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A330933
Starts of runs of 4 consecutive Niven numbers in base 2 (A049445).
19
6222, 33102, 53262, 66702, 94830, 221550, 268302, 284910, 295182, 300750, 316590, 364110, 379950, 427470, 533950, 554190, 570030, 590862, 617550, 633390, 696750, 791790, 807630, 855150, 870990, 902670, 934350, 1081422, 1140270, 1282830, 1314510, 1330350, 1343502
OFFSET
1,1
COMMENTS
Cai proved that there are infinitely many runs of 4 consecutive Niven numbers in base 2.
Grundman proved that there are no runs of 5 or more consecutive Niven numbers in base 2.
REFERENCES
József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 382.
LINKS
Tianxin Cai, On 2-Niven numbers and 3-Niven numbers, Fibonacci Quarterly, Vol. 34, No. 2 (1996), pp. 118-120.
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
6222 is a term since 6222, 6223, 6224 and 6225 are all Niven numbers in base 2.
MATHEMATICA
binNivenQ[n_] := Divisible[n, Total @ IntegerDigits[n, 2]]; bin = binNivenQ /@ Range[4]; seq = {}; Do[bin = Join[Rest[bin], {binNivenQ[k]}]; If[And @@ bin, AppendTo[seq, k - 3]], {k, 4, 10^6}]; seq
PROG
(Magma) f:=func<n|n mod &+Intseq(n, 2) eq 0>; a:=[]; for k in [1..1400000] 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
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Jan 03 2020
STATUS
approved