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A330932 Starts of runs of 3 consecutive Niven numbers in base 2 (A049445). 21
623, 846, 2358, 4206, 4878, 6127, 6222, 6223, 12438, 16974, 21006, 27070, 31295, 33102, 33103, 35343, 37134, 37630, 37638, 40703, 43263, 45550, 48190, 49230, 52590, 53262, 53263, 56110, 59630, 66198, 66702, 66703, 67878, 69310, 69487, 72655, 74766, 77230, 77958 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Cai proved that there are infinitely many runs of 4 consecutive Niven numbers in base 2. Therefore this sequence is infinite.

REFERENCES

József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 382.

LINKS

Amiram Eldar, Table of n, a(n) for n = 1..10000

Tianxin Cai, On 2-Niven numbers and 3-Niven numbers, Fibonacci Quarterly, Vol. 34, No. 2 (1996), pp. 118-120.

Wikipedia, Harshad number.

Brad Wilson Construction of 2n consecutive n-Niven numbers, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.

EXAMPLE

623 is a term since 623, 624 and 625 are all Niven numbers in base 2.

MATHEMATICA

binNivenQ[n_] := Divisible[n, Total @ IntegerDigits[n, 2]]; bin = binNivenQ /@ Range[3]; seq = {}; Do[bin = Join[Rest[bin], {binNivenQ[k]}]; If[And @@ bin, AppendTo[seq, k - 2]], {k, 3, 8*10^4}]; seq

PROG

(Magma) f:=func<n|n mod &+Intseq(n, 2) eq 0>; a:=[]; for k in [1..80000] 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

CROSSREFS

Cf. A049445, A154701, A328210, A328214, A330931, A330933.

Sequence in context: A321675 A345556 A345810 * A255086 A158373 A265119

Adjacent sequences: A330929 A330930 A330931 * A330933 A330934 A330935

KEYWORD

nonn,base

AUTHOR

Amiram Eldar, Jan 03 2020

STATUS

approved

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Last modified February 5 08:15 EST 2023. Contains 360082 sequences. (Running on oeis4.)