A083284 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.

4, 527, 779, 869, 899, 1079, 1157, 1271, 1679, 4187, 6497, 6887, 24287, 24881, 25019, 29591, 35237, 37127, 37769, 38807, 39269, 39911, 41309, 43361, 44831, 45347, 46001, 46127, 47261, 48509, 48929, 51809, 52907, 54389, 55481, 55751, 55961

Offset: 1

Author

Jason Earls, Jun 03 2003

Keywords

• base
• easy
• nonn

Comments

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.

Examples

			a(3) = 779 because 779=19*41 and 781=11*71.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A078972.