A364217 Numbers k such that k and k+1 are both Jacobsthal-Niven numbers (A364216).

1, 2, 3, 8, 11, 14, 15, 27, 32, 42, 43, 44, 45, 51, 56, 75, 86, 87, 92, 95, 99, 104, 125, 128, 135, 144, 155, 171, 176, 182, 183, 195, 204, 264, 267, 275, 287, 305, 344, 363, 375, 387, 428, 444, 455, 474, 497, 512, 524, 535, 544, 545, 552, 555, 581, 605, 623, 639

Offset: 1

Author

Amiram Eldar, Jul 14 2023

Keywords

• nonn
• base
• easy

Comments

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.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A364215.

Subsequence of A364216.

Subsequences: A364218, A364219, A364220, A364221.

Similar sequences: A330927, A328205, A328209, A328213, A330931, A331086, A333427, A334309, A331820, A342427, A344342, A351715, A351720, A352090, A352108, A352321, A352343, A352509.

Programs

Mathematica

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]

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, ", ")));}
lista(640, 2)