This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A330931 #24 Jan 05 2025 19:51:41 %S A330931 1,20,68,80,115,155,184,204,260,272,284,320,344,355,395,404,424,464, %T A330931 555,564,595,623,624,636,664,675,804,835,846,847,864,875,888,904,972, %U A330931 1028,1040,1075,1088,1124,1164,1182,1211,1224,1239,1266,1280,1304,1315,1424 %N A330931 Numbers k such that both k and k + 1 are Niven numbers in base 2 (A049445). %C A330931 Cai proved that there are infinitely many runs of 4 consecutive Niven numbers in base 2. Therefore this sequence is infinite. %D A330931 József Sándor and Borislav Crstici, Handbook of Number theory II, Kluwer Academic Publishers, 2004, Chapter 4, p. 382. %H A330931 Amiram Eldar, <a href="/A330931/b330931.txt">Table of n, a(n) for n = 1..10000</a> %H A330931 Tianxin Cai, <a href="https://web.archive.org/web/2024*/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 A330931 Wikipedia, <a href="https://en.wikipedia.org/wiki/Harshad_number">Harshad number</a>. %H A330931 Brad Wilson, <a href="https://web.archive.org/web/2024*/https://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 A330931 20 is a term since 20 and 20 + 1 = 21 are both Niven numbers in base 2. %t A330931 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 A330931 (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 A330931 (Python) %o A330931 def sbd(n): return sum(map(int, str(bin(n)[2:]))) %o A330931 def niv2(n): return n%sbd(n) == 0 %o A330931 def aupto(nn): return [k for k in range(1, nn+1) if niv2(k) and niv2(k+1)] %o A330931 print(aupto(1424)) # _Michael S. Branicky_, Jan 20 2021 %Y A330931 Cf. A049445, A328205, A328209, A328213, A330713, A330927, A330932, A330933. %K A330931 nonn,base %O A330931 1,2 %A A330931 _Amiram Eldar_, Jan 03 2020