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a(n) is the least number such that bigomega(a(n)) = bigomega(R(a(n)))/n, where R(n) is the digit reverse of n and bigomega(n) is the number of prime divisors of n counted with multiplicity.
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%I #24 Mar 31 2017 05:30:36

%S 2,19,29,61,23,487,821,2557,2749,4201,25747,61843,211643,614881,

%T 238673,272059,270131,405719,8239969,42037283,44922991,44166589,

%U 67719697,61277761,253442899,2321783617,4471131827,4640384471,40782386369,4055086219,25135138387,63558323051

%N a(n) is the least number such that bigomega(a(n)) = bigomega(R(a(n)))/n, where R(n) is the digit reverse of n and bigomega(n) is the number of prime divisors of n counted with multiplicity.

%C All values are prime.

%F Solutions of the equation A001222(n) = A001222(A004086(n))/n.

%e bigomega(2) = 1. Its digit reverse is again 2 and bigomega(2) = 1 = 1 * 1;

%e bigomega(19) = 1 and bigomega(91) = 2 = 2 * 1;

%e bigomega(29) = 1 and bigomega(92) = 3 = 3 * 1;

%e bigomega(61) = 1 and bigomega(16) = 4 = 4 * 1; etc.

%p with(numtheory): R:=proc(w) local x, y, z; x:=w; y:=0;for z from 1 to ilog10(x)+1 do y:=10*y+(x mod 10); x:=trunc(x/10); od; y; end:

%p P:=proc(q) local k,n; for k from 1 to q do for n from 2 to q do

%p if k*bigomega(n)=bigomega(R(n)) then print(n); break; fi; od; od; end: P(10^9);

%t rev[n_] := FromDigits@ Reverse@ IntegerDigits@ n;a[n_] := Block[{k, v}, For[k=2, Mod[v = PrimeOmega@ rev@ k, n] > 0 || v != n PrimeOmega@ k, k++]; k]; Array[a, 12] (* _Giovanni Resta_, Mar 29 2017 *)

%o (PARI) genit(maxx)={for(num=1,maxx,soln=0;forprime(p=2,9E20,f=Vecrev(Str(p));g=eval(concat(f));if(bigomega(g)==num,print(num," ",p);soln=1;break));if(soln<1,print("increase max prime")));} \\ _Bill McEachen_, Mar 30 2017

%Y Cf. A001222, A004086, A284495, A284497, A284498.

%K nonn,base

%O 1,1

%A _Paolo P. Lava_, Mar 28 2017

%E a(23)-a(32) from _Giovanni Resta_, Mar 29 2017

%E a(26) corrected by _Bill McEachen_, Mar 30 2017