A375090 Omit the trailing 4 from the terms of the Lucas-Lehmer sequence A003010.

0, 1, 19, 3763, 141631795, 200595654682274611, 402386166774103602282563565610210099, 1619146272111567178177755907012051366495859012549915851432930874097578803

Offset: 0

Author

Stefano Spezia, Jul 29 2024

Keywords

• nonn
• base

Comments

The next term a(8) has 146 digits.

Proposition: A003010(n) has only a trailing 4.

Proof: the case n = 0 is trivial since a(0) = 4 has only a digit. For n > 0, we need to prove that the terms have at least a trailing 4 and that this is unique. A003010(n) == 4 (mod 10) because A003010(0) = 4 and 4^2 - 2 = 14 == 4 (mod 10). To prove that this is unique, we prove that the tenth digit of A003010(n) is never equal to 4 for n > 0, or equivalently, that it is odd. Considering that (10*h + 4)^2 = 100*h^2 + 80*h + 16 and 80*h + 16 has the tenth digit of the form 8*k + 1 mod 10, it follows that the tenth digit of A003010(n-1)^2 is odd, and therefore, also that of A003010(n-1)^2 - 2 = A003010(n). QED.

Links

• Stefano Spezia, Table of n, a(n) for n = 0..10

Crossrefs

Cf. A003010, A017077.

Programs

Mathematica

A003010[0]=4; A003010[n_]:=A003010[n-1]^2-2; Array[A003010, 9, 0]; a[n_]:=(A003010[n]-4)/10; Array[a,8,0]

Formula

a(n) = (A003010(n) - 4)/10.

a(n) = 10*a(n-1)^2 + 8*a(n-1) + 1 for n > 0.