A330931 Numbers k such that both k and k + 1 are Niven numbers in base 2 (A049445).

1, 20, 68, 80, 115, 155, 184, 204, 260, 272, 284, 320, 344, 355, 395, 404, 424, 464, 555, 564, 595, 623, 624, 636, 664, 675, 804, 835, 846, 847, 864, 875, 888, 904, 972, 1028, 1040, 1075, 1088, 1124, 1164, 1182, 1211, 1224, 1239, 1266, 1280, 1304, 1315, 1424

Offset

1,2

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

20 is a term since 20 and 20 + 1 = 21 are both Niven numbers in base 2.

Mathematica

binNivenQ[n_] := Divisible[n, Total @ IntegerDigits[n, 2]]; bnq1 = binNivenQ[1]; seq = {}; Do[bnq2 = binNivenQ[k]; If[bnq1 && bnq2, AppendTo[seq, k - 1]]; bnq1 = bnq2, {k, 2, 10^4}]; seq

Prog

(Magma) f:=func<n|n mod &+Intseq(n, 2) eq 0>; a:=[]; for k in [1..1500] 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
(Python)
def sbd(n): return sum(map(int, str(bin(n)[2:])))
def niv2(n): return n%sbd(n) == 0
def aupto(nn): return [k for k in range(1, nn+1) if niv2(k) and niv2(k+1)]
print(aupto(1424)) # Michael S. Branicky, Jan 20 2021

Crossrefs

Cf. A049445, A328205, A328209, A328213, A330713, A330927, A330932, A330933.

Sequence in context: A228839 A331096 A304253 * A158444 A235281 A303620

Adjacent sequences: A330928 A330929 A330930 * A330932 A330933 A330934

Keyword

nonn,base

Author

Amiram Eldar, Jan 03 2020

Status

approved