A166711 Permutation of the integers: two positives, one negative.

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

Offset: 0

Author

Jaume Oliver Lafont, Oct 18 2009

Keywords

• sign
• easy

Comments

Setting m=2 in

log(m) = Sum_{n>0} (n mod m - (n-1) 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+1/2-1)+(1/3+1/4-1/2)+(1/5+1/6-1/3)+...

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<=m-1} x^k ) at x=1 (see comment to A061347.)

Links

• G. C. Greubel, Table of n, a(n) for n = 0..10000
• Wikipedia, Riemann series theorem
• Index entries for linear recurrences with constant coefficients, signature (0,0,2,0,0,-1).

Crossrefs

Cf. A001057, A002162, A038608. Signed and shifted version of A009947.

Programs

Mathematica

LinearRecurrence[{0, 0, 2, 0, 0, -1}, {0, 1, 2, -1, 3, 4}, 100] (* G. C. Greubel, May 24 2016 *)
Join[{0},With[{nn=50},Riffle[Range[nn],Range[-1,-nn/2,-1],3]]] (* Harvey P. Dale, May 15 2023 *)

PARI

a(n)=(2*(n+1)\3)*(1-3/2*!(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

Formula

G.f.: (x*(1+2*x-x^2+x^3)/((1-x)^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(n-3)-a(n-6), n>=6.

a(n) = (n+1)/3 +2*A049347(n)/3 -(-1)^n*A076118(n+1). - R. J. Mathar, Oct 30 2009

Extensions

Corrected by Jaume Oliver Lafont, Oct 22 2009

frac keyword removed by Jaume Oliver Lafont, Nov 02 2009