OFFSET
1,1
COMMENTS
Write n as product of primes raised to powers, let D(n) = number of digits in product, l(n) = number of digits in n; sequence gives n such that D(n)>l(n).
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J.-P. Delahaye, Les chasseurs de nombres premiers.
R. G. E. Pinch, Economical numbers., arXiv:math/9802046 [math.NT], 1998.
Eric Weisstein's World of Mathematics, Wasteful Number.
Wikipedia, Extravagant number.
EXAMPLE
For n = 125 = 5^3, l(n) = 3 but D(n) = 2. So 125 is not a term of this sequence. [clarified by Derek Orr, Jan 30 2015]
MATHEMATICA
Cases[Range[115], n_ /; Length[Flatten[IntegerDigits[FactorInteger[n] /. 1 -> Sequence[]]]] > Length[IntegerDigits[n]]] (* Jean-François Alcover, Mar 21 2011 *)
PROG
(Haskell)
a046760 n = a046760_list !! (n-1)
a046760_list = filter (\n -> a050252 n > a055642 n) [1..]
-- Reinhard Zumkeller, Aug 02 2013
(PARI) for(n=1, 100, s=""; F=factor(n); for(i=1, #F[, 1], s=concat(s, Str(F[i, 1])); s=concat(s, Str(F[i, 2]))); c=0; for(j=1, #F[, 2], if(F[j, 2]==1, c++)); if(#digits(n)<#s-c, print1(n, ", "))) \\ Derek Orr, Jan 30 2015
(Python)
from itertools import count, islice
from sympy import factorint
def A046760_gen(): # generator of terms
return (n for n in count(1) if len(str(n)) < sum(len(str(p))+(len(str(e)) if e > 1 else 0) for p, e in factorint(n).items()))
CROSSREFS
KEYWORD
nonn,nice,base,easy
AUTHOR
STATUS
approved