A053755 a(n) = 4*n^2 + 1.
1, 5, 17, 37, 65, 101, 145, 197, 257, 325, 401, 485, 577, 677, 785, 901, 1025, 1157, 1297, 1445, 1601, 1765, 1937, 2117, 2305, 2501, 2705, 2917, 3137, 3365, 3601, 3845, 4097, 4357, 4625, 4901, 5185, 5477, 5777, 6085, 6401, 6725, 7057
Offset: 0
References
- Donald E. Knuth, The Art of Computer Programming, Addison-Wesley, Reading, MA, 1997, Vol. 1, exercise 1.2.1 Nr. 11, p. 19.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..1000
- Tom M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
- Roland Bacher, Counting Packings of Generic Subsets in Finite Groups, Electr. J. Combinatorics, 19 (2012), #P7. - From _N. J. A. Sloane_, Feb 06 2013
- Kival Ngaokrajang, Illustration of initial terms.
- Robert G. Wilson v, Cover of the March 1964 issue of Scientific American.
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Crossrefs
Column 2 of array A188647.
Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Programs
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GAP
List([0..45],n->4*n^2+1); # Muniru A Asiru, Nov 01 2018
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Haskell
a053755 = (+ 1) . (* 4) . (^ 2) -- Reinhard Zumkeller, Apr 20 2015
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Magma
m:=50; R
:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+2*x+5*x^2)/((1-x)^3))); /* or */ I:=[1,5]; [n le 2 select I[n] else 2*Self(n-1)-Self(n-2)+8: n in [1..50]]; // Vincenzo Librandi, Jun 26 2013 -
Maple
with (combinat):seq(fibonacci(3,2*n), n=0..42); # Zerinvary Lajos, Apr 21 2008
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Mathematica
f[n_] := 4n^2 +1; Array[f, 40] (* Vladimir Joseph Stephan Orlovsky, Sep 02 2008 *) CoefficientList[Series[(1 + 2 x + 5 x^2) / (1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Jun 26 2013 *) LinearRecurrence[{3,-3,1},{1,5,17},50] (* Harvey P. Dale, Dec 28 2021 *)
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PARI
for(x=0,100,print1(4*x^2+1",")) \\ Cino Hilliard, Aug 26 2006
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Python
for n in range(0,50): print(4*n**2+1, end=', ') # Stefano Spezia, Nov 01 2018
Formula
a(n) = A000466(n) + 2. - Zak Seidov, Jan 16 2007
From R. J. Mathar, Apr 28 2008: (Start)
O.g.f.: (1 + 2*x + 5*x^2)/(1-x)^3.
a(n) = 3a(n-1) - 3a(n-2) + a(n-3). (End)
Equals binomial transform of [1, 4, 8, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008
a(n+1) = denominator of Sum_{k=0..n} (-1)^n*(2*n + 1)^3/((2*n + 1)^4 + 4), see Knuth reference. - Reinhard Zumkeller, Apr 11 2010
a(n) = 8*n + a(n-1) - 4. with a(0)=1. - Vincenzo Librandi, Aug 06 2010
a(n) = ((2*n - 1)^2 + (2*n + 1)^2)/2. - J. M. Bergot, May 31 2012
a(n) = 2*a(n-1) - a(n-2) + 8 with a(0)=1, a(1)=5. - Vincenzo Librandi, Jun 26 2013
a(n+1) = a(n) + A017113(n), a(0) = 1. - Altug Alkan, Sep 26 2015
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/2)*coth(Pi/2))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/2)*csch(Pi/2))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/2)*sinh(Pi/sqrt(2)).
Product_{n>=1} (1 - 1/a(n)) = (Pi/2)*csch(Pi/2). (End)
E.g.f.: exp(x)*(1 + 2*x)^2. - Stefano Spezia, Jun 10 2021
Extensions
Equation corrected, and examples that were based on a different offset removed, by R. J. Mathar, Mar 18 2010
Comments