A348299 Indices of 0 in A348295: numbers m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = 0.

0, 4, 24, 28, 140, 144, 164, 168, 816, 820, 840, 844, 956, 960, 980, 984, 4756, 4760, 4780, 4784, 4896, 4900, 4920, 4924, 5572, 5576, 5596, 5600, 5712, 5716, 5736, 5740, 27720, 27724, 27744, 27748, 27860, 27864, 27884, 27888, 28536, 28540, 28560, 28564, 28676, 28680, 28700, 28704

Offset: 1

Author

Jianing Song, Oct 10 2021

Keywords

• nonn

Comments

Conjecture: sequence is infinite.

Conjecture: all terms are divisible by 4. There are no counterexamples <= 10^9.

Examples

			24 is a term: A097508(k) is even for k = 1, 2, 5, 6, 7, 10, 11, 12, 15, 16, 20, 21 and odd for k = 3, 4, 8, 9, 13, 14, 17, 18, 19, 22, 23, 24, so Sum_{k=1..24} (-1)^A097508(k) = 0.
		

Links

• Jianing Song, Table of n, a(n) for n = 1..2048 (all terms <= 10^9)

Crossrefs

Cf. A348295, A097508, A001109.

Programs

Mathematica

Join[{0}, Position[Accumulate@ Table[(-1)^Floor[k*(Sqrt[2] - 1)], {k, 1, 30000}], 0] // Flatten] (* Amiram Eldar, Oct 11 2021 *)

PARI

list(lim) = my(Sum=-1, v=[]); for(k=0, lim, Sum+=(-1)^(sqrtint(2*k^2)-k); if(Sum==0, v=concat(v, k))); v

Formula

Conjecture: for n >= 2, a(2n-1) = ceiling(a(n) * (3+2*sqrt(2))), a(2n) = a(2n-1) + 4. This is correct for the first 2048 terms (all terms <= 10^9).

Conjectured explicit formula: if the binary expansion of n-1 is n-1 = 2^(e_0) + 2^(e_1) + ... + 2^(e_k), then a(n) = 4*(A001109(1+(e_0)) + A001109(1+(e_1)) + ... + A001109(1+(e_k))). For example, since 28-1 = 27 = 2^0 + 2^1 + 2^3 + 2^4, a(28) = 4*(A001109(1) + A001109(2) + A001109(4) + A001109(5)) = 5600.