OFFSET

1,1

COMMENTS

Let a(n) be composite, and gp(n) be the greatest prime less than a(n), with final digit d(n). Then sp(n), the smallest prime greater than a(n) is in the sequence if and only if d(n) divides (sp(n) - a(n)), in which case (sp(n) - a(n))/d(n) is the integer m, and a(n + m) = sp(n).

If a(n) is prime, with final digit d'(n) then sp(n) (the next prime after a(n)) is in the sequence if and only if sp(n) = a(n)+ d(n) + r * d'(n) for some r >= 1, in which case

a(n + r + 1) = sp(n).

Primes appearing in the sequence are 3, 5, 31, 47, 79, 107, 137, 157, 181, 191, 193, 197, ... If a prime occurs at a(n), a(n+2) = a(n+1) + d(n) only if there is no prime between a(n) and a(n+1).

Primes in the sequence whose final digit is not contributed to a subsequent term include 3, 5, 31, 107, 197, ...

Primes not appearing in the sequence but which contribute a final digit include 2, 7, 13, 17, 23, 29, 37, ...

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

FORMULA

a(n+1) = a(n) + d(n) where d(n) = A007652(gp(n)); gp(n) = greatest prime < a(n).

EXAMPLE

a(2) = a(1) + gp(1) = 3 + 2 = 5.

a(59) = 242 (composite), gp(59) = 241, and d(59) = 1. sp(59) = 251 is in the sequence because (sp(59) - a(59))/d(59) = (251 - 242)/1 = 9 (= m). Therefore a(59 + 9) = a(68) = 251.

a(40) = 181 (prime), d'(40) = 1, gp(40) = 179, d(40) = 9. Then sp(40) = 191 is in the sequence because with r = 1,

a(40) + d(40) + r*d'(40) = 181 + 9 + 1*1 = 191 = a(40+1+1) = a(42).

MAPLE

A[1]:= 3:

for i from 2 to 100 do

A[i]:= A[i-1] + (prevprime(A[i-1]) mod 10)

od:

seq(A[i], i=1..100); # Robert Israel, Aug 13 2019

MATHEMATICA

NestList[# + Mod[NextPrime[#, -1], 10] &, 3, 68] (* Michael De Vlieger, Aug 19 2017 *)

PROG

(PARI) lista(nn) = {print1(a = 3, ", "); for (n=2, nn, a = a + precprime(a-1) % 10; print1(a, ", "); ); } \\ Michel Marcus, Aug 19 2017

CROSSREFS

KEYWORD

nonn,base

AUTHOR

David James Sycamore, Aug 07 2017

STATUS

approved