A123737 Partial sums of (-1)^floor(n*sqrt(2)).

-1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 1, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, -2, -1, -2, -3, -2, -1, -2, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0, 1, 2, 1, 0, 1, 0, -1, 0, 1, 0, -1, 0, -1, -2, -1, 0, -1, 0, 1, 0, -1, 0, 1, 0

Offset: 1

Author

T. D. Noe, Oct 11 2006

Keywords

• easy
• sign

Comments

Conjecture: A001652(n) is the index of the first occurrence of n in sequence A123737, A001108(n) is the index of the first occurrence of -n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015

Links

• T. D. Noe, Table of n, a(n) for n = 1..10000
• Henk Bruin and Robbert Fokkink, On the records and zeros of a deterministic random walk, arXiv:2503.11734 [math.DS], 2025. See p. 2.
• Kevin O'Bryant, Bruce Reznick and Monika Serbinowska, Almost alternating sums, arXiv:math/0308087 [math.NT], 2003-2205; Amer. Math. Monthly, Vol. 113 (October 2006), pp. 673-688.

Crossrefs

Cf. A123724 (sum for 2^(1/3)), A123738 (sum for Pi), A123739 (sum for e).

Cf. A001652, A001951, A228639.

Programs

Magma

[&+[(-1)^Floor(j*Sqrt(2)): j in [1..n]]: n in [1..130]]; // G. C. Greubel, Sep 05 2019

Maple

ListTools:-PartialSums([seq((-1)^floor(n*sqrt(2)),n=1..100)]); # Robert Israel, Jun 02 2015

Mathematica

Rest[FoldList[Plus,0,(-1)^Floor[Sqrt[2]*Range[120]]]]
Accumulate[(-1)^Floor[Range[120]Sqrt[2]]] (* Harvey P. Dale, Jan 16 2012 *)
(* The positions of the first occurrences of n and -n in this sequence: *) stab = Rest[FoldList[Plus,0,(-1)^Floor[Sqrt[2]*Range[1000000]]]]; Print[Table[FirstPosition[stab,n][[1]],{n,1,8}]]; Print[Table[FirstPosition[stab,-n][[1]],{n,1,8}]]; (* Vaclav Kotesovec, Jun 02 2015 *)

PARI

a(n)=sum(i=1,n,(-1)^sqrtint(2*i^2)) \\ Charles R Greathouse IV, Feb 07 2013

Sage

[sum((-1)^floor(j*sqrt(2)) for j in (1..n)) for n in (1..130)] # G. C. Greubel, Sep 05 2019

Formula

O'Bryant, Reznick, & Serbinowska show that |a(n)| <= k log n + 1, with k = 1/(2 log (1 + sqrt(2))), and further -a(n) > k log n + 0.78 infinitely often. - Charles R Greathouse IV, Feb 07 2013