A171621 -id:A171621 - cp's OEIS frontend

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

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

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 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

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

A171638 Denominator of 1/(n-2)^2 - 1/(n+2)^2.

Original entry on oeis.org

0, 25, 9, 441, 64, 2025, 225, 5929, 576, 13689, 1225, 27225, 2304, 48841, 3969, 81225, 6400, 127449, 9801, 190969, 14400, 275625, 20449, 385641, 28224, 525625, 38025, 700569, 50176, 915849, 65025, 1177225, 82944, 1490841

Offset: 2

Paul Curtz, Dec 13 2009

Fifth column of an array of denominators related to the energies of the hydrogen spectrum, mentioned in A171522. At n=2, the defining formula has a pole and is replaced by 0 to conform with A171621 and A099761.

G. C. Greubel, Table of n, a(n) for n = 2..5000
Index entries for linear recurrences with constant coefficients, signature (0,5,0,-10,0,10,0,-5,0,1).

Cf. A099761, A171621.

[0] cat [Denominator((1/(n-2)^2 -1/(n+2)^2)): n in [3..350]]; // Bruno Berselli, Apr 05 2011

A061037 := proc(n) 1/4-1/n^2 ; numer(%) ; end proc:
A171621 := proc(n) if n mod 4 = 2 then 4*A061037(n) ; else A061037(n) ; end if; end proc:
A171638 := proc(n) A171621(n)^2 ; end proc:
seq(A171638(n),n=2..90) ; # R. J. Mathar, Apr 02 2011

Table[If[n == 2, 0, Denominator[1/(n-2)^2 - 1/(n+2)^2]], {n, 2, 50}] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,5,0,-10,0,10,0,-5,0,1},{0,25,9,441,64,2025,225,5929,576,13689},50] (* Harvey P. Dale, Sep 07 2021 *)

for(n=2,100, print1(if(n==2,0, denominator(1/(n-2)^2 - 1/(n+2)^2)), ", ")) \\ G. C. Greubel, Sep 20 2018

a(n) = (A171621(n))^2.
a(2*n+2) = A099761(n).
G.f.: -((x(25+9*x+316*x^2+19*x^3+70*x^4-5*x^5-36*x^6+x^7+9*x^8))/((-1+x)^5 (1+x)^5)). - Harvey P. Dale, Sep 07 2021
Sum_{n>=3} 1/a(n) = 19*Pi^2/192 - 115/144. - Amiram Eldar, Aug 14 2022

cp's OEIS Frontend

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

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A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

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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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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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A171638 Denominator of 1/(n-2)^2 - 1/(n+2)^2.

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