1,1

Let p and p+2 be twin primes. Then Lucas(p) = 1 + p*A006206(p) and Lucas(p+2) = 1 + (p+2)*A006206(p+2). It follows from Lucas(n) + Lucas(n+1) = Lucas(n+2) that p = (Lucas(p+1) - 2*A006206(p+2))/(A006206(p+2) - A006206(p))

For i = 3, 4, 5, 6, 7, 8, 9, 10, 11, : ((Lucas(i+1) - 2*A006206(i+2))/(A006206(i+2) - A006206(i))) = (3, 7, 5, 19/3, 31/4, 9, 87/10, 149/14, 11, 135/11, 663/50, 1094/77, 1787/120, 2939/181, 17, 7849/434, 12799/672, 20894/1041, 34031/1622, 55469/2514, 45131/1962, 146921/6115, 238915/9554, 194252/7465, 631347/23386, 1025917/36617, 29, 2706059/90178, 4393211/141710, 3565643/111405, 11573003/350702, ) - Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jan 31 2006

It is conjectured that a(n+4) = A001359(n+2) for all n.

Cf. A000204, A006206, A001359.

Sequence in context: A122001 A019809 A021270 this_sequence A094009 A088514 A066677

Adjacent sequences: A113907 A113908 A113909 this_sequence A113911 A113912 A113913

nonn

Creighton Dement (creighton.k.dement(AT)uni-oldenburg.de), Jan 29 2006