A061252 a(n) = 16^n - 15^n.

0, 1, 31, 721, 14911, 289201, 5386591, 97576081, 1732076671, 30276117361, 522861237151, 8942430185041, 151728638820031, 2557404559011121, 42864668012537311, 715027614225987601, 11878335717996660991

Offset

0,3

Comments

Number of ways to assign truth values to n quaternary conjunctions connected by disjunctions such that the proposition is true. For example, a(2) = 31, since for the proposition '(a & b & c & d) v (e & f & g & h)' there are 31 assignments that make the proposition true. - Ori Milstein, Dec 31 2022

Equivalently, the number of length-n words over the alphabet {0,1,..,15} with at least one letter = 15. - Joerg Arndt, Jan 01 2023

Links

Table of n, a(n) for n=0..16.

Index entries for linear recurrences with constant coefficients, signature (31,-240).

Formula

a(0)=0, a(n) = 15*a(n-1) + 16^(n-1). - Vincenzo Librandi, Feb 09 2011

a(0)=0, a(1)=1, a(n) = 31*a(n-1) - 240*a(n-2). - Vincenzo Librandi, Feb 09 2011

a(n) = A001025(n) - A001024(n). - Michel Marcus, Aug 26 2013

Mathematica

Table[16^n-15^n, {n, 0, 20}] (* or  *) LinearRecurrence[{31, -240}, {0, 1}, 20] (* Harvey P. Dale, Jan 23 2021 *)

Prog

(PARI) a(n) = 16^n - 15^n; \\ Michel Marcus, Aug 26 2013

Crossrefs

Base 8: A016177, 4: A005061, 2: A000225, 10: A016189.

Sequence in context: A014930 A196988 A358179 * A096049 A166488 A051587

Adjacent sequences: A061249 A061250 A061251 * A061253 A061254 A061255

Keyword

nonn,easy

Author

Frank Ellermann, Jun 05 2001

Status

approved