A062828 a(n) = gcd(2n, n(n+1)/2).

1, 1, 6, 2, 5, 3, 14, 4, 9, 5, 22, 6, 13, 7, 30, 8, 17, 9, 38, 10, 21, 11, 46, 12, 25, 13, 54, 14, 29, 15, 62, 16, 33, 17, 70, 18, 37, 19, 78, 20, 41, 21, 86, 22, 45, 23, 94, 24, 49, 25, 102, 26, 53, 27, 110, 28, 57, 29, 118, 30, 61, 31, 126, 32, 65, 33, 134, 34, 69, 35, 142

Offset

1,3

Comments

Same as A123168 (without the initial 0) interleaved with the natural numbers. - Wesley Ivan Hurt, Apr 01 2022

Links

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Formula

a(4n+1) = 4n+1, a(4n+2) = 2n+1, a(4n+3) = 8n+6, a(4n+4) = 2n+2. - Ralf Stephan, Jun 10 2005

G.f.: x*(1 + x + 6*x^2 + 2*x^3 + 3*x^4 + x^5 + 2*x^6) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ). - R. J. Mathar, Jul 25 2013

From Wesley Ivan Hurt, Apr 01 2022: (Start)

a(n) = n*(2-(-1)^n-sin(n*Pi/2))/2.

a(n) = 2*a(n-4) - a(n-8). (End)

Maple

A062828 := proc(n)
    igcd(2*n, n*(n+1)/2) ;
end proc: # R. J. Mathar, Jul 25 2013

Mathematica

Table[GCD[2n, (n(n+1))/2], {n, 120}] (* or *) LinearRecurrence[ {0, 0, 0, 2, 0, 0, 0, -1}, {1, 1, 6, 2, 5, 3, 14, 4}, 120] (* Harvey P. Dale, Apr 09 2018 *)

Prog

(PARI) j=[]; for(n=1, 150, j=concat(j, gcd(2*n, n*(n+1)/2))); j
(PARI) a(n)=if(n%2, n*if(n%4>2, 2, 1), n/2) \\ Charles R Greathouse IV, Jul 07 2013

Crossrefs

Cf. A123168.

Sequence in context: A007320 A199734 A007321 * A124457 A258102 A309449

Adjacent sequences: A062825 A062826 A062827 * A062829 A062830 A062831

Keyword

easy,nonn

Author

Jason Earls, Jul 20 2001

Status

approved