A177002 -id:A177002 - cp's OEIS frontend

A227316 a(n) = n(n+1) if n == 0 or 1 (mod 4), otherwise a(n) = n(n+1)/2.

0, 2, 3, 6, 20, 30, 21, 28, 72, 90, 55, 66, 156, 182, 105, 120, 272, 306, 171, 190, 420, 462, 253, 276, 600, 650, 351, 378, 812, 870, 465, 496, 1056, 1122, 595, 630, 1332, 1406, 741, 780, 1640, 1722, 903, 946, 1980, 2070, 1081, 1128

Offset: 0

Paul Curtz, Jul 06 2013

			a(0) = 2*0 = 0, a(1) = 2*1 = 2, a(2) = 1*3 = 3, a(3) = 1*6 = 6, a(4) = 2*10 = 20.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A000217, A002378, A130658, A169642 (first bisection), A176743, A109043, A227380.

[(3+(-1)^Floor(n/2))*n*(n+1)/4: n in [0..50]]; // Bruno Berselli, Jul 10 2013

a[n_] := n*(n+1)/4*GCD[n-1, 4]*GCD[n, 4]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jul 10 2013 *)
Table[If[Mod[n,4]<2,n(n+1),(n(n+1))/2],{n,0,50}] (* or *) LinearRecurrence[ {3,-6,10,-12,12,-10,6,-3,1},{0,2,3,6,20,30,21,28,72},50] (* Harvey P. Dale, Aug 26 2016 *)

a(n) = A130658(n+2)*A000217(n), a(-n-1) = A130658(n)*A000217(n).
a(2n) = A169642(n), a(2n+1) = 2*(2*n+1)*A026741(n+1).
a(n) = A176743(n-2)*A176743(n-1).
a(n) = A177002(n+2)*A064038(n+1).
a(n) = 3*a(n-1) -6*a(n-2) +10*a(n-3) -12*a(n-4) +12*a(n-5) -10*(n-6) +6*(n-7) -3*a(n-8) +a(n-9) = 3*a(n-4) -3*a(n-8) +a(n-12).
G.f.: x*(2-3*x+9*x^2+3*x^5+x^6)/((1-x)^3*(1+x^2)^3). - Bruno Berselli, Jul 10 2013
a(n) = (3+(-1)^floor(n/2))*n*(n+1)/4. - Bruno Berselli, Jul 10 2013
Sum_{n>=1} 1/a(n) = 1 + log(2)/2. - Amiram Eldar, Aug 12 2022

A227168 a(n) = gcd(2n, n(n+1)/2)^2.

Original entry on oeis.org

1, 1, 36, 4, 25, 9, 196, 16, 81, 25, 484, 36, 169, 49, 900, 64, 289, 81, 1444, 100, 441, 121, 2116, 144, 625, 169, 2916, 196, 841, 225, 3844, 256, 1089, 289, 4900, 324, 1369, 361, 6084, 400

Offset: 1

Paul Curtz, Jul 03 2013

a(n) is defined as A062828(n)^2 for n >= 1. If we extend the sequence to n=0 and negative n by use of the recurrence that relates a(n) to a(n+12), a(n+8) and a(n+4), we obtain a(0)=0, a(-1)=4 and a(-n) = A176743(n-2)^2 for n >= 2.
Define c(n) = a(n+2) - a(n-2) for c >= 0. Because a(n) is a shuffle of three interleaved 2nd-order polynomials, c(n) is a shuffle of three interleaved 1st-order polynomials: c(n) = 4* A062828(n)*(periodically repeated 1, 8, 1, 1).
The sequence a(n) is case p=0 of the family A062828(n)*A062828(n+p):
0, 1, 1, 36,  4, 25,  9, 196, ... = a(n).
0, 1, 6, 12, 10, 15, 42,  56, ... = A130658(n)*A000217(n) = A177002(n-1)*A064038(n+1).
0, 6, 2, 30,  6, 70, 12, 126, ... = 2*A198148(n)
0, 2, 5, 18, 28, 20, 27,  70, ... = A177002(n+2)*A160050(n+1) = A014695(n+2)*A000096(n).

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A062828, A016814, A017138, A005565, A006752, A061037, A061038.

[GCD(2*n, n*(n+1)/2)^2: n in [1..50]]; // G. C. Greubel, Sep 20 2018

A227168 := proc(n)
    A062828(n)^2 ;
end proc: # R. J. Mathar, Jul 25 2013

a[n_] := GCD[2*n, n*(n + 1)/2]^2; Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Jul 03 2013 *)

a(n)=if(n%2, n*if(n%4>2, 2, 1), n/2)^2 \\ Charles R Greathouse IV, Jul 07 2013

a(n) = A062828(n)^2.
a(4n) = (4*n+1)^2; a(2n+1) = (n+1)^2; a(4n+2) = 4*(4*n+3)^2.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) * (period 4: repeat 4, 1, 1, 4) = A061038(n).
A005565(n-3) = a(n+1) * A061037(n). - Corrected by R. J. Mathar, Jul 25 2013
a(n) = A130658(n-1)^2 * A181318(n). - Corrected by R. J. Mathar, Aug 01 2013
G.f.: -x*(1 + x + 36*x^2 + 4*x^3 + 22*x^4 + 6*x^5 + 88*x^6 + 4*x^7 + 9*x^8 + x^9 + 4*x^10) / ( (x-1)^3*(1+x)^3*(x^2+1)^3 ). - R. J. Mathar, Jul 20 2013
Sum_{n>=1} 1/a(n) = 47*Pi^2/192 + 3*G/8, where G is Catalan's constant (A006752). - Amiram Eldar, Aug 21 2022

cp's OEIS Frontend

A227316 a(n) = n(n+1) if n == 0 or 1 (mod 4), otherwise a(n) = n(n+1)/2.

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A227168 a(n) = gcd(2n, n(n+1)/2)^2.

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cp's OEIS Frontend

A227316 a(n) = n(n+1) if n == 0 or 1 (mod 4), otherwise a(n) = n(n+1)/2.

Views

Author

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Examples

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Programs

Magma

Mathematica

Formula

A227168 a(n) = gcd(2*n, n*(n+1)/2)^2.

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Author

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Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A227168 a(n) = gcd(2n, n(n+1)/2)^2.