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

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

Robert Israel, Table of n, a(n) for n = 1..10000

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

Adjacent sequences: A087322 A087323 A087324 * A087326 A087327 A087328

Keyword

base,nonn,changed

Author

Amarnath Murthy, Sep 04 2003

Extensions

More terms from David Wasserman, May 06 2005

Status

approved