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%I #23 Sep 13 2017 02:15:18
%S 14,28,34,39,46,55,56,62,68,82,92,94,95,98,111,112,117,124,136,142,
%T 145,146,155,158,164,178,183,184,188,194,196,203,205,206,219,221,224,
%U 226,248,254,259,272,274,275,284,291,292,295,299,302,305,316,323,327,328
%N Composite numbers m for which L(p(i)/p(j)) = 1 for all i, j, where p(k) are the prime factors of m and L(x/y) is the Legendre symbol of x and y, defined to be 1 if x is a quadratic residue (mod y) and -1 if x is a quadratic non-residue (mod y).
%C L(x/y) = L(y/x) for primes x, y, and either x = 4r+1 for some r, or y = 4r+1 for some r.
%C We observe pairs of the form (a(n), a(n)+1) = (55, 56), (94, 95), (111, 112), (145, 146), (183, 184), (205, 206), (274, 275), (291, 292), (327, 328), ..., .
%C From _Robert G. Wilson v_, Nov 16 2016: (Start)
%C We observe triples of the form (a(n), a(n)+1, a(n)+2) with a(n): 542, 543, 655, 1262, 1411, 1646, 1982, 2305, 2306, 2942, 3025, ..., .
%C We observe quadruples of the form (a(n), a(n)+1, a(n)+2, a(n)+3) with a(n): 542, 2305, 7022, 10081, 19981, 35821, 41372, 50011, 50941, ..., .
%C We observe quintuples of the form (a(n), a(n)+1, a(n)+2, a(n)+3, a(n)+4) with a(n): 85631, ..., .
%C Number of terms less than 10^k: 0, 14, 168, 1591, 14175, 127791, ...
%C Number of twin terms less than 10^k: 0, 2, 31, 250, 1737, 13604, ...
%C (End)
%H Robert G. Wilson v, <a href="/A276525/b276525.txt">Table of n, a(n) for n = 1..10000</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LegendreSymbol.html">Legendre Symbol</a>
%e 14 is in the sequence because the prime factors are 2 and 7 => L(2,7)= L(7,2)= 1.
%e 137678 is in the sequence because the prime factors are 2, 23, 41 and 73 => L(2,23) = L(2,41) = L(2,73) = L(23,41) = L(23,73) = L(41,73) = L(73,41) = L(73,23) = L(73,2) = L(41,23) = L(41,2) = L(23,2) = 1.
%p with(numtheory): nn:=400:
%p for n from 1 to nn do:
%p x:=factorset(n):n0:=nops(x):
%p if n0>1
%p then
%p ii:=0:
%p for i from 1 to n0-1 while(ii=0) do:
%p for j from 2 to n0 while(ii=0) do:
%p p:=legendre(x[i],x[j]):q:=legendre(x[j],x[i]):
%p if p=-1 or q=-1 then ii:=1:
%p else fi:
%p od:od:
%p if ii=0 then printf(`%d, `,n):
%p else fi:fi:
%p od:
%t fQ[n_] := If[ CompositeQ@ n, Block[{pf = Transpose[ FactorInteger[n]][[1]]}, lng = Length@ pf; Union[ Flatten[ Table[ JacobiSymbol[pf[[i]], pf[[j]]], {i, lng}, {j, lng}]]] == {0, 1}], False]; Select[ Range@ 330, fQ] (* _Robert G. Wilson v_, Nov 16 2016 *)
%Y Cf. A002144, A002145.
%K nonn
%O 1,1
%A _Michel Lagneau_, Nov 16 2016
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