

A267856


LB numbers: positive integers of the form m = a*10^k+b (with a > 0 and b < 10^k) satisfying two properties: 1) the set of prime factors of m is the union of the sets of prime factors of a and b; and 2) A001222(m) = A001222(a) + A001222(b).


2



250, 375, 648, 972, 2430, 2500, 3750, 6480, 6750, 9375, 9720, 24300, 25000, 36450, 37500, 60750, 64800, 67500, 84672, 93750, 97200, 243000, 250000, 364500, 375000, 546750, 607500, 648000, 675000, 846720, 937500, 972000, 2430000, 2500000, 3645000, 3750000, 5467500
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,1


COMMENTS

The name LB comes from the Stanton reference below. These are the initials of a student who mistakenly tries to find the prime factorization of 972 by finding the prime factorizations of 9 and 72 respectively and then multiplying these together. The result is incorrect but close to correct since the same primes and the same number of primes are involved.
A001222(n) = Bigomega(n) gives the number of primes divisors of n counted with multiplicity.
This sequence is infinite: if m is a term, then so is 10*m.
The next term that is not of the form 10*m for a term already listed is 8346672.
Every term is composite with at least two distinct prime factors.


LINKS



EXAMPLE

We have 972 = 9*10^2 + 72, 972 = 2^2*3^5, 9 = 3^2, 72 = 2^3*3^2. Thus, counting multiplicity there are seven primes dividing 972, two primes dividing 9 and five primes dividing 72. Since the set of primes dividing n is {2,3}, which is the union of the sets of primes dividing 9 ({3}) and 72 ({2,3}), and 7 = 5 + 2, we have that 972 is a term in the sequence.


MATHEMATICA

f[n_] := Flatten@ Apply[Table[#1, {#2}] &, FactorInteger@ n, 1]; Select[Range@ 100000, Function[n, AnyTrue[Map[Flatten, Map[f, Map[FromDigits, Map[Function[k, TakeDrop[#, k]], Range[Length@ #  1]] &@ IntegerDigits@ #, {2}] &@ n, {2}], {1}], Length@ # == PrimeOmega@ n && Union@ # == First /@ FactorInteger@ n &]]] (* Michael De Vlieger, Jan 29 2016, Version 10.2 *)


PROG

(PARI) isok(n) = {nb = #Str(n); spf = Set(factor(n)[, 1]~); nbpfr = bigomega(n); for (k=1, nb1, a = n\10^k; b = n  10^k*a; if (b && (bigomega(a)+ bigomega(b) == nbpfr) && (setunion(factor(a)[, 1]~, factor(b)[, 1]~) == spf), return (1)); ); } \\ Michel Marcus, Jan 30 2016


CROSSREFS



KEYWORD

nonn,base


AUTHOR



EXTENSIONS



STATUS

approved



