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

Author

Amiram Eldar, Jan 03 2020

Keywords

• nonn
• base

Comments

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

Examples

			20 is a term since 20 and 20 + 1 = 21 are both 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

• 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.

Crossrefs

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

Programs

Magma

f:=func; 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

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

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