OFFSET
1,3
COMMENTS
Conjecture: a(n) < n^2 for all n > 1.
From Robert Dougherty-Bliss, Sep 17 2019: (Start)
a(n) > n for n >= 10, as shown in the linked proof below.
The conjecture a(n) < n^2 holds for all 1 < n < N if each gap between consecutive squares < N contains a prime. In particular, the conjecture is true if Legendre's conjecture is true.
(End)
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
Robert Dougherty-Bliss, Proof that a(n) > n for n >= 10
EXAMPLE
The next prime after a(10) = 11 is 13:
- 13 in base 11 is "12",
- reading backwards we obtain "21" = 2*11 + 1 = 23,
- hence a(11) = 23.
MATHEMATICA
nxt[{n_, p_}]:={n+1, FromDigits[Reverse[IntegerDigits[NextPrime[ p], n+1]], n+1]}; NestList[nxt, {1, 1}, 60][[All, 2]] (* Harvey P. Dale, Jul 29 2021 *)
PROG
(PARI) for (n=1, 55, print1 (v=if(n==1, 1, fromdigits(Vecrev(digits(nextprime(1+v), n)), n)) ", "))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Rémy Sigrist, Sep 14 2019
EXTENSIONS
Name amended by Felix Fröhlich, Sep 16 2019
STATUS
approved