%I
%S 12,20,110,510,131052,12751220,10000095,2162049150,124324220,1,
%T 920067411130599,43494229746440272890,
%U 12100324200007455010742303399999999999999999990,4201420328711160916072939999999999999999999999999999999999999996
%N Initial term of a series of exactly n consecutive Harshad or Niven numbers (a Harshad number is such that is divided by the sum of its digits).
%C Cooper and Kennedy (1993) proved that this sequence contains 20 terms. [From _Sergio Pimentel_, Sep 18 2008]
%C a(16) = 50757686696033684694106416498959861492*10^280  9 and a(17) = 14107593985876801556467795907102490773681*10^280  10.  _Max Alekseyev_, Apr 07 2013
%D J.M. De Koninck, Ces nombres qui nous fascinent, Entry 110, p. 39, Ellipses, Paris 2008.
%H C. N. Cooper and R. E. Kennedy (1993). <a href="http://www.fq.math.ca/Scanned/312/cooper.pdf">On consecutive Niven numbers</a>. Fibonacci Quart, 21, 146151.
%H H. G. Grundman (1994). <a href="http://www.fq.math.ca/Scanned/322/grundman.pdf">Sequences of consecutive nNiven numbers</a>. Fibonacci Quarterly 32 (2): 174175.
%H B. Wilson (1997). <a href="http://www.fq.math.ca/Scanned/352/wilson.pdf">Construction of 2n Consecutive nNiven Numbers</a>. Fibonacci Quarterly, 35, 122128.
%H C. Rivera, <a href="http://www.primepuzzles.net/puzzles/puzz_129.htm">Puzzle 129. Earliest sets of K consecutive Harshad Numbers</a>
%e a(3)=110 since 110 is divisible by 2, 111 is divisible by 3, 112 is divisible by 4 but 113 is not divisible by 5.
%Y Cf. A005349.
%K fini,hard,nonn,base
%O 1,1
%A _Carlos Rivera_, Mar 12 2001
%E a(8) is found by _Jud McCranie_, Nov 13 2001
%E a(11)a(13) are found by _Giovanni Resta_, Feb 21 2008
%E a(14),a(16)a(17) from _Max Alekseyev_, Apr 07 2013
