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a(n) = gcd(reverse(prime(n)), reverse(prime(n+1))).
1

%I #14 Jan 31 2020 15:05:26

%S 1,1,1,1,1,1,1,1,4,1,1,1,2,2,1,5,1,4,1,1,1,1,2,1,1,1,1,1,1,1,1,1,1,1,

%T 1,1,1,1,1,1,1,1,1,1,1,1,14,2,2,2,4,2,2,8,2,2,2,4,2,2,2,1,1,1,1,1,1,1,

%U 1,1,1,1,1,1,1,1,1,13,8,2,2,2,2,2,2,8,2,2,4,4,2,2,2,2,1,5,5,25,5,5,5,5,5,5,25,5,5,5,1,2,2,4,4,4

%N a(n) = gcd(reverse(prime(n)), reverse(prime(n+1))).

%C Greatest common divisor of two consecutive primes after each prime is written backward.

%H Harvey P. Dale, <a href="/A103163/b103163.txt">Table of n, a(n) for n = 1..1000</a>

%F a(n) = gcd(A004087(n), A004087(n+1)).

%e Neither of these common divisors are divisible by 3 or by 10 or by 11.

%p A103163 := proc(n)

%p p := ithprime(n) ;

%p q := nextprime(p) ;

%p igcd(digrev(p),digrev(q)) ;

%p end proc:

%p seq(A103163(n),n=1..114) ; # _R. J. Mathar_, Sep 22 2018

%t rd[x_] :=FromDigits[Reversed[IntegerDigits[x]]]; Table[GCD[rd[Prime[w]], rd[Prime[w+1]]], {w, 1, 1000}]

%t GCD@@#&/@(Partition[IntegerReverse/@Prime[Range[120]],2,1]) (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jan 31 2020 *)

%Y Cf. A004087, A000040.

%K base,nonn

%O 1,9

%A _Labos Elemer_, Jan 27 2005

%E Edited by _Jon E. Schoenfield_, Oct 26 2019