OFFSET
1,2
COMMENTS
Cooper and Kennedy proved that there are infinitely many runs of 20 consecutive Niven numbers. Therefore this sequence is infinite.
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.
EXAMPLE
1 is a term since 1 and 1 + 1 = 2 are both Niven numbers.
MATHEMATICA
nivenQ[n_] := Divisible[n, Total @ IntegerDigits[n]]; nq1 = nivenQ[1]; seq = {}; Do[nq2 = nivenQ[k]; If[nq1 && nq2, AppendTo[seq, k - 1]]; nq1 = nq2, {k, 2, 2000}]; seq
SequencePosition[Table[If[Divisible[n, Total[IntegerDigits[n]]], 1, 0], {n, 2000}], {1, 1}][[;; , 1]] (* Harvey P. Dale, Dec 24 2023 *)
PROG
(Magma) f:=func<n|n mod &+Intseq(n) eq 0>; a:=[]; for k in [1..2000] do if forall{m:m in [0..1]|f(k+m)} then Append(~a, k); end if; end for; a; // Marius A. Burtea, Jan 03 2020
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Jan 03 2020
STATUS
editing