A347365 a(n) = n * (2-(-1)^n), or zero together with first differences of even triangular numbers halved (A074378).

0, 3, 2, 9, 4, 15, 6, 21, 8, 27, 10, 33, 12, 39, 14, 45, 16, 51, 18, 57, 20, 63, 22, 69, 24, 75, 26, 81, 28, 87, 30, 93, 32, 99, 34, 105, 36, 111, 38, 117, 40, 123, 42, 129, 44, 135, 46, 141, 48, 147, 50, 153, 52, 159, 54, 165, 56, 171, 58, 177, 60, 183, 62, 189, 64

Offset

0,2

Comments

This sequence and A165998 form a complementary pair as solutions of alternating sequences a(n) + b(n) = 4*n (A008586), and a(n)*b(n) = 3*n^2 (A033428).

This is the particular case of the two integer sequences x(n)=2n and y(n)=n, where more generally, x(n) + y(n) = 2*a(n) and x(n)*y(n) = (a(n) + b(n)) * (a(n) - b(n)) give the two conjugate binomials a(n) = x(n) + (-1)^n*y(n) and b(n) = x(n) - (-1)^n*y(n) as solutions over the integer domain.

a(n) is also A005843 and A016945 interleaved.

For every integer k: a(n*k) = n*k is multiplicative for nonnegative even integers n and a(n*k) = n*a(k) for nonnegative odd integers n.

For every nonnegative odd integer k, the k-th difference of a(k*n)/k = (2n+1)*(-1)^n + 2 = A166519(n), and 1 for all nonnegative even integers.

a(6n+1)/3 = 6n+1, and a(6n+5)/3 = 6n+5, related to Collatz Conjecture.

Half-periods of a(k) mod n is A026741(n).

Links

Table of n, a(n) for n=0..64.

Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Formula

G.f.: x*(3 + 2*x + 3*x^2)/(1 - x^2)^2.

E.g.f.: x*(3*cosh(x) + sinh(x)).

Dirichlet g.f.: 2^(-s) * (3*2^s - 4) * zeta(s-1) = (3 - 4/2^s) * zeta(s-1) = (3 - 1/2^(s-2)) * zeta(s-1).

a(n) = n*(2-(-1)^n) = 3*n / (2+(-1)^n).

a(n) = 3*n if n odd, a(n) = n if n even, implies a(a(2n)) = 2n, a(a(2n+1)) = 9*a(2n+1).

a(n) = 3*b(n), if n odd and a(n) = b(n)/3, if n even, with b(n) = A165998(n).

a(n) = a(a(2k*n)/(2k)) = a((2k+1)*n) / (2k+1), since a(2*k*n) / (2*k) = n.

a(n) = 4*n - A165998(n).

a(n+1) = a(n) + A086970(n+1)*(-1)^n.

a(n) = 2*A014682(n) - A000035(n).

a(n) = n*A010684(n). - Michel Marcus, Sep 13 2021

For positive integers k and n, a(n) = A(n,1) = n * (A(n,k)/n)^(1/k), where the k-th nesting composition A(n,k) = a(a(...a(a(n))...)) = n * ( a(n) / n )^k, and d.g.f. of A(n,k) = (2^(1-s) + (1-2^(1-s))*3^k) * zeta(s-1). - Federico Provvedi, Sep 18 2021

a(n+1) = A165998(n)*(1 + 1/n). - Federico Provvedi, Sep 19 2021

Mathematica

Table[n(2-(-1)^n), {n, 0, 99}] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {0, 3, 2, 9}, 100] (* or *)
If[EvenQ@#, #, 3#]&/@Range[0, 99]  (* or *)
Drop[Flatten@Transpose[{2#, 6#+3}&@Range[0, Quotient[#, 2]]], -Boole@EvenQ@#]&@(10^2)

Prog

(Sage) (x*(3+2*x+3*x^2)/(1-x^2)^2).series(x, 65).coefficients(x, sparse=False) # Stefano Spezia, Aug 30 2021
(PARI) a(n) = n*(2-(-1)^n); \\ Michel Marcus, Sep 13 2021

Crossrefs

Cf. A074378, A165998, A008586, A033428, A237420, A193356, A016921, A016969, A166519, A014682, A005843, A016945, A010684, A008585, A000027, A026741.

Sequence in context: A320273 A266636 A182652 * A251555 A090880 A258439

Adjacent sequences: A347362 A347363 A347364 * A347366 A347367 A347368

Keyword

nonn,easy

Author

Federico Provvedi, Aug 29 2021

Status

approved