A154701 Numbers k such that k, k + 1 and k + 2 are 3 consecutive Harshad numbers.

1, 2, 3, 4, 5, 6, 7, 8, 110, 510, 511, 1010, 1014, 1015, 2022, 2023, 2464, 3030, 3031, 4912, 5054, 5831, 7360, 8203, 9854, 10010, 10094, 10307, 10308, 11645, 12102, 12103, 12255, 12256, 13110, 13111, 13116, 13880, 14704, 15134, 17152, 17575, 18238, 19600, 19682

Offset: 1

Author

Avik Roy (avik_3.1416(AT)yahoo.co.in), Jan 14 2009, Jan 15 2009

Keywords

• nonn
• base

Comments

Harshad numbers are also known as Niven numbers.

Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite. - Amiram Eldar, Jan 03 2020

Examples

			110 is a term since 110 is divisible by 1 + 1 + 0 = 2, 111 is divisible by 1 + 1 + 1 = 3, and 112 is divisible by 1 + 1 + 2 = 4.
		

References

• Jean-Marie De Koninck, Those Fascinating Numbers, American Mathematical Society, 2009, p. 36, entry 110.

Links

• Robert Israel, Table of n, a(n) for n = 1..10000
• Curtis Cooper and Robert E. Kennedy, On consecutive Niven numbers, Fibonacci Quarterly, Vol. 21, No. 2 (1993), pp. 146-151.
• Helen G. Grundman, Sequences of consecutive Niven numbers, Fibonacci Quarterly, Vol. 32, No. 2 (1994), pp. 174-175.
• Wikipedia, Harshad number
• Brad Wilson, Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Crossrefs

A subset of A005349.

Cf. A060159, A141769, A328210, A328214, A330927, A330928, A330929, A330930, A330932.

Programs

C

#include 
#include 
int is_harshad(int n){
  int i,j,count=0;
  i=n;
  while(i>0){
    count=count+i%10;
    i=i/10;
  }
  return n%count==0?1:0;
}
main(){
  int k;
  clrscr();
  for(k=1;k<=30000;k++)
    if(is_harshad(k)&&is_harshad(k+1)&&is_harshad(k+2))
      printf("%d,",k);
  getch();
  return 0;
}

Magma

f:=func; a:=[]; for k in [1..20000] do  if forall{m:m in [0..2]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

Maple

Res:= NULL: count:= 0:
state:= 1:
L:= [1]:
for n from 2 while count < 100 do
  L[1]:=L[1]+1;
  for k from 1 while L[k]=10 do L[k]:= 0;
    if k = nops(L) then L:= [0$nops(L),1]; break
    else L[k+1]:= L[k+1]+1 fi
  od:
  s:= convert(L,`+`);
  if n mod s = 0 then
     state:= min(state+1,3);
     if state = 3 then count:= count+1; Res:= Res, n-2; fi
  else state:= 0
  fi
od:
Res; # Robert Israel, Feb 01 2019

Mathematica

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[3]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 2]], {k, 3, 2*10^4}]; seq (* Amiram Eldar, Jan 03 2020 *)

Python

from itertools import count, islice
def agen(): # generator of terms
    h1, h2, h3 = 1, 2, 3
    while True:
        if h3 - h1 == 2: yield h1
        h1, h2, h3 = h2, h3, next(k for k in count(h3+1) if k%sum(map(int, str(k))) == 0)
print(list(islice(agen(), 45))) # Michael S. Branicky, Mar 17 2024