A204454 Odd numbers not divisible by 11.

1, 3, 5, 7, 9, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 123, 125, 127, 129, 131

Offset

1,2

Comments

Up to a(45) this sequence coincides with A029740, but 101 is not in A029740.

This sequence is the fourth member of the family of sequences of odd numbers not divisible by a given odd prime p. For p = 3, 5, and 7 these sequences are A007310, A045572, and A162699, respectively. The formula is

a(p;n) = 2*n+1 + 2*floor((n-(p+1)/2)/(p-1)), n>=1, p an odd prime. If one puts a(p;0):=0, the o.g.f. is

G(p;x) = (x/((1-x^(p-1))*(1-x)))*(1 + 2*sum(x^k,k=1..(p-3)/2) + 4*x^((p-1)/2) + 2*sum(x^((p-1)/2+k),k=1..(p-3)/2) + x^(p-1)).

See the array A204456 with the coefficients of the numerator polynomials of these o.g.f.s.

This sequence gives also the numbers relatively prime to 2 and 11.

Another formula is a(p;n) = 2*n-1 + 2*floor(( n-(p-3)/2)/(p-1)), n>=1. From the rows of the array A204456 for the o.g.f. one can show first: a(p;n) = n + sum(floor((n+p-3-k)/(p-1)),k=1..(p-3)/2) + 3*floor((n+(p-3)/2)/(p-1)) + sum(floor((n+(p-3)/2-k)/(p-1)),k=1..(p-1)/2), p an odd prime, n>=1. - Wolfdieter Lang, Jan 26 2012

Recurrences for odd p: a(p;n) = a(p;n-(p-1)) + 2*p. For first differences: a(p;n) = a(p;n-1) + a(p;n-p+1) - a(p;n-p), n>=p, and inputs a(p;0):=-1 (here not 0) and a(p;k) for k=1,...,p-1. See the formula sections of the A-numbers for the instances p = 3, 5, and 7 for the contributions from Zak Seidov and R. J. Mathar. From this recurrence follows the o.g.f. (starting with x^1) directly. Above it has been found from the formula for a(p;n). Here the coefficients of the numerator polynomial of the o.g.f. (besides the 1s for x^1 and x^p) arise as first differences of the input members of the {a(p;n)} sequence. - Wolfdieter Lang, Jan 27 2012

Numbers coprime to 22. The asymptotic density of this sequence is 5/11. - Amiram Eldar, Oct 20 2020

Links

Table of n, a(n) for n=1..60.

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

Formula

a(n) = 2*n+1 + 2*floor((n-6)/10), n>=1. Note that this is -1 for n=0, but the following o.g.f. uses a(0)=0.

O.g.f: x*(1+2*x+2*x^2+2*x^3+2*x^4+4*x^5+2*x^6+2*x^7+2*x^8+2*x^9+x^10)/((1-x^10)*(1-x)). See the comment above for p=11.

a(n) = n + sum(floor((n+9-k)/10),k=1..4) + 3*floor((n+4)/10) + sum(floor((n+4-k)/10),k=1..5) = n + (n-1) + 2*floor((n+4)/10), n>=1. See the line m=5, p=11 of the array A204456, and the general formula given in a comment above. - Wolfdieter Lang, Jan 26 2012

Recurrences: a(n) = a(n-10) + 2*11. First differences: a(n) = a(n-1) + a(n-10) - a(n-11), n>=11, and inputs a(p;0):=-1 ( here not 0) and a(p;k) for k=1,...,10. See the general comment above. - Wolfdieter Lang, Jan 27 2012

Example

2*floor((n-6)/10), n>=0, is the sequence (the exponent of a number indicates how many times this number appears consecutively): (-2)^6 0^10 2^10 4^10 ... By adding these numbers to 2*n+1, n>=0, one obtains -1 for n=0 and a(n) for n>=1. The o.g.f is computed from this sum, but adjusted such that one obtains a vanishing a(0).
Recurrences: 31 = a(15) = a(5) + 2*11 = 9 + 22. a(15) = a(14) + a(5) - a(4) = 29 + 9 - 7 = 31. - Wolfdieter Lang, Jan 27 2012

Mathematica

Complement[Range[1, 131, 2], 11Range[11]] (* Alonso del Arte, Jan 24 2012 *)

Prog

(PARI) a(n)=2*n+(n-6)\10*2+1 \\ Charles R Greathouse IV, Jan 24 2012

Crossrefs

Cf. A007310, A045572, A162699, A204456, A204457, A204458.

Sequence in context: A063081 A337431 A067031 * A029740 A063425 A215806

Adjacent sequences: A204451 A204452 A204453 * A204455 A204456 A204457

Keyword

nonn,easy

Author

Wolfdieter Lang, Jan 24 2012

Status

approved