

A087475


a(n) = n^2 + 4.


32



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
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OFFSET

0,1


COMMENTS

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^(n1))). (End)
a(n) = A156798(n)/A002522(n).  Reinhard Zumkeller, Feb 16 2009
Given sequences of the form S(n) = N*S(n1) + S(n2) starting (1, N,...), and having convergents with discriminant (N^2 + 4), S(p) == (a(n))^((p1)/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^((p1)/2)) mod p.  Gary W. Adamson, Feb 23 2009
a(n) = A156701(n) / A053755(n).  Reinhard Zumkeller, Feb 13 2009
Number of units of a(n) belongs to a periodic sequence: 4, 5, 8, 3, 0, 9, 0, 3, 8, 5. We conclude that a(n) and a(n+10) have the same number of units.  Mohamed Bouhamida, Sep 05 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 nth 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..k1), the falling factorials. See the T. Koshy reference, pp. 263264 (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. 330331. See also the first comment above.
a(n)^3 = A155965(n)^2 + A155966(n)^2.  Vincenzo Librandi, Feb 22 2012


REFERENCES

Manfred R. Schroeder, "Fractals, Chaos, Power Laws"; W.H. Freeman & Co, 1991, p. 330331.
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]


LINKS

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


FORMULA

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) = a(n1) + 2*n1 (with a(0)=4).  Vincenzo Librandi, Nov 22 2010
G.f.: (4  7*x + 5*x^2)/(1  3*x + 3*x^2  x^3).  Colin Barker, Jan 06 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. (End)


EXAMPLE

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.


MATHEMATICA

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


PROG

(PARI) a(n)=n^2+4 \\ Charles R Greathouse IV, Jun 10 2011
(Scala) (0 to 49).map(n => n * n + 4) // Alonso del Arte, May 29 2019


CROSSREFS

Cf. A005563, A046092, A001082, A002378, A036666, A062717, A028347, A155965, A155966.
Sequence in context: A174398 A030978 A101948 * A019526 A242014 A145488
Adjacent sequences: A087472 A087473 A087474 * A087476 A087477 A087478


KEYWORD

nonn,easy


AUTHOR

Gary W. Adamson, Sep 09 2003


STATUS

approved



