OFFSET
1,1
COMMENTS
An integer m is a Rhonda number to base b if the product of its digits in base b equals b*(sum of prime factors of m (taken with multiplicity)).
Does every Rhonda number to base 10 contain at least one 5? - Howard Berman (howard_berman(AT)hotmail.com), Oct 22 2008
Yes, every Rhonda number m to base 10 contains at least one 5 and also one even digit, otherwise A007954(m) mod 10 > 0. - Reinhard Zumkeller, Dec 01 2012
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..1000 (first 180 terms from Harvey P. Dale)
Kevin Brown, Infinitely Many Rhondas.
Giovanni Resta, Rhonda numbers the 64507 base-10 Rhonda numbers up to 10^12.
Walter Schneider, Rhonda Numbers.
Eric Weisstein's World of Mathematics, Rhonda Number.
EXAMPLE
1568 has prime factorization 2^5 * 7^2. Sum of prime factors = 2*5 + 7*2 = 24. Product of digits of 1568 = 1*5*6*8 = 240 = 10*24, hence 1568 is a Rhonda number to base 10.
MATHEMATICA
Select[Range[200000], 10Total[Times@@@FactorInteger[#]]==Times@@ IntegerDigits[ #]&] (* Harvey P. Dale, Oct 16 2011 *)
PROG
(Haskell)
import Data.List (unfoldr); import Data.Tuple (swap)
a099542 n = a099542_list !! (n-1)
a099542_list = filter (rhonda 10) [1..]
rhonda b x = a001414 x * b == product (unfoldr
(\z -> if z == 0 then Nothing else Just $ swap $ divMod z b) x)
-- Reinhard Zumkeller, Mar 05 2015, Dec 01 2012
CROSSREFS
KEYWORD
base,nice,nonn
AUTHOR
Mark Hudson (mrmarkhudson(AT)hotmail.com), Oct 21 2004
STATUS
approved