A095180 Reverse digits of primes, append to sequence if result is a prime.

2, 3, 5, 7, 11, 31, 71, 13, 73, 17, 37, 97, 79, 101, 701, 311, 131, 941, 151, 751, 761, 971, 181, 191, 991, 113, 313, 733, 743, 353, 953, 373, 383, 983, 107, 907, 727, 337, 937, 347, 157, 757, 167, 967, 787, 797, 709, 919, 929, 739, 149, 359, 769, 179, 389, 199

Offset: 1

Author

Cino Hilliard, Jun 21 2004

Keywords

• base
• easy
• nonn

Comments

Conjecture: the Benford law limit is 2=Sum[N[Log[10, 1 + 1/d[[n]]]], {n, 1, Length[d]}]^2/(( #totalprimes/#totalPrimes)). At 50000 primes total it is 2.05931. - Roger L. Bagula and Gary W. Adamson, Jul 02 2008

Presumably this does not satisfy Benford's law. - N. J. A. Sloane, Feb 09 2017

Examples

			The prime 107 in reverse is 701 which is prime.
		

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..25000 (terms 1..206 from Roger L. Bagula and Gary W. Adamson)
• Eric Weisstein's World of Mathematics, Benford's Law.
• Index entries for sequences related to Benford's law

Crossrefs

Cf. A007500.

Cf. A004087, A010051.

Programs

Haskell

a095180 n = a095180_list !! (n-1)
a095180_list =filter ((== 1) . a010051) a004087_list
-- Reinhard Zumkeller, Oct 14 2011

Mathematica

b = Flatten[Table[If[PrimeQ[Sum[IntegerDigits[Prime[n]][[i]]*10^(i - 1), {i, 1, Length[IntegerDigits[Prime[n]]]}]], Sum[IntegerDigits[Prime[n]][[i]]*10^(i - 1), {i, 1, Length[IntegerDigits[Prime[n]]]}], {}], {n, 1,1000}]] (* Roger L. Bagula and Gary W. Adamson, Jul 02 2008 *)
Select[FromDigits[Reverse[IntegerDigits[#]]]&/@Prime[Range[300]],PrimeQ] (* Harvey P. Dale, May 05 2015 *)

PARI

r(n) = forprime(x=1,n,y=eval(rev(x));if(isprime(y),print1(y","))) \ Get the reverse of the input string rev(str) = { local(tmp,j,s); tmp = Vec(Str(str)); s=""; forstep(j=length(tmp),1,-1, s=concat(s,tmp[j])); return(s) }