A295838 Largest value corresponding to a string of n printable ASCII characters.

0, 126, 32382, 8289918, 2122219134, 543288098430, 139081753198206, 35604928818740862, 9114861777597660798, 2333404615065001164414, 597351581456640298090110

Offset: 0

Author

Abilene Sukin, Nov 28 2017

Keywords

• nonn
• easy
• base

Comments

The n-th term of this sequence is the result of forming an ASCII (American Standard Code for Information Exchange) text string of n characters using the (printable) character with the largest binary value and then converting the binary value of the string to base 10. a(n) is therefore a measure of the largest possible size of an ASCII string with n printable characters. This sequence uses standard 7-bit ASCII; A175824 is the same sequence using 8-bit Extended ASCII.

Conjecture: For every a(n) there exists a sequence of primes (p(1), p(2), p(3), ...) such that for each term a(n) there exists a set of primes that when added to the term result in another prime. For example, a(2)=126 and 126 + {5,11,13,23,37,41,47,...} all are primes.

Corollary 1: If it is the case that the size of the set of prime numbers is countably infinite, then the cardinality of the sequence that contains the sequence of primes p(1), p(2), ... that produce a new prime for every a(n) is uncountably infinite ... [ (a(1)+p(1), a(1)+p(2), a(1)+p(3), ...), (a(2)+p(1)', a(2)+p(2)', a(2)+p(3)'), ...)

Examples

			The lexicographically last 2-character printable ASCII string is "~~", which is 7E7E in hexadecimal or 32382 in decimal, thus a(2) = 32382.
		

Links

• Iain Fox, Table of n, a(n) for n = 0..415
• Index entries for linear recurrences with constant coefficients, signature (257,-256).

Crossrefs

Cf. A175824.

Programs

Magma

[(42/85)*(256^n - 1) : n in [0..20]]; // Wesley Ivan Hurt, Nov 29 2017

Maple

A295838:=n->(42/85)*(256^n - 1): seq(A295838(n), n=0..20); # Wesley Ivan Hurt, Nov 29 2017

Mathematica

Array[(42/85) (256^# - 1) &, 11, 0] (* Michael De Vlieger, Dec 11 2017 *)
CoefficientList[Series[126 x/((1 - 256 x) (1 - x)), {x, 0, 10}], x] (* Michael De Vlieger, Dec 11 2017 *)

PARI

a(n) = 42/85*(256^n-1) \\ Iain Fox, Nov 28 2017

PARI

first(n) = Vec(126*x/((1-256*x)*(1-x)) + O(x^n), -n) \\ Iain Fox, Nov 28 2017

Formula

a(n) = (42/85)*(256^n - 1).

From Iain Fox, Nov 28 2017: (Start)

G.f.: 126*x/((1-256*x)*(1-x)).

E.g.f.: 42/85*(e^(256*x)-e^x).

a(n) = 42/85 * A175824(n).

(End)

For n > 0, a(n) = 256*a(n-1) + 126. - Jon E. Schoenfield, Nov 29 2017

For n > 1, a(n) = 257*a(n-1) - 256*a(n-2). - Iain Fox, Jan 02 2018