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A087325
Numbers k such that k and its 10's complement both have the same prime signature.
2
3, 5, 7, 11, 14, 15, 17, 26, 29, 30, 35, 38, 41, 47, 50, 53, 59, 62, 65, 70, 71, 74, 83, 85, 86, 89, 94, 97, 110, 111, 113, 122, 129, 132, 134, 137, 140, 150, 153, 158, 170, 173, 174, 179, 183, 185, 186, 187, 191, 195, 201, 206, 209, 212, 215, 219, 221, 227, 236
OFFSET
1,1
COMMENTS
Conjecture: (1) Sequence is infinite. (2) For every prime signature there corresponds a term in this sequence.
From Robert Israel, Jul 02 2024: (Start)
Conjecture (2) is false: k and its 10's complement can't both have prime signature p^m where m is even.
If k is a term, then so is 10 * k.
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)
LINKS
EXAMPLE
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.
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.
MAPLE
ps:= n -> sort(ifactors(n)[2][.., 2]):
tc:= n -> 10^(1+ilog10(n))-n:
select(n -> ps(n) = ps(tc(n)), [$1..1000]); # Robert Israel, Jul 02 2024
CROSSREFS
Cf. A087324, A089186. Contains A068811.
Sequence in context: A373349 A123677 A133954 * A072151 A332028 A280018
KEYWORD
base,nonn
AUTHOR
Amarnath Murthy, Sep 04 2003
EXTENSIONS
More terms from David Wasserman, May 06 2005
STATUS
approved