

A114112


a(1)=1, a(2)=2; thereafter a(n) = n+1 if n odd, n1 if n even.


2



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

1,2


COMMENTS

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 to n1} 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, n1 if n even.
The induction hypotheses are: for i<n, a(2i)=2i+1, a(2i)=2i1, 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 2i1, so the two smallest candidates for a(2i) are 2i1 and 2i+1. Since 2i1 does not divide 2i^2i1, a(n)=2i1. 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


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..2000


FORMULA

G.f.: x*(x^42*x^3+x^2+x+1)/((1x)*(1x^2)).  N. J. A. Sloane, Mar 12 2018
The g.f. for the b(n) sequence is x*(x^33*x^2+2*x+2)/(1x)*(1x^2)).  Conjectured (correctly) by Colin Barker, Mar 04 2018)


MATHEMATICA

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 *)


CROSSREFS

All of A014681, A103889, A113981, A114112, A114285 are essentially the same sequence.  N. J. A. Sloane, Mar 12 2018
Cf. A114113 (partial sums).
See A084265 for the partial sums of the b(n) sequence.
Sequence in context: A110458 A217559 A217560 * A113981 A234519 A289726
Adjacent sequences: A114109 A114110 A114111 * A114113 A114114 A114115


KEYWORD

easy,nonn


AUTHOR

Leroy Quet, Nov 13 2005


EXTENSIONS

More terms from Robert G. Wilson v, Nov 18 2005
Entry edited (with simpler definition) by N. J. A. Sloane, Mar 12 2018


STATUS

approved



