A309940 - OEIS

%I #36 Jul 29 2021 16:44:09

%S 1,1,2,3,1,2,3,5,7,11,23,62,31,128,173,59,173,315,263,193,177,74,233,

%T 561,347,299,281,94,293,220,193,166,71,172,1159,428,899,1277,1241,391,

%U 1157,1245,115,1718,533,1621,1397,365,1183,1873,2127,2588,2539,317,61

%N a(1) = 1, thereafter a(n) is the next prime after a(n-1), but written backwards in base n, then converted back to decimal.

%C Conjecture: a(n) < n^2 for all n > 1.

%C From _Robert Dougherty-Bliss_, Sep 17 2019: (Start)

%C a(n) > n for n >= 10, as shown in the linked proof below.

%C 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.

%C (End)

%H Rémy Sigrist, <a href="/A309940/b309940.txt">Table of n, a(n) for n = 1..10000</a>

%H Robert Dougherty-Bliss, <a href="/A309940/a309940.txt">Proof that a(n) > n for n >= 10</a>

%e The next prime after a(10) = 11 is 13:

%e - 13 in base 11 is "12",

%e - reading backwards we obtain "21" = 2*11 + 1 = 23,

%e - hence a(11) = 23.

%t 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 *)

%o (PARI) for (n=1, 55, print1 (v=if(n==1, 1, fromdigits(Vecrev(digits(nextprime(1+v), n)), n)) ", "))

%Y Cf. A326344.

%Y The distinct numbers are listed in increasing order in A328076. See also A328257.

%K nonn,base

%O 1,3

%A _Rémy Sigrist_, Sep 14 2019

%E Name amended by _Felix Fröhlich_, Sep 16 2019