

A166711


Permutation of the integers: two positives, one negative.


4



0, 1, 2, 1, 3, 4, 2, 5, 6, 3, 7, 8, 4, 9, 10, 5, 11, 12, 6, 13, 14, 7, 15, 16, 8, 17, 18, 9, 19, 20, 10, 21, 22, 11, 23, 24, 12, 25, 26, 13, 27, 28, 14, 29, 30, 15, 31, 32, 16, 33, 34, 17, 35, 36, 18, 37, 38, 19, 39, 40, 20, 41, 42, 21, 43, 44, 22, 45, 46
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OFFSET

0,3


COMMENTS

Setting m=2 in
log(m) = Sum_{n>0} (n mod m  (n1) mod m)/n [1]
yields the sum
log(2) = (1 1/2) +(1/3 1/4) +(1/5 1/6)+...
Substituting every 1/d by 1/d  2/d we obtain
log(2) = (1+1/21)+(1/3+1/41/2)+(1/5+1/61/3)+...
a(n) is the sequence of denominators of this modified sum with unit numerators, so
Sum_{k>0} 1/a(k) = log(2)
Substituting 1/d by 2/d + 1/d would yield another permutation (one positive, one negative, one positive) with the same sum of inverses.
Similar sequences (m positives, one negative) may be obtained for the logarithm of any integer m>0. A001057 is the case m=1, with sum of inverses log(1).
Equation [1] is a result of expanding log( Sum_{0<=k<=m1} x^k ) at x=1 (see comment to A061347.)


LINKS

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,1).
Wikipedia, Riemann series theorem [From Jaume Oliver Lafont, Nov 01 2009]


FORMULA

G.f.: (x*(1+2*xx^2+x^3)/((1x)^2*(1+x+x^2)^2)).
a(0)=0, a(1)=1, a(2)=2, a(3)=1, a(4)=3, a(5)=4, a(n)=2*a(n3)a(n6), n>=6.
a(n) = (n+1)/3 +2*A049347(n)/3 (1)^n*A076118(n+1).  R. J. Mathar, Oct 30 2009


MATHEMATICA

LinearRecurrence[{0, 0, 2, 0, 0, 1}, {0, 1, 2, 1, 3, 4}, 100] (* G. C. Greubel, May 24 2016 *)


PROG

(PARI) a(n)=(2*(n+1)\3)*(13/2*!(n%3))
(PARI) a(n)=if(n>=0, [ n\3, 2*(n\3)+1, 2*(n\3)+2][n%3+1]) \\ Jaume Oliver Lafont, Nov 14 2009


CROSSREFS

Cf. A001057, A002162, A038608. Signed and shifted version of A009947.
Sequence in context: A117384 A125160 A009947 * A026249 A130527 A026366
Adjacent sequences: A166708 A166709 A166710 * A166712 A166713 A166714


KEYWORD

sign


AUTHOR

Jaume Oliver Lafont, Oct 18 2009


EXTENSIONS

Corrected by Jaume Oliver Lafont, Oct 22 2009
frac keyword removed by Jaume Oliver Lafont, Nov 02 2009


STATUS

approved



