A330928 Starts of runs of 5 consecutive Niven (or harshad) numbers (A005349).

1, 2, 3, 4, 5, 6, 131052, 491424, 1275140, 1310412, 1474224, 1614623, 1912700, 2031132, 2142014, 2457024, 2550260, 3229223, 3931224, 4422624, 4914024, 5405424, 5654912, 5920222, 7013180, 7125325, 7371024, 8073023, 8347710, 9424832, 10000095, 10000096, 10000097

Offset: 1

Author

Amiram Eldar, Jan 03 2020

Keywords

• nonn
• base

Comments

Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite.

Examples

			131052 is a term since 131052 is divisible by 1 + 3 + 1 + 0 + 5 + 2 = 12, 131053 is divisible by 13, 131054 is divisible by 14, 131055 is divisible by 15, and 131056 is divisible by 16.
		

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.
• 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, A060159; A330927, A154701, A141769, A330929, A330930 (same for 2, 3, 4, 6, 7 consecutive harshad numbers).

Programs

Magma

f:=func; a:=[]; for k in [1..11000000] do  if forall{m:m in [0..4]|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[5]; seq = {}; Do[niv = Join[Rest[niv], {nivenQ[k]}]; If[And @@ niv, AppendTo[seq, k - 4]], {k, 5, 10^7}]; seq
SequencePosition[Table[If[Divisible[n,Total[IntegerDigits[n]]],1,0],{n,10^7+200}],{1,1,1,1,1}][[;;,1]] (* Harvey P. Dale, Dec 24 2023 *)

PARI

{first( N=50, LEN=5, L=List())= for(n=1,oo, n+=LEN; for(m=1,LEN, n--%sumdigits(n) && next(2)); listput(L,n); N--|| break);L} \\ M. F. Hasler, Jan 03 2022

Formula

This A330928 = { A005349(k) | A005349(k+4) = A005349(k)+4 }. - M. F. Hasler, Jan 03 2022