

A087475


a(n) = n^2 + 4.


30



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".
Sequence allows us to find X values of the equation: X^3  4*X^2 = Y^2. To find Y values: b(n)=n*(n^2 + 4).  Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 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 (bhmd95(AT)yahoo.fr), 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(xj,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


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

a[n_] := n^2 + 4; (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)
Range[0, 50]^2 + 4 (* Harvey P. Dale, Jan 05 2011 *)
LinearRecurrence[{3, 3, 1}, {4, 5, 8}, 50] (* Vincenzo Librandi, Feb 22 2012 *)


PROG

(Sage) [lucas_number1(3, n, 4) for n in xrange(0, 51)] # Zerinvary Lajos, May 16 2009
(PARI) a(n)=n^2+4 \\ Charles R Greathouse IV, Jun 10 2011
(MAGMA) I:=[4, 5, 8]; [n le 3 select I[n] else 3*Self(n1)3*Self(n2)+1*Self(n3): n in [1..40]]; // Vincenzo Librandi, Feb 22 2012


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



