A282092 Numbers m such that there exists at least one integer k < m such that m^2+1 and k^2+1 have the same prime factors.
7, 18, 117, 239, 378, 843, 2207, 2943, 4443, 4662, 6072, 8307, 8708, 9872, 31561, 103682, 271443, 853932, 1021693, 3539232, 3699356, 6349657, 6907607, 7042807, 7249325, 9335094, 12623932, 12752043, 12813848, 22211431, 33385282, 42483057, 52374157, 105026693
Offset: 1
Keywords
Examples
7 is in the sequence because of the pair (m, k) = (7, 3), 7^2+1 = 2*5^2 and 3^2+1 = 2*5 with the same prime factors 2 and 5.
Crossrefs
Programs
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Mathematica
Select[Range@ 5000, Function[m, Total@ Boole@ Table[Function[w, And[SameQ[First@ w, #], SameQ[Last@ w, #]] &@ Union@ Flatten@ w]@ Map[FactorInteger[#][[All, 1]] &, {m^2 + 1, k^2 + 1}], {k, m - 1}] > 0]] (* Michael De Vlieger, Feb 07 2017 *)
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PARI
isok(n)=ok = 0; vn = factor(n^2+1)[,1]; for (k=1, n-1, if (factor(k^2+1)[,1] == vn, ok = 1; break);); ok; \\ Michel Marcus, Feb 09 2017
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PARI
squeeze(f)=factorback(f)\2 list(lim)=my(v=List(),m=Map(),t); for(n=1,lim, t=squeeze(factor(n^2+1)[,1]); if(mapisdefined(m,t), listput(v,n), mapput(m,t,0))); Vec(v) \\ Charles R Greathouse IV, Feb 12 2017
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Perl
use ntheory qw(:all); for (my ($m, %t) = 1 ; ; ++$m) { my $k = vecprod(map{$_->[0]}factor_exp($m**2+1)); push @{$t{$k}}, $m; if (@{$t{$k}} >= 2) { print'('.join(', ',reverse(@{$t{$k}})).")\n"; } } # Daniel Suteu, Feb 08 2017
Extensions
a(15)-a(29) from Daniel Suteu, Feb 08 2017
a(30) from Daniel Suteu, Feb 10 2017
a(31)-a(34) from Joerg Arndt, Feb 11 2017
Comments