5 | y^2 + 1. - _Robert Israel_, Apr 10 2017
%H Giovanni Resta, <a href="/A284477/b284477.txt">Table of n, a(n) for n = 1..68</a> (terms with x < 1.5*10^8)
%e The pair (843, 377) is in the sequence because the prime factors of 843^2 + 1 and 377^2 + 1 are 2, 5, 61 and 233.
%p A:= NULL:
%p for x from 2 to 10^5 do
%p P:= numtheory:-factorset(x^2+1);
%p if not assigned(R[P]) then R[P]:= x
%p else A:= A, op(map(t -> (x,t), [R[P]]));
%p R[P]:= R[P],x
%p fi
%p od:
%p A; # _Robert Israel_, Apr 10 2017
%t d[n_] := First /@ FactorInteger[n]; Flatten@ Reap[ Do[ dx = d[x^2+1]; Do[ If[ dx == d[y^2+1], Sow[{x, y}]], {y, x-1}], {x, 1, 10^4}]][[2, 1]]
%o (PARI) upto(n) = {my(l = List(), res=List()); for(i=1, n, f = factor(i^2+1)[, 1]; listput(l, [f, i])); listsort(l); for(i=1, n-1, if(l[i][1]==l[i+1][1], listput(res, [l[i+1][2], l[i][2]]))); listsort(res); res} \\ _David A. Corneth_, Mar 28 2017
%Y Cf. A002496, A089122, A128428.
%K nonn
%O 1,1
%A _Michel Lagneau_, Mar 27 2017
%E a(29)-a(34) from _Giovanni Resta_, Mar 27 2017
%E a(35)-a(42) from _David A. Corneth_, Mar 28 2017
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