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
Amiram Eldar, Table of n, a(n) for n = 1..100
Robert O. Stanton, The Remarkable Number 648, Mathematics Magazine, Vol. 66, No. 1 (Feb., 1993), pp. 48-55.
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, 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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Tom Edgar, Jan 28 2016
EXTENSIONS
More terms from Michel Marcus, Jan 30 2016
STATUS
approved