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%I #22 Mar 17 2024 07:33:34
%S 1,20,68,80,115,155,184,204,260,272,284,320,344,355,395,404,424,464,
%T 555,564,595,623,624,636,664,675,804,835,846,847,864,875,888,904,972,
%U 1028,1040,1075,1088,1124,1164,1182,1211,1224,1239,1266,1280,1304,1315,1424
%N Numbers k such that both k and k + 1 are Niven numbers in base 2 (A049445).
%C Cai proved that there are infinitely many runs of 4 consecutive Niven numbers in base 2. Therefore this sequence is infinite.
%D József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 382.
%H Amiram Eldar, <a href="/A330931/b330931.txt">Table of n, a(n) for n = 1..10000</a>
%H Tianxin Cai, <a href="https://www.fq.math.ca/Scanned/34-2/cai1.pdf">On 2-Niven numbers and 3-Niven numbers</a>, Fibonacci Quarterly, Vol. 34, No. 2 (1996), pp. 118-120.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Harshad_number">Harshad number</a>.
%H Brad Wilson, <a href="http://www.fq.math.ca/Scanned/35-2/wilson.pdf">Construction of 2n consecutive n-Niven numbers</a>, Fibonacci Quarterly, Vol. 35, No. 2 (1997), pp. 122-128.
%e 20 is a term since 20 and 20 + 1 = 21 are both Niven numbers in base 2.
%t 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
%o (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
%o (Python)
%o def sbd(n): return sum(map(int, str(bin(n)[2:])))
%o def niv2(n): return n%sbd(n) == 0
%o def aupto(nn): return [k for k in range(1, nn+1) if niv2(k) and niv2(k+1)]
%o print(aupto(1424)) # _Michael S. Branicky_, Jan 20 2021
%Y Cf. A049445, A328205, A328209, A328213, A330713, A330927, A330932, A330933.
%K nonn,base
%O 1,2
%A _Amiram Eldar_, Jan 03 2020
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