A087279 Nonnegative numbers whose distance to the nearest positive square equals exactly 1.

0, 2, 3, 5, 8, 10, 15, 17, 24, 26, 35, 37, 48, 50, 63, 65, 80, 82, 99, 101, 120, 122, 143, 145, 168, 170, 195, 197, 224, 226, 255, 257, 288, 290, 323, 325, 360, 362, 399, 401, 440, 442, 483, 485, 528, 530, 575, 577, 624, 626, 675, 677, 728, 730, 783, 785, 840

Offset: 1

Author

Reinhard Zumkeller, Aug 28 2003

Keywords

• nonn
• easy

Comments

Union of A005563 and A002522\{1}: a(2*k+1) = (k+1)^2 - 1 = A005563(k); a(2*k) = k^2 + 1 = A002522(k); positive square + 1 or positive square - 1.

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Crossrefs

Cf. A000290, A004526, A000035, A087278.

Cf. A005563, A002522.

Programs

Haskell

a087279 n = a087279_list !! (n-1)
a087279_list = 0 : 2 : f (drop 2 a000290_list)
   where f (x:xs) = x-1 : x+1 : f xs
-- Reinhard Zumkeller, Nov 01 2013

Magma

&cat[[n^2-1,n^2+1]: n in [1..30]]; // Bruno Berselli, Apr 21 2011

Mathematica

Union[(r = Range[30]^2) - 1, r + 1] (* Jean-François Alcover, Oct 25 2013 *)
Flatten[#+{1,-1}&/@(Range[30]^2)]//Union (* Harvey P. Dale, Oct 15 2016 *)

PARI

a(n)=if(n%2,(n+1)^2/4-1,n^2/4+1) \\ Charles R Greathouse IV, Apr 25 2012

Python

def A087279(n): return ((n+(b:=n&1))**2>>2)+1-(b<<1) # Chai Wah Wu, Aug 03 2022

Formula

a(1) = 0; a(2*k+1) = a(2*k) + 2*k-1; a(2*k) = a(2*k-1) + 2.

a(n-1) = floor((n+1)/2)^2+(-1)^(n mod 2).

From Bruno Berselli, Apr 21 2011: (Start)

G.f.: x^2*(2+x-2*x^2+x^3)/((1+x)^2*(1-x)^3).

a(n) = (2*n*(n+1) - (2*n-7)*(-1)^n+1)/8. (End)

From Amiram Eldar, Sep 14 2022: (Start)

Sum_{n>=2} 1/a(n) = coth(Pi)*Pi/2 + 1/4.

Sum_{n>=2} (-1)^n/a(n) = coth(Pi)*Pi/2 - 5/4. (End)

Extensions

Franklin T. Adams-Watters pointed out on Jun 26 2007 that there were problems with the first couple of terms. I have made some changes, so now the definition matches the sequence. But some of the comments may need further minor adjustments. - N. J. A. Sloane, Jun 01 2008