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%I #8 Jul 15 2023 05:53:16
%S 1,2,3,8,11,14,15,27,32,42,43,44,45,51,56,75,86,87,92,95,99,104,125,
%T 128,135,144,155,171,176,182,183,195,204,264,267,275,287,305,344,363,
%U 375,387,428,444,455,474,497,512,524,535,544,545,552,555,581,605,623,639
%N Numbers k such that k and k+1 are both Jacobsthal-Niven numbers (A364216).
%C A001045(2*n+1) = A007583(n) = (2^(2*n+1) + 1)/3 is a term for n >= 0, since its representation is 2*n 1's, so A364215(A001045(2*n+1)) = 1 divides A001045(2*n+1), and the representation of A001045(2*n+1) + 1 = (2^(2*n+1) + 4)/3 is max(2*n-1, 0) 0's between 2 1's, so A364215(A001045(2*n+1) + 1) = 2 which divides (2^(2*n+1) + 4)/3.
%H Amiram Eldar, <a href="/A364217/b364217.txt">Table of n, a(n) for n = 1..10000</a>
%t consecJacobsthalNiven[kmax_, len_] := Module[{m = 1, c = Table[False, {len}], s = {}}, Do[c = Join[Rest[c], {Divisible[k, DigitCount[m, 2, 1]]}]; While[m++; OddQ[IntegerExponent[m, 2]]]; If[And @@ c, AppendTo[s, k - len + 1]], {k, 1, kmax}]; s]; consecJacobsthalNiven[640, 2]
%o (PARI) lista(kmax, len) = {my(m = 1, c = vector(len)); for(k = 1, kmax, c = concat(vecextract(c, "^1"), !(k % sumdigits(m, 2))); until(valuation(m, 2)%2 == 0, m++); if(vecsum(c) == len, print1(k-len+1, ", ")));}
%o lista(640, 2)
%Y Cf. A364215.
%Y Subsequence of A364216.
%Y Subsequences: A364218, A364219, A364220, A364221.
%Y Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720, A352090, A352108, A352321, A352343, A352509.
%K nonn,base,easy
%O 1,2
%A _Amiram Eldar_, Jul 14 2023
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