A168077 -id:A168077 - cp's OEIS frontend

A181318 a(n) = A060819(n)^2.

0, 1, 1, 9, 1, 25, 9, 49, 4, 81, 25, 121, 9, 169, 49, 225, 16, 289, 81, 361, 25, 441, 121, 529, 36, 625, 169, 729, 49, 841, 225, 961, 64, 1089, 289, 1225, 81, 1369, 361, 1521, 100, 1681, 441, 1849, 121, 2025, 529, 2209, 144, 2401, 625, 2601, 169, 2809, 729

Offset: 0

Paul Curtz, Jan 26 2011

The first sequence, p=0, of the family A060819(n)*A060819(n+p).
Hence array
p=0:  0, 1, 1,  9,  1, 25,  9,  49,    a(n)=A060819(n)^2,
p=1:  0, 1, 3,  3,  5, 15, 21,  14,     A064038(n),
p=2:  0, 3, 1, 15,  3, 35,  6,  63,     A198148(n),
p=3:  0, 1, 5,  9,  7, 10, 27,  35,     A160050(n),
p=4:  0, 5, 3, 21,  2, 45, 15,  77,     A061037(n),
p=5:  0, 3, 7,  6,  9, 25, 33,  21,     A178242(n),
p=6:  0, 7, 2, 27,  5, 55,  9,  91,     A217366(n),
p=7:  0, 2, 9, 15, 11, 15, 39,  49,     A217367(n),
p=8:  0, 9, 5, 33,  3, 65, 21, 105,     A180082(n).
Compare columns 2, 3 and 5, columns 4 and 7 and columns 6 and 8.
From Peter Bala, Feb 19 2019: (Start)
We make some general remarks about the sequence a(n) = numerator(n^2/(n^2 + k^2)) = (n/gcd(n,k))^2 for k a fixed positive integer (we suppress the dependence of a(n) on k). The present sequence corresponds to the case k = 4.
a(n) is a quasi-polynomial in n. In fact, a(n) = n^2/b(n) where b(n) = gcd(n^2,k^2) is a purely periodic sequence in n.
In addition to being multiplicative these sequences are also strong divisibility sequences, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. In particular, it follows that a(n) is a divisibility sequence: if n divides m then a(n) divides a(m).
By the multiplicativeness and strong divisibility property of the sequence a(n) it follows that if gcd(n,m) = 1 then a( a(n)*a(m) ) = a(a(n)) * a(a(m)), a( a(a(n))*a(a(m)) ) = a(a(a(n))) * a(a(a(m))) and so on.
The sequence a(n) has the rational generating function Sum_{d divides k} f(d)*F(x^d), where F(x) = x*(1 + x)/(1 - x)^3 = x + 4*x^2 + 9*x^3 + 16*x^4 + ... is the o.g.f. for the squares A000290, and where f(n) is the Dirichlet inverse of the Jordan totient function J_2(n) - see A007434. The function f(n) is multiplicative and is defined on prime powers p^k by f(p^k) = (1 - p^2). See A046970. Cf. A060819. (End)
a(n-4) is the constant needed to complete the n-polygonal numbers into squares (see A377851); a(-1) = 1, which completes the triangle numbers, is not shown in the data. - Jonathan Dushoff, Nov 12 2024

G. C. Greubel, Table of n, a(n) for n = 0..10000
Wikipedia, Completing the square.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A181829, A046970, A007434, A060819, A168077, A377851.

[n^2/GCD(n,4)^2: n in [0..100]]; // G. C. Greubel, Sep 19 2018

a:=n->n^2/gcd(n,4)^2: seq(a(n),n=0..60); # Muniru A Asiru, Feb 20 2019

Table[n^2/GCD[n,4]^2, {n,0,100}] (* G. C. Greubel, Sep 19 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,1,1,9,1,25,9,49,4,81,25,121},60] (* Harvey P. Dale, Jan 18 2025 *)

a(n)=n^2/gcd(n,4)^2 \\ Charles R Greathouse IV, Dec 21 2011

[n^2/gcd(n, 4)^2 for n in (0..100)] # G. C. Greubel, Feb 20 2019

a(2*n) = A168077(n), a(2*n+1) = A016754(n).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
G.f.: x*(1 + x + 9*x^2 + x^3 + 22*x^4 + 6*x^5 + 22*x^6 + x^7 + 9*x^8 + x^9 + x^10)/(1-x^4)^3. - R. J. Mathar, Mar 10 2011
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 16)) = n^2/(gcd(n^2,16)) = (n/gcd(n,4))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 16, 1, 4, 1, 16, ...] is a purely periodic sequence of period 4.
a(n) is a quasi-polynomial in n: a(4*n) = n^2; a(4*n + 1) = (4*n + 1)^2; a(4*n + 2) = (2*n + 1)^2; a(4*n + 3) = (4*n + 3)^2.
O.g.f.: Sum_{d divides 4} A046970(d)*x^d*(1 + x^d)/(1 - x^d)^3 = x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3 - 3*x^4*(1 + x^4)/(1 - x^4)^3. (End)
Sum_{n>=1} 1/a(n) = 5*Pi^2/12. - Amiram Eldar, Aug 12 2022
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 4^max(0, e-2), and a(p^e) = p^(2*e) for p > 2.
Dirichlet g.f.: zeta(s-2)*(1 - 3/2^s - 3/4^s).
Sum_{k=1..n} a(k) ~ (37/192) * n^3. (End)
a(n) = (37 - 27*(-1)^n - 3*(-1)^(n*(n-1)/2) - 3*(-1)^(n*(n+1)/2)) * n^2/64. - Vaclav Kotesovec, Nov 14 2024

Edited by Jean-François Alcover, Oct 01 2012 and Jan 15 2013

More terms from Michel Marcus, Jun 09 2014

A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

Original entry on oeis.org

1, 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 845, 226, 965, 257, 1093, 290, 1229, 325, 1373, 362, 1525, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677

Offset: 0

Paul Curtz, Aug 15 2015

Using (n+sqrt(4+n^2))/2, after the integer 1 for n=0, the reduced metallic means are b(1) = (1+sqrt(5))/2, b(2) = 1+sqrt(2), b(3) = (3+sqrt(13))/2, b(4) = 2+sqrt(5), b(5) = (5+sqrt(29))/2, b(6) = 3+sqrt(10), b(7) = (7+sqrt(53))/2, b(8) = 4+sqrt(17), b(9) = (9+sqrt(85))/2, b(10) = 5+sqrt(26), b(11) = (11+sqrt(125))/2 = (11+5*sqrt(5))/2, ... . The last value yields the radicals in a(n) or A013946.
b(2) = 2.41, b(3) = 3.30, b(4) = 4.24, b(5) = 5.19 are "good" approximations of fractal dimensions corresponding to dimensions 3, 4, 5, 6: 2.48, 3.38, 4.33 and 5.45 based on models. See "Arbres DLA dans les espaces de dimension supérieure: la théorie des peaux entropiques" in Queiros-Condé et al. link. DLA: beginning of the title of the Witten et al. link.
Consider the symmetric array of the half extended Rydberg-Ritz spectrum of the hydrogen atom:
   0,    1/0,     1/0,     1/0,     1/0,      1/0,      1/0,     1/0, ...
-1/0,      0,     3/4,     8/9,   15/16,    24/25,    35/36,   48/49, ...
-1/0,   -3/4,       0,    5/36,    3/16,   21/100,      2/9,  45/196, ...
-1/0,   -8/9,   -5/36,       0,   7/144,   16/225,     1/12,  40/441, ...
-1/0, -15/16,   -3/16,  -7/144,       0,    9/400,    5/144,  33/784, ...
-1/0, -24/25, -21/100, -16/225,  -9/400,        0,   11/900, 24/1225, ...
-1/0, -35/36,    -2/9,   -1/12,  -5/144,  -11/900,        0, 13/1764, ...
-1/0, -48/49, -45/196, -40/441, -33/784, -24/1225, -13/1764,       0, ... .
The numerators are almost A165795(n).
Successive rows: A000007(n)/A057427(n), A005563(n-1)/A000290(n), A061037(n)/A061038(n), A061039(n)/A061040(n), A061041(n)/A061042(n), A061043(n)/A061044(n), A061045(n)/A061046(n), A061047(n)/A061048(n), A061049(n)/A061050(n).
A144433(n) or A195161(n+1) are the numerators of the second upper diagonal (denominators: A171522(n)).
c(n+1) = a(n) + a(n+1) = 6, 7, 15, 18, 34, 39, 63, 70, 102, 111, ... .
c(n+3) - c(n+1) = 9, 11, 19, 21, 29, 31, ... = A090771(n+2).
The final digit of a(n) is neither 4 nor 8. - Paul Curtz, Jan 30 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Diogo Queiros-Condé, Jean Chaline, and Jacques Dubois, Le monde des fractales La Nature trans-échelles, 478p., ellipses, Paris, 2015, page 220.
T. A. Witten, Jr. and L. M. Sander, Diffusion-Limited Aggregation, a Kinetic Critical Phenomenom, Phys. Rev. Lett., Vol. 47 (Nov 09 1981), pp. 1400-1403.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000007, A000290, A001622, A002522, A005563, A010685, A010698, A013946, A014176, A057427, A061035-A061050, A078370, A087475, A090771, A098316, A098317, A098318, A144433, A168077, A171621, A176398, A176439, A176458, A176522, A176537, A195161.

Cf. A069894.

[Numerator(1+n^2/4): n in [0..60]]; // Vincenzo Librandi, Aug 15 2015

A261327:=n->numer((4 + n^2)/4); seq(A261327(n), n=0..60); # Wesley Ivan Hurt, Aug 15 2015

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 60] (* Vincenzo Librandi, Aug 15 2015 *)
a[n_] := (n^2 + 4) / 4^Mod[n + 1, 2]; Table[a[n], {n, 0, 52}] (* Peter Luschny, Mar 18 2022 *)

vector(60, n, n--; numerator(1+n^2/4)) \\ Michel Marcus, Aug 15 2015

Vec((1+5*x-x^2-2*x^3+2*x^4+5*x^5)/(1-x^2)^3 + O(x^60)) \\ Colin Barker, Aug 15 2015

a(n)=if(n%2,n^2+4,(n/2)^2+1) \\ Charles R Greathouse IV, Oct 16 2015

[(n*n+4)//4**((n+1)%2) for n in range(60)] # Gennady Eremin, Mar 18 2022

[numerator(1+n^2/4) for n in (0..60)] # G. C. Greubel, Feb 09 2019

a(n) = numerator(1 + n^2/4). (Previous name.) See A010685 (denominators).
a(2*k) = 1 + k^2.
a(2*k+1) = 5 + 4*k*(k+1).
a(2*k+1) = 4*a(2*k) + 4*k + 1.
a(4*k+2) = A069894(k). - Paul Curtz, Jan 30 2019
a(-n) = a(n).
a(n+2) = a(n) + A144433(n) (or A195161(n+1)).
a(n) = A168077(n) + period 2: repeat 1, 4.
a(n) = A171621(n) + period 2: repeat 2, 8.
From Colin Barker, Aug 15 2015: (Start)
a(n) = (5 - 3*(-1)^n)*(4 + n^2)/8.
a(n) = n^2/4 + 1 for n even;
a(n) = n^2 + 4 for n odd.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.
G.f.: (1 + 5*x - x^2 - 2*x^3 + 2*x^4 + 5*x^5)/ (1 - x^2)^3. (End)
E.g.f.: (5/8)*(x^2 + x + 4)*exp(x) - (3/8)*(x^2 - x + 4)*exp(-x). - Robert Israel, Aug 18 2015
Sum_{n>=0} 1/a(n) = (4*coth(Pi)+tanh(Pi))*Pi/8 + 1/2. - Amiram Eldar, Mar 22 2022

New name by Peter Luschny, Mar 18 2022

A229525 Sum of coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0 for a,b,c = 1,-1,-1, k = 1,2,3...

Original entry on oeis.org

11, 5, 31, 11, 59, 19, 95, 29, 139, 41, 191, 55, 251, 71, 319, 89, 395, 109, 479, 131, 571, 155, 671, 181, 779, 209, 895, 239, 1019, 271, 1151, 305, 1291, 341, 1439, 379, 1595, 419, 1759, 461, 1931, 505, 2111, 551, 2299, 599, 2495, 649, 2699, 701, 2911, 755

Offset: 1

Russell Walsmith, Sep 26 2013

The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula Q = ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. For a,b,c = 1,-1,-1 and k = 1,2,3..., the coefficients given by Q are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions and summing a+b+c gives the sequence.

The negative of the n-th term is the n+4th term of the c coefficient sequence A229526.

			For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions, 25, 45, -11 and 25 + 45 -11 = 59 = a[5].

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CL-Chemy III: Hyper-Quadratics
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

The a coefficients are A168077, b coefficients are A171621, c coefficients are A229526.

Vec(-x*(x^5-x^4-4*x^3-2*x^2+5*x+11)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014

ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3... n.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5-x^4-4*x^3-2*x^2+5*x+11) / ((x-1)^3*(x+1)^3). - Colin Barker, Nov 02 2014
a(n) = -(-5+3*(-1)^n)*(4+6*n+n^2)/8. - Colin Barker, Nov 03 2014

More terms from Colin Barker, Nov 02 2014

A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).

Original entry on oeis.org

0, 0, 1, 1, 3, 3, 15, 6, 14, 10, 45, 15, 33, 21, 91, 28, 60, 36, 153, 45, 95, 55, 231, 66, 138, 78, 325, 91, 189, 105, 435, 120, 248, 136, 561, 153, 315, 171, 703, 190, 390, 210, 861, 231, 473, 253, 1035, 276, 564, 300, 1225, 325

Offset: 0

Paul Curtz, Mar 12 2012

a(n) is divisible by the n-th term of the sequence 3, 3, 1, 1, 3, 3 (periodically repeated with period 6).
a(n) is divisible by b(floor((n-1)/3)), where b(n) = 1, 3, 2, 3, 7, 3, 5, 3, 13, 3, 8, 3, 19, 3,... , n>=0, is defined by inserting a 3 after each entry of A165355.
(n+1)*(n+2)*(n+3)/2=3*A000292(n+1) is divisible by a(n+2), so there is an integer sequence c(n)= 3*A000292(n+1)/a(n+2) = 3, 12, 10, 20, 7, 28, 18,... with c(2*n)=A123167(n+1) and c(n)/A109613(n+2)=A176895(n).
The sequence of denominators of a(n+2)/n has period length 8: 1, 2, 1, 4, 1, 1, 1, 4.
A table T(k,c) = a(1+c*(1+2k)) of (2*k+1)-sections starts as follows:
0  1  1   3   3   15...
0  3  6  45  21   60...
0 15 15  60  55  325...
0 14 28 231 105  315...
0 45 45 189 171 1035...
The table of T'(k,c) = T(k,c)/(2k+1), columns c>=0, looks as follows, construction similar to A165943:
0 1 1  3  3  15  6  14   k=0
0 1 2 15  7  20 15  77   k=1
0 3 3 12 11  65 24  63   k=2
0 2 4 33 15  45 33 175   k=3
0 5 5 21 19 115 42 112   k=4
0 3 6 51 23  70 51 273   k=5
The entries T'(k,c) are divisible by A060819(c).
Differences are T'(2,c)-T'(0,c) = T'(4,c)-T'(2,c) = 0, 2, 2, 9, 8, 50, 18, 49, 32, ... which is A168077(c) multiplied by the c-th term of the period-4 sequence 2, 2, 2, 1.
Differences are T'(3,c)- T'(1,c) = T'(5,c)-T'(3,c) = 0, 1, 2, 18, 8, 25, 18, 98, 32,... which is  A168077(c) multiplied by the period-4 sequence 2, 1, 2, 2.
The reduced fractions T'(0,c)/T'(1,c) = 1, 1/2, 1/5, 3/7, 3/4, 2/5, 2/11, 5/13, 5/7, 3/8, 3/17, 7/19, .., c>=1, have a numerator sequence A026741(floor(c/2)+1). The denominator sequence is f(c) = 1, 2, 5, 7, 4, 5,.. = A001651(c+1)/A130658(c+1), with f(2*c+1) +f(2*c+2) = 3, 12, 9, 24 .. =3*A022998(c).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000217, A014634, A026741, A033991, A064038, A060819, A165355, A208950.

A185138 := proc(n)
        if n mod 4 = 0 then
                return n/4*(n-1) ;
        elif n mod 2 = 1 then
                return (n-1)*(n+1)/8 ;
        else
                return (n-1)*n/2 ;
        end if;
end proc: # R. J. Mathar, Apr 05 2012

Clear[b];b[1] = 0; b[2] = 0; b[3] = 1; b[4] = 1; b[5] = 3; b[6] = 3; b[7] = 15;b[8] = 6;b[n_Integer] := b[n] = ((-2 + n) (-4 (-4 + n) (-3 + n) (-2 + n) (8 + n (-9 + 2 n)) b[-3 + n] + (-5 + n) ((-3 +n) ((-4 + n) (211 + 2 n (-215 + n (147 + n (-41 + 4 n)))) - 4 (-1 + n) (19 + n (-13 + 2 n)) b[-2 + n]) - 4 (-4 + n)^2 (8 + n (-9 + 2 n)) b[-1 + n])))/(4 (-5 + n) (-4 + n) (-3 + n)^2 (19 + n (-13 + 2 n)))
a = Table[b[n], {n, 1, 52}] (* Roger L. Bagula, Mar 14 2012 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{0,0,1,1,3,3,15,6,14,10,45,15},60] (* Harvey P. Dale, Nov 23 2015 *)

x='x+O('x^50); concat([0,0], Vec(-x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3))) \\ G. C. Greubel, Jun 23 2017

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*n) = A064038(2*n), a(2*n+1) = A000217(n).
a(n) = 3*A208950(n)/A109613(n).
a(n+1) = A060819(n) * A026741(n+2)(floor(n/2)).
G.f.: -x^2*(3*x^8+x^7+5*x^6+3*x^5+12*x^4+3*x^3+3*x^2+x+1)/ ((x-1)^3*(x+1)^3*(x^2+1)^3). - R. J. Mathar, Mar 22 2012
a(n) = (4*n^2-3*n-1+(2*n^2-3*n+1)*(-1)^n + n*(n-1)*(1+(-1)^n)*(-1)^((2*n-3-(-1)^n)/4))/16. - Luce ETIENNE, May 13 2016
Sum_{n>=2} 1/a(n) = 2 - Pi/4 + 7*log(2)/2. - Amiram Eldar, Aug 12 2022

A229526 The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...

Original entry on oeis.org

5, 1, 1, -1, -11, -5, -31, -11, -59, -19, -95, -29, -139, -41, -191, -55, -251, -71, -319, -89, -395, -109, -479, -131, -571, -155, -671, -181, -779, -209, -895, -239, -1019, -271, -1151, -305, -1291, -341, -1439, -379, -1595, -419, -1759, -461, -1931, -505

Offset: 1

Russell Walsmith, Sep 27 2013

The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. Let a,b,c = 1,-1,-1 and k = 1,2,3... Then the coefficients given by this last equation are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions, the c coefficients are the sequence above.

The n-th term = the (positive) n-4th term of A229525.

			For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions gives 25, 45, -11 and -11 = a[5].

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CL-Chemy III: Hyper-Quadratics

The a coefficients are A168077, b coefficients are A171621, the sum of a, b and c coefficients is A229525.

Vec(-x*(x^5+x^4-4*x^3-14*x^2+x+5)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014

ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3..n.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5+x^4-4*x^3-14*x^2+x+5) / ((x-1)^3*(x+1)^3). - Colin Barker, Nov 02 2014
a(n) = (-5+3*(-1)^n)*(-4-2*n+n^2)/8. - Colin Barker, Nov 03 2014

More terms from Colin Barker, Nov 02 2014

A168251 a(n) = n^2 if n is odd, n^2*2^(n-2) if n is even.

Original entry on oeis.org

0, 1, 4, 9, 64, 25, 576, 49, 4096, 81, 25600, 121, 147456, 169, 802816, 225, 4194304, 289, 21233664, 361, 104857600, 441, 507510784, 529, 2415919104, 625, 11341398016, 729, 52613349376, 841, 241591910400, 961, 1099511627776, 1089, 4964982194176, 1225

Offset: 0

Paul Curtz, Nov 21 2009

This is the main diagonal of the following array defined by T(n,2k+1) = A168077(k) for odd column indices and T(n,2k) = A168077(2k)*2^n for even column indices:
0, 1, 1, 9, 4, 25, ... A168077
0, 1, 2, 9, 8, 25, ... A129194
0, 1, 4, 9, 16,25, ... A000290
0, 1, 8, 9, 32,25, ...
0, 1, 16,9, 64,25, ... A154615

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0, 15, 0, -87, 0, 245, 0, -348, 0, 240, 0, -64).

[(n^2)*2^((n-2)*(1+(-1)^n) div 2): n in [0..40]]; // Vincenzo Librandi, Jul 17 2016

A168251 := proc(n)
        if type(n,'even') then
                n^2*2^n/4 ;
        else
                n^2 ;
        end if;
end proc: # R. J. Mathar, Sep 20 2011

Table[(n^2)*2^((n - 2)*(1 + (-1)^n)/2), {n, 0, 50}] (* G. C. Greubel, Jul 16 2016 *)
Table[If[OddQ[n],n^2,n^2 2^(n-2)],{n,0,50}] (* or *) LinearRecurrence[{0,15,0,-87,0,245,0,-348,0,240,0,-64},{0,1,4,9,64,25,576,49,4096,81,25600,121},41] (* Harvey P. Dale, May 14 2022 *)

a(2n) = A128782(n). a(2n+1) = A016754(n).
a(n) = +15*a(n-2) -87*a(n-4) +245*a(n-6) -348*a(n-8) +240*a(n-10) - 64*a(n-12).
G.f.: x*(1 + 4*x - 6*x^2 + 4*x^3 - 23*x^4 - 36*x^5 + 212*x^6 + 44*x^7 - 336*x^8 - 16*x^9 - 64*x^10) / ( (1-x)^3*(2*x+1)^3*(1-2*x)^3*(1+x)^3 ). - R. J. Mathar, Sep 20 2011
a(n) = (n^2)*2^((n-2)*(1+(-1)^n)/2). - Luce ETIENNE, Feb 03 2015

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

A251091 a(n) = n^2 / gcd(n+2, 4).

Original entry on oeis.org

0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121, 72, 169, 49, 225, 128, 289, 81, 361, 200, 441, 121, 529, 288, 625, 169, 729, 392, 841, 225, 961, 512, 1089, 289, 1225, 648, 1369, 361, 1521, 800, 1681, 441, 1849, 968, 2025, 529, 2209, 1152, 2401, 625, 2601, 1352

Offset: 0

Paul Curtz, May 08 2015

A061038(n), which appears in 4*a(n) formula, is a permutation of n^2.
Origin. In December 2010, I wrote in my 192-page Exercise Book no. 5, page 41, the array (difference table of the first row):
   1     0,   1/3,     1,   9/5,    8/3,   25/7,    9/2,   49/9, ...
  -1,  1/3,   2/3,   4/5, 13/15,  19/21,  13/14,  17/18,  43/45, ...
Numerators are listed in A176126, denominators are in A064038, and denominator - numerator = 2, 2, 1, 1,... (A014695).
  4/3, 1/3,  2/15,  1/15, 4/105,   1/42,   1/63,   1/90,  4/495, ...
  -1, -1/5, -1/15, -1/35, -1/70, -1/126, -1/210, -1/330, -1/495, ...
where the denominators of the second row are listed in A000332.
Also for those of the inverse binomial transform
  1, -1, 4/3, -1, 4/5, -2/3, 4/7, -1/2, 4/9, -2/5, 4/11, -1/3, ... ?
a(n) is the (n+1)-th term of the numerators of the first row.

			a(0) = 0/2, a(1) = 1/1, a(2) = 4/4, a(3) = 9/1.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A000290, A000332, A016754, A026741, A060819, A061038, A109008, A139098, A168077, A176895, A181900.

[(1-(1/16)*(1+(-1)^n)*(5-(-1)^(n div 2)) )*n^2: n in [0..60]]; // Vincenzo Librandi, Jun 12 2015

seq(seq((4*i+j-1)^2/[2,1,4,1][j],j=1..4),i=0..30); # Robert Israel, May 14 2015

f[n_] := Switch[ Mod[n, 4], 0, n^2/2, 1, n^2, 2, n^2/4, 3, n^2]; Array[f, 50, 0] (* or *) Table[(4 i + j - 1)^2/{2, 1, 4, 1}[[j]], {i, 0, 12}, {j, 4}] // Flatten (* after Robert Israel *) (* or *) LinearRecurrence[{0, 0, 0, 3, 0, 0, 0, -3, 0, 0, 0, 1}, {0, 1, 1, 9, 8, 25, 9, 49, 32, 81, 25, 121}, 53] (* or *) CoefficientList[ Series[-((x (1 + x (1 + x (9 + x (8 + x (22 + x (6 + x (22 + x (8 + x (9 + x + x^2))))))))))/(-1 + x^4)^3), {x, 0, 52}], x] (* Robert G. Wilson v, May 19 2015 *)

concat(0, Vec(-x*(x^10 + x^9 + 9*x^8 + 8*x^7 + 22*x^6 + 6*x^5 + 22*x^4 + 8*x^3 + 9*x^2 + x + 1) / ((x-1)^3*(x+1)^3*(x^2+1)^3) + O(x^100))) \\ Colin Barker, May 14 2015

a(n) = n^2/(period 4: repeat 2, 1, 4, 1).
a(4n) = 8*n^2, a(2n+1) = a(4n+2) = (2*n+1)^2.
a(n+4) = a(n) + 8*A060819(n).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12), n>11.
4*a(n) = (period 4: repeat 2, 1, 4, 1) * A061038(n).
G.f.: -x*(x^10+x^9+9*x^8+8*x^7+22*x^6+6*x^5+22*x^4+8*x^3+9*x^2+x+1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, May 14 2015
a(2n) = A181900(n), a(2n+1) = A016754(n). [Bruno Berselli, May 14 2015]
a(n) = ( 1 - (1/16)*(1+(-1)^n)*(5-(-1)^(n/2)) )*n^2. - Bruno Berselli, May 14 2015
Sum_{n>=1} 1/a(n) = 13*Pi^2/48. - Amiram Eldar, Aug 12 2022

Missing term (1521) inserted in the sequence by Colin Barker, May 14 2015
Definition uses a formula by Jean-François Alcover, Jul 01 2015
Keyword:mult added by Andrew Howroyd, Aug 06 2018

cp's OEIS Frontend

A181318 a(n) = A060819(n)^2.

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A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

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A229525 Sum of coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0 for a,b,c = 1,-1,-1, k = 1,2,3...

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A185138 a(4*n) = n*(4*n-1); a(2*n+1) = n*(n+1)/2; a(4*n+2) = (2*n+1)*(4*n+1).

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A229526 The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...

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A168251 a(n) = n^2 if n is odd, n^2*2^(n-2) if n is even.

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A185138 a(4n) = n(4n-1); a(2n+1) = n(n+1)/2; a(4n+2) = (2n+1)(4*n+1).