A017138 -id:A017138 - cp's OEIS frontend

A226008 a(0) = 0; for n>0, a(n) = denominator(1/4 - 4/n^2).

0, 4, 4, 36, 1, 100, 36, 196, 16, 324, 100, 484, 9, 676, 196, 900, 64, 1156, 324, 1444, 25, 1764, 484, 2116, 144, 2500, 676, 2916, 49, 3364, 900, 3844, 256, 4356, 1156, 4900, 81, 5476, 1444, 6084, 400, 6724, 1764, 7396, 121, 8100

Offset: 0

Paul Curtz, May 22 2013

Numerators are in A225948.

Repeated terms of A016826 are in the positions 1, 2, 3, 6, 5, 10, ... (A043547).

			a(0) = (-1+1)^2 = 0, a(1) = (-3+5)^2 = 4, a(2) = (-1+3)^2 = 4.

Cf. A016754, A016802, A016826, A017090, A017138, A154615, A225948 (numerators).

Cf. A225975 (associated square roots).

[0] cat [Denominator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 23 2013

Join[{0},Table[Denominator[1/4 - 4/n^2], {n, 49}]] (* Alonso del Arte, May 22 2013 *)

a(n)    = 3*a(n-8) -3*a(n-16) +a(n-24).
a(8n)   = A016802(n), a(8n+4) = A016754(n).
a(4n)   = A154615(n).
a(4n+1) = A017090(n).
a(4n+2) = a(2n+1) = A016826(n); a(2n) = A061038(n).
a(4n+3) = A017138(n).
From Bruno Berselli, May 23 2013: (Start)
G.f.: x*(4 +4*x +36*x^2 +x^3 +100*x^4 +36*x^5 +196*x^6 +16*x^7 +312*x^8 +88*x^9 +376*x^10 +6*x^11 +376*x^12 +88*x^13 +312*x^14 +16*x^15 +196*x^16 +36*x^17 +100*x^18 +x^19 +36*x^20 +4*x^21 +4*x^22)/(1-x^8)^3.
a(n) = n^2*(6*cos(3*Pi*n/4)+6*cos(Pi*n/4)-54*cos(Pi*n/2)-219*(-1)^n+293)/128.
a(n+9) = a(n+1)*((n+9)/(n+1))^2. (End)
Sum_{n>=1} 1/a(n) = 19*Pi^2/96. - Amiram Eldar, Aug 14 2022

Edited by Bruno Berselli, May 23 2013

A017139 a(n) = (8*n + 6)^3.

Original entry on oeis.org

216, 2744, 10648, 27000, 54872, 97336, 157464, 238328, 343000, 474552, 636056, 830584, 1061208, 1331000, 1643032, 2000376, 2406104, 2863288, 3375000, 3944312, 4574296, 5268024, 6028568, 6859000, 7762392, 8741816, 9800344, 10941048, 12167000, 13481272, 14886936

Offset: 0

N. J. A. Sloane

4*n + 3 = (8*n + 6) / 2 is never a square, as 3 is not a quadratic residue modulo 4. Using this, we can show that each term has an even square part and an even squarefree part, neither part being a power of 2. (Less than 2% of integers have this property - see A339245.) - Peter Munn, Dec 14 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Quadratic Residue.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

A000578, A016839, A017137 are used in a formula defining this sequence.
Subsequence of A339245.
Cf. A002117, A017138, A017140.

[(8*n+6)^3: n in [0..35]]; // Vincenzo Librandi, Jul 22 2011

Table[(8*n+6)^3,{n,0,5!}] (* Vladimir Joseph Stephan Orlovsky, Mar 17 2010 *)
LinearRecurrence[{4,-6,4,-1},{216,2744,10648,27000},30] (* Harvey P. Dale, Dec 11 2012 *)

From R. J. Mathar, Mar 22 2010: (Start)
G.f.: 8*(27 + 235*x + 121*x^2 + x^3)/(x-1)^4.
a(n) = 8*A016839(n). (End)
a(0)=216, a(1)=2744, a(2)=10648, a(3)=27000, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 11 2012
a(n) = A017137(n)^3 = A000578(A017137(n)). - Peter Munn, Dec 20 2020
Sum_{n>=0} 1/a(n) = 7*zeta(3)/128 - Pi^2/512. - Amiram Eldar, Apr 26 2023

More terms from Vladimir Joseph Stephan Orlovsky, Mar 17 2010

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

A226008 a(0) = 0; for n>0, a(n) = denominator(1/4 - 4/n^2).

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A017139 a(n) = (8*n + 6)^3.

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

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cp's OEIS Frontend

A226008 a(0) = 0; for n>0, a(n) = denominator(1/4 - 4/n^2).

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Magma

Mathematica

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Extensions

A017139 a(n) = (8*n + 6)^3.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Links

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Magma

Maple

Mathematica

PARI

Formula

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