cp's OEIS Frontend

A087325 Numbers k such that k and its 10's complement both have the same prime signature.

%I A087325 #19 Jul 03 2024 18:47:02
%S A087325 3,5,7,11,14,15,17,26,29,30,35,38,41,47,50,53,59,62,65,70,71,74,83,85,
%T A087325 86,89,94,97,110,111,113,122,129,132,134,137,140,150,153,158,170,173,
%U A087325 174,179,183,185,186,187,191,195,201,206,209,212,215,219,221,227,236
%N A087325 Numbers k such that k and its 10's complement both have the same prime signature.
%C A087325 Conjecture: (1) Sequence is infinite. (2) For every prime signature there corresponds a term in this sequence.
%C A087325 From _Robert Israel_, Jul 02 2024: (Start)
%C A087325 Conjecture (2) is false: k and its 10's complement can't both have prime signature p^m where m is even.
%C A087325 If k is a term, then so is 10 * k.
%C A087325 It appears that the first term with m prime factors, counted with multiplicity, is 3 * 10^((m-1)/2) if m is odd and 132 * 10^((m-4)/2) if m >= 4 is even. (End)
%H A087325 Robert Israel, <a href="/A087325/b087325.txt">Table of n, a(n) for n = 1..10000</a>
%e A087325 35 is a member as 35= 5*7 and its 10's complement (100-35) = 65 = 13*5 both have the prime signature p*q.
%e A087325 35 is a member as 35 = 5*7 and its 10's complement (100-35) = 65 = 13*5 both have the prime signature p*q.
%p A087325 ps:= n -> sort(ifactors(n)[2][..,2]):
%p A087325 tc:= n -> 10^(1+ilog10(n))-n:
%p A087325 select(n -> ps(n) = ps(tc(n)), [$1..1000]); # _Robert Israel_, Jul 02 2024
%Y A087325 Cf. A087324, A089186. Contains A068811.
%K A087325 base,nonn
%O A087325 1,1
%A A087325 _Amarnath Murthy_, Sep 04 2003
%E A087325 More terms from _David Wasserman_, May 06 2005