A293990 a(n) = (3*n + ((n-2) mod 4))/2.

1, 3, 3, 5, 7, 9, 9, 11, 13, 15, 15, 17, 19, 21, 21, 23, 25, 27, 27, 29, 31, 33, 33, 35, 37, 39, 39, 41, 43, 45, 45, 47, 49, 51, 51, 53, 55, 57, 57, 59, 61, 63, 63, 65, 67, 69, 69, 71, 73, 75, 75, 77, 79, 81, 81, 83, 85, 87, 87, 89, 91, 93, 93

Offset: 0

Author

Dimitris Valianatos, Oct 21 2017

Keywords

• nonn
• easy

Comments

The product (2/3) * (4/3) * (6/5) * (6/7) * (8/9) * (10/9) * (12/11) * (12/13) * ... = Pi/(2*sqrt(3)). The denominators are a(n) for n >= 1 and numerators are a(n-1) + A093148(n) for n >= 1 -> [2, 4, 6, 6, 8, 10, 12, 12, ...].

Let r(n) = (a(n)-1)/(a(n)+1) if a(n) mod 4 = 1, (a(n)+1)/(a(n)-1) otherwise; then Product_{n>=1} r(n) = (2/1) * (2/1) * (2/3) * (4/3) * (4/5) * (4/5) * (6/5) * (6/7) * ... = Pi*sqrt(3)/2 = 2.72069904635132677...

The odd numbers of partial sums this sequence, are identified with the A003215 sequence. Also the prime numbers that appear in partial sums in this sequence, are identified with the A002407 sequence.

Links

• G. C. Greubel, Table of n, a(n) for n = 0..5000
• Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Crossrefs

Cf. A007310, A047298, A063305, A093148, A134967, A157932, A162330, A168329, A214549, A285869, A003215, A002407.

Programs

Magma

[(3*n+((n-2) mod 4))/2 : n in [0..100]]; // Wesley Ivan Hurt, Oct 29 2017

Maple

A293990:=n->(3*n+((n-2) mod 4))/2: seq(A293990(n), n=0..100); # Wesley Ivan Hurt, Oct 29 2017

Mathematica

Table[(3*n + Mod[(n - 2), 4])/2, {n, 0, 100}] (* Wesley Ivan Hurt, Oct 29 2017 *)
f[n_] := (3n + Mod[n - 2, 4])/2; Array[f, 65, 0] (* or *)
LinearRecurrence[{1, 0, 0, 1, -1}, {1, 3, 3, 5, 7}, 65] (* or *)
CoefficientList[ Series[(x^4 + 2x^3 + 2x + 1)/((x - 1)^2 (x^3 + x^2 + x + 1)), {x, 0, 64}], x] (* Robert G. Wilson v, Nov 28 2017 *)

PARI

a(n) = (3*n + (n-2)%4) / 2

PARI

Vec(x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^30)) \\ Colin Barker, Oct 21 2017

PARI

first(n) = my(start=[1,3,3,5,7,9,9,11]); if(n<=8, return(start)); my(res=vector(n)); for (i=1, 8, res[i] = start[i]); for(i = 1, n-8 ,res[i+8] = res[i] + 12); res \\ David A. Corneth, Oct 21 2017

Formula

Sum_{n>=0} 1/a(n)^2 = 5*Pi^2/36 = 1.3707783890401886970... = 10*A086729.

(a(n) - n) * (-1)^(n+1) = A134967(n) for n >= 0.

a(n) - n = A162330(n) for n >= 0.

a(n) - n = A285869(n+1) for n >= 0.

a(n) + a(n+1) = A157932(n+2) for n >= 0.

a(n) + (2*n+1) = A047298(n+1) for n >= 0.

From Colin Barker, Oct 21 2017: (Start)

G.f.: x*(1 + 2*x + 2*x^3 + x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).

a(n) = a(n-1) + a(n-4) - a(n-5) for n > 5.

(End)

a(n + 8) = a(n) + 12. - David A. Corneth, Oct 21 2017

a(4*k+4) * a(4*k+3) - a(4*k+2) * a(4*k+1) = 2*A063305(k+3) for k >= 0.

Sum_{n>=0} 1/(a(n) + a(n+2))^2 = (4*Pi^2 - 27) / 108 = (A214549 - 1) / 4.