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%I #34 Feb 24 2020 08:46:15
%S 250,375,648,972,2430,2500,3750,6480,6750,9375,9720,24300,25000,36450,
%T 37500,60750,64800,67500,84672,93750,97200,243000,250000,364500,
%U 375000,546750,607500,648000,675000,846720,937500,972000,2430000,2500000,3645000,3750000,5467500
%N 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).
%C 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.
%C A001222(n) = Bigomega(n) gives the number of primes divisors of n counted with multiplicity.
%C This sequence is infinite: if m is a term, then so is 10*m.
%C The next term that is not of the form 10*m for a term already listed is 8346672.
%C Every term is composite with at least two distinct prime factors.
%H Amiram Eldar, <a href="/A267856/b267856.txt">Table of n, a(n) for n = 1..100</a>
%H Robert O. Stanton, <a href="http://www.jstor.org/stable/2690478">The Remarkable Number 648</a>, Mathematics Magazine, Vol. 66, No. 1 (Feb., 1993), pp. 48-55.
%e 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.
%t 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 *)
%o (PARI) isok(n) = {nb = #Str(n); spf = Set(factor(n)[,1]~); nbpfr = bigomega(n); for (k=1, nb-1, 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
%Y Cf. A001222.
%K nonn,base
%O 1,1
%A _Tom Edgar_, Jan 28 2016
%E More terms from _Michel Marcus_, Jan 30 2016
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