|
%I #11 Dec 16 2019 10:18:10
%S 4,527,779,869,899,1079,1157,1271,1679,4187,6497,6887,24287,24881,
%T 25019,29591,35237,37127,37769,38807,39269,39911,41309,43361,44831,
%U 45347,46001,46127,47261,48509,48929,51809,52907,54389,55481,55751,55961
%N Numbers m such that m and m+2 are both brilliant numbers, where brilliant numbers are semiprimes whose prime factors have an equal number of decimal digits, or whose prime factors are equal.
%C The only consecutive brilliant numbers are {9, 10} and {14, 15}; and for m > 14 there are no brilliant constellations of the form {m, m+(2k+1)} or equivalently {n, 2k+m+1} with k >= 0. Proof: One of m and 2k+m+1 will be even. And there are no even brilliant numbers > 14 since they must have the form 2*p where p is a prime having only one digit.
%H Amiram Eldar, <a href="/A083284/b083284.txt">Table of n, a(n) for n = 1..10000</a>
%e a(3) = 779 because 779=19*41 and 781=11*71.
%Y Cf. A078972.
%K base,easy,nonn
%O 1,1
%A _Jason Earls_, Jun 03 2003
|
