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 A330931 Numbers k such that both k and k + 1 are Niven numbers in base 2 (A049445). 16
 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 (list; graph; refs; listen; history; text; internal format)
 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; 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

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Last modified July 28 15:08 EDT 2021. Contains 346335 sequences. (Running on oeis4.)