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A330927 Numbers k such that both k and k + 1 are Niven numbers. 18
1, 2, 3, 4, 5, 6, 7, 8, 9, 20, 80, 110, 111, 132, 152, 200, 209, 224, 399, 407, 440, 480, 510, 511, 512, 629, 644, 735, 800, 803, 935, 999, 1010, 1011, 1014, 1015, 1016, 1100, 1140, 1160, 1232, 1274, 1304, 1386, 1416, 1455, 1520, 1547, 1651, 1679, 1728, 1853 (list; graph; refs; listen; history; text; internal format)
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

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

Cf. A005349, A060159, A141769, A154701, A328205, A328209, A328213, A330713, A330928, A330929, A330930, A330931.

Sequence in context: A290951 A114800 A079170 * A319387 A076105 A094280

Adjacent sequences:  A330924 A330925 A330926 * A330928 A330929 A330930

KEYWORD

nonn,base

AUTHOR

Amiram Eldar, Jan 03 2020

STATUS

approved

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Last modified August 3 21:19 EDT 2021. Contains 346441 sequences. (Running on oeis4.)