A155966 -id:A155966 - cp's OEIS frontend

A087475 a(n) = n^2 + 4.

4, 5, 8, 13, 20, 29, 40, 53, 68, 85, 104, 125, 148, 173, 200, 229, 260, 293, 328, 365, 404, 445, 488, 533, 580, 629, 680, 733, 788, 845, 904, 965, 1028, 1093, 1160, 1229, 1300, 1373, 1448, 1525, 1604, 1685, 1768, 1853, 1940, 2029, 2120, 2213, 2308, 2405, 2504

Offset: 0

Gary W. Adamson, Sep 09 2003

Schroeder, p. 330, states "For positive n, these winding numbers are precisely those whose continued fraction expansion is periodic and has period length 1".
Positive X values of solutions to the equation X^3 - 4*X^2 = Y^2. To find Y values: b(n) = n*(n^2 + 4). - Mohamed Bouhamida, Nov 06 2007
From Artur Jasinski, Oct 03 2008: (Start)
General formula for cotangent recurrences type:
a(n+1) = a(n)^3 + 3*a(n) and a(1)=k is
a(n) = floor(((k + sqrt(k^2 + 4))/2)^(3^(n-1))). (End)
Given sequences of the form S(n) = N*S(n-1) + S(n-2) starting (1, N, ...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p-1)/2) mod p, for n>0, p = odd prime. Example: with N = 2 we have the Pell series (1, 2, 5, 12, 29, 70, 169, ...) with P(7) = 169. Then 169 == 8^3 mod 7, with a(2) = 8. Cf. Schroeder, "Number Theory in Science and Communication", p. 90, for N = 1: F(p) == 5^((p-1)/2) mod p. - Gary W. Adamson, Feb 23 2009
The only two real solutions of the form f(x) = A*x^p with positive p that satisfy f^(n)(x) = f^[-1](x), x >= 0, n >= 1, with f^(n) the n-th derivative and f^[-1] the compositional inverse of f, are obtained for p = p1(n) = (n + sqrt(a(n)))/2 and p = p2(n) = (n - sqrt(a(n)))/2, n >= 1, and A = A(n) = (fallfac(p,n))^(-p/(p+1)), for p = p1(n) and p = p2(n), respectively. Here fallfac(x, k) := product(x - j, j = 0..k-1), the falling factorials. See the T. Koshy reference, pp. 263-264 (there is also a solution for negative p if n is even; see the corresponding comment in A002522). - Wolfdieter Lang, Oct 21 2010, Oct 28 2010
(n + sqrt(a(n)))/2 = [n;n,n,...], with the regular continued fraction with period length 1. For a simple proof see, e.g., the Schroeder reference, pp. 330-331. See also the first comment above.

			a(2) = 8, discriminant of algebraic representation of barover(2) = [2,2,2,...] = sqrt 2 - 1 = 0.41421356... = ((sqrt 8) - 2)/2. a(3) = 13, discriminant of barover(3) = [3,3,3,...] = 0.3027756... = ((sqrt 13) - 3)/2.

Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330-331.
Manfred R. Schroeder, "Number Theory in Science and Communication", Springer Verlag, 5th ed., 2009. [From Gary W. Adamson, Feb 23 2009]
Thomas Koshy, "Fibonacci and Lucas Numbers with Applications", John Wiley and Sons, New York, 2001. [From Wolfdieter Lang, Oct 21 2010]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Near-Square Prime.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A001082, A002378, A002522, A005563, A028347, A036666, A046092, A053755, A062717, A155965, A155966, A156701, A156798.

Range[0, 50]^2 + 4 (* Harvey P. Dale, Jan 05 2011 *)

a(n)=n^2+4 \\ Charles R Greathouse IV, Jun 10 2011

(0 to 49).map(n => n * n + 4) // Alonso del Arte, May 29 2019

n^2 + 4 are discriminant terms in the formula for Positive Silver Mean Constants, defined as barover(n), = (sqrt (n^2 + 4) - n)/2. Such constants barover(n) = C have the property: 1/C - C = n.
a(n) = A156701(n) / A053755(n). - Reinhard Zumkeller, Feb 13 2009
a(n) = A156798(n)/A002522(n). - Reinhard Zumkeller, Feb 16 2009
a(n) = a(n-1) + 2*n-1 (with a(0)=4). - Vincenzo Librandi, Nov 22 2010
G.f.: (4 - 7*x + 5*x^2)/(1 - x)^3. - Colin Barker, Jan 06 2012
a(n)^3 = A155965(n)^2 + A155966(n)^2. - Vincenzo Librandi, Feb 22 2012
From Amiram Eldar, Jul 13 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + 2*Pi*coth(2*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + 2*Pi*cosech(2*Pi))/8 = A371803. (End)
E.g.f.: exp(x)*(4 + x + x^2). - Stefano Spezia, Jul 08 2023
From Amiram Eldar, Feb 05 2024: (Start)
Product_{n>=0} (1 - 1/a(n)) = sqrt(3)*sinh(sqrt(3)*Pi)/(2*sinh(2*Pi)).
Product_{n>=0} (1 + 1/a(n)) = sqrt(5)*sinh(sqrt(5)*Pi)/(2*sinh(2*Pi)). (End)

A255843 a(n) = 2*n^2 + 4.

Original entry on oeis.org

4, 6, 12, 22, 36, 54, 76, 102, 132, 166, 204, 246, 292, 342, 396, 454, 516, 582, 652, 726, 804, 886, 972, 1062, 1156, 1254, 1356, 1462, 1572, 1686, 1804, 1926, 2052, 2182, 2316, 2454, 2596, 2742, 2892, 3046, 3204, 3366, 3532, 3702, 3876, 4054, 4236, 4422

Offset: 0

Avi Friedlich, Mar 08 2015

This is the case k=2 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2.

Equivalently, numbers m such that 2*m - 8 is a square.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A059100.
Cf. unsigned A147973: numbers of the form 2*m^2-4.
Cf. sequences of the form 2*m^2+2*k: A005893 (k=1), this sequence (k=2), A255844 (k=3), A155966 (k=4), A255845 (k=5), A255842 (k=6), A255846 (k=7), A255847 (k=8), A255848 (k=9).

Magma
```
[2*n^2+4: n in [0..50]];
```
Mathematica
```
Table[2 n^2 + 4, {n, 0, 50}]
```
PARI
```
vector(50, n, n--; 2*n^2+4)
```
Sage
```
[2*n^2+4 for n in (0..50)]
```

G.f.: 2*(2 - 3*x + 3*x^2)/(1 - x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*A059100(n).
a(n) = a(n-1) + 4n - 2. - Bob Selcoe, Mar 25 2020
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(2)*Pi*coth(sqrt(2)*Pi))/8.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(2)*Pi*cosech(sqrt(2)*Pi))/8. (End)
E.g.f.: 2*exp(x)*(2 + x + x^2). - Stefano Spezia, Aug 07 2024

Edited by Bruno Berselli, Mar 13 2015

A155965 a(n) = n*(n^2+4).

Original entry on oeis.org

0, 5, 16, 39, 80, 145, 240, 371, 544, 765, 1040, 1375, 1776, 2249, 2800, 3435, 4160, 4981, 5904, 6935, 8080, 9345, 10736, 12259, 13920, 15725, 17680, 19791, 22064, 24505, 27120, 29915, 32896, 36069, 39440, 43015, 46800, 50801, 55024, 59475, 64160

Offset: 0

Vincenzo Librandi, Jan 31 2009

The identity (n^3+4*n)^2 + (2*n^2+8)^2 = (n^2+4)^3 can be written as a(n)^2 + A155966(n)^2 = A087475(n)^3.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A006003, A087475, A155966.

Table[n (n^2 + 4), {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, May 04 2011 *)
CoefficientList[Series[x (5 - 4 x + 5 x^2)/(1 - x)^4, {x, 0, 60}], x] (* Vincenzo Librandi, May 03 2014 *)
LinearRecurrence[{4,-6,4,-1},{0,5,16,39},50] (* Harvey P. Dale, Jan 23 2019 *)

a(n)=n*(n^2+4) \\ Charles R Greathouse IV, Jan 11 2012

[lucas_number1(4,n,-2) for n in range(0, 41)] # Zerinvary Lajos, May 16 2009

G.f.: x*(5 - 4*x + 5*x^2)/(1 - x)^4. - Vincenzo Librandi, May 03 2014
a(n) = 4*a(n-1) - 6*a(n-2) +4*a(n-3) - a(n-4) for n>3. - Vincenzo Librandi, May 03 2014
a(n) = A006003(n-1) + A006003(n+1). - Lechoslaw Ratajczak, Oct 31 2021

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A087475 a(n) = n^2 + 4.

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A255843 a(n) = 2*n^2 + 4.

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A155965 a(n) = n*(n^2+4).

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