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A114112
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a(1)=1, a(2)=2; thereafter a(n) = n+1 if n odd, n-1 if n even.
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7
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1, 2, 4, 3, 6, 5, 8, 7, 10, 9, 12, 11, 14, 13, 16, 15, 18, 17, 20, 19, 22, 21, 24, 23, 26, 25, 28, 27, 30, 29, 32, 31, 34, 33, 36, 35, 38, 37, 40, 39, 42, 41, 44, 43, 46, 45, 48, 47, 50, 49, 52, 51, 54, 53, 56, 55, 58, 57, 60, 59, 62, 61, 64, 63, 66, 65, 68, 67, 70, 69, 72, 71
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OFFSET
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1,2
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COMMENTS
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a(1)=1; for n>1, a(n) is the smallest positive integer not occurring earlier in the sequence such that a(n) does not divide Sum_{k=1..n-1} a(k). - Leroy Quet, Nov 13 2005 (This was the original definition. A simple induction argument shows that this is the same as the present definition. - N. J. A. Sloane, Mar 12 2018)
Define b(1)=2; for n>1, b(n) is the smallest number not yet in the sequence which shares a prime factor with the sum of all preceding terms. Then a simple induction argument shows that the b(n) sequence is the same as the present sequence with the first term omitted. - David James Sycamore, Feb 26 2018
Here are the details of the two induction arguments (Start)
For a(n), let A(n) = a(1)+...+a(n). The claim is that for n>2 a(n)=n+1 if n odd, n-1 if n even.
The induction hypotheses are: for i<n, a(2i)=2i+1, a(2i)=2i-1, A(2i)=i(2i+1), A(2i) = 2i^2+3i+2. This implies that when looking for a(2i), we have seen all the numbers 1 through 2i except 2i-1, so the two smallest candidates for a(2i) are 2i-1 and 2i+1. Since 2i-1 does not divide 2i^2-i-1, a(n)=2i-1. When looking for a(2i+1), we have seen all the numbers 1 through 2i already, so the two smallest candidates for a(2i+1) are 2i+1 and 2i+2. But 2i+1 does divide A(2i) and 2i+2 does not, so a(2i+1)=2i+2. QED
For b(n), the argument is very similar, except that the missing numbers when looking for b(n) are slightly different, and (setting B(n) = b(1)+...b(n)), we have B(2i)=(i+1)(2i+1), B(2i+1)=(i+2)(2i+1). - N. J. A. Sloane, Mar 14 2018
When sequence a(n) is increasing, then the Cesàro means sequence c(n) = (a(1)+...+a(n))/n is also increasing, but the converse is false. This sequence is a such an example where c(n) is increasing, while a(n) is not increasing (Arnaudiès et al.). See proof in A354008. - Bernard Schott, May 11 2022
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REFERENCES
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J. M. Arnaudiès, P. Delezoide et H. Fraysse, Exercices résolus d'Analyse du cours de mathématiques - 2, Dunod, Exercice 10, pp. 14-16.
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LINKS
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FORMULA
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G.f.: x*(x^4-2*x^3+x^2+x+1)/((1-x)*(1-x^2)). - N. J. A. Sloane, Mar 12 2018
The g.f. for the b(n) sequence is x*(x^3-3*x^2+2*x+2)/(1-x)*(1-x^2)). - Conjectured (correctly) by Colin Barker, Mar 04 2018)
E.g.f.: 1 - x + x^2/2 + (x - 1)*cosh(x) + (x + 1)*sinh(x). - Stefano Spezia, Sep 02 2022
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MATHEMATICA
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a[1] = 1; a[n_] := a[n] = Block[{k = 1, s, t = Table[ a[i], {i, n - 1}]}, s = Plus @@ t; While[ Position[t, k] != {} || Mod[s, k] == 0, k++ ]; k]; Array[a, 72] (* Robert G. Wilson v, Nov 18 2005 *)
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PROG
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(PARI) a(n) = if (n<=2, n, if (n%2, n+1, n-1)); \\ Michel Marcus, May 16 2022
(Python)
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CROSSREFS
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See A084265 for the partial sums of the b(n) sequence.
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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