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A204454
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Odd numbers not divisible by 11.
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4
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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
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OFFSET
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1,2
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COMMENTS
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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
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LINKS
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Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,1,-1).
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FORMULA
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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
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EXAMPLE
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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
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MATHEMATICA
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Complement[Range[1, 131, 2], 11Range[11]] (* Alonso del Arte, Jan 24 2012 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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