A157932 -id:A157932 - cp's OEIS frontend

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

Dimitris Valianatos, Oct 21 2017

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.

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).

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

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

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

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
```

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

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

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.

A345211 Numbers with the same number of odd / even, refactorable divisors.

Original entry on oeis.org

2, 4, 6, 10, 14, 18, 20, 22, 26, 28, 30, 34, 38, 42, 44, 46, 50, 52, 54, 58, 62, 66, 68, 70, 74, 76, 78, 82, 86, 90, 92, 94, 98, 100, 102, 106, 110, 114, 116, 118, 122, 124, 126, 130, 134, 138, 140, 142, 146, 148, 150, 154, 158, 162, 164, 166, 170, 172, 174, 178, 182, 186, 188

Offset: 1

Wesley Ivan Hurt, Jun 10 2021

nonn

All the terms are even. Conjecture: The sequence a(n)/2 is the complement of A157932 (verified for a(n) < 4*10^7). - Amiram Eldar, Oct 11 2023

			18 is in the sequence since it has 2 odd and 2 even, refactorable divisors: 1, 2, 9, and 18.

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

Cf. A033950, A157932, A336041.

q[n_] := DivisorSum[n, (-1)^# &, Divisible[#, DivisorSigma[0, #]] &] == 0; Select[Range[2, 200, 2], q] (* Amiram Eldar, Oct 11 2023 *)

is(n) = (n%2)! && sumdiv(n, d, (-1)^d * !(d % numdiv(d))) == 0; \\ Amiram Eldar, Oct 11 2023

cp's OEIS Frontend

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

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A345211 Numbers with the same number of odd / even, refactorable divisors.

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