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A099542
Rhonda numbers to base 10.
22
1568, 2835, 4752, 5265, 5439, 5664, 5824, 5832, 8526, 12985, 15625, 15698, 19435, 25284, 25662, 33475, 34935, 35581, 45951, 47265, 47594, 52374, 53176, 53742, 54479, 55272, 56356, 56718, 95232, 118465, 133857, 148653, 154462, 161785
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)
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.
FORMULA
A007954(a(n)) = 10 * A001414(a(n)).
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
Cf. Rhonda numbers to other bases: A100968 (base 4), A100969 (base 6), A100970 (base 8), A100973 (base 9), A100971 (base 12), A100972 (base 14), A100974 (base 15), A100975 (base 16), A255735 (base 18), A255732 (base 20), A255736 (base 30), A255731 (base 60), see also A255880.
Column k=5 of A291925.
Sequence in context: A035890 A252109 A045276 * A232672 A232036 A035765
KEYWORD
base,nice,nonn
AUTHOR
Mark Hudson (mrmarkhudson(AT)hotmail.com), Oct 21 2004
STATUS
approved