A141769 Beginning of a run of 4 consecutive Niven (or Harshad) numbers.

1, 2, 3, 4, 5, 6, 7, 510, 1014, 2022, 3030, 10307, 12102, 12255, 13110, 60398, 61215, 93040, 100302, 101310, 110175, 122415, 127533, 131052, 131053, 196447, 201102, 202110, 220335, 223167, 245725, 255045, 280824, 306015, 311232, 318800, 325600, 372112, 455422

Offset: 1

Author

Sergio Pimentel, Sep 15 2008

Keywords

• base
• nonn

Comments

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

			510 is in the sequence because 510, 511, 512 and 513 are all Niven numbers.
		

References

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

Links

• Amiram Eldar, 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.
• Eric Weisstein's World of Mathematics, Harshad Number.
• Wikipedia, Harshad number.
• Brad Wilson Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

Crossrefs

Cf. A005349, A330927, A154701, A330928, A330929, A330930, A060159 (start of run of 1, 2, ..., 7, exactly n consecutive Harshad numbers).

Cf. A330933, A328211, A328215 (analog for base 2, Zeckendorf- resp. Fibonacci-Niven variants).

Programs

Magma

f:=func; a:=[]; for k in [1..500000] do  if forall{m:m in [0..3]|f(k+m)} then Append(~a,k); end if; end for; a; // Marius A. Burtea, Jan 03 2020

Mathematica

nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; niv = nivenQ /@ Range[4]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 3]], {k, 4, 5*10^5}]; seq (* Amiram Eldar, Jan 03 2020 *)

PARI

{A141769_first( N=50, L=4, a=List())= for(n=1,oo, n+=L; for(m=1,L, n--%sumdigits(n) && next(2)); listput(a,n); N--|| break);a} \\ M. F. Hasler, Jan 03 2022

Python

from itertools import count, islice
def agen(): # generator of terms
    h1, h2, h3, h4 = 1, 2, 3, 4
    while True:
        if h4 - h1 == 3: yield h1
        h1, h2, h3, h4, = h2, h3, h4, next(k for k in count(h4+1) if k%sum(map(int, str(k))) == 0)
print(list(islice(agen(), 40))) # Michael S. Branicky, Mar 17 2024

Formula

This A141769 = { A005349(k) | A005349(k+3) = A005349(k)+3 }. - M. F. Hasler, Jan 03 2022

Extensions

More terms from Amiram Eldar, Jan 03 2020