cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A058331 a(n) = 2*n^2 + 1.

Original entry on oeis.org

1, 3, 9, 19, 33, 51, 73, 99, 129, 163, 201, 243, 289, 339, 393, 451, 513, 579, 649, 723, 801, 883, 969, 1059, 1153, 1251, 1353, 1459, 1569, 1683, 1801, 1923, 2049, 2179, 2313, 2451, 2593, 2739, 2889, 3043, 3201, 3363, 3529, 3699, 3873, 4051
Offset: 0

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Author

Erich Friedman, Dec 12 2000

Keywords

Comments

Maximal number of regions in the plane that can be formed with n hyperbolas.
Also the number of different 2 X 2 determinants with integer entries from 0 to n.
Number of lattice points in an n-dimensional ball of radius sqrt(2). - David W. Wilson, May 03 2001
Equals A112295(unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Oct 07 2007
Binomial transform of A166926. - Gary W. Adamson, May 03 2008
a(n) = longest side a of all integer-sided triangles with sides a <= b <= c and inradius n >= 1. Triangle has sides (2n^2 + 1, 2n^2 + 2, 4n^2 + 1).
{a(k): 0 <= k < 3} = divisors of 9. - Reinhard Zumkeller, Jun 17 2009
Number of ways to partition a 3*n X 2 grid into 3 connected equal-area regions. - R. H. Hardin, Oct 31 2009
Let A be the Hessenberg matrix of order n defined by: A[1, j] = 1, A[i, i] := 2, (i > 1), A[i, i - 1] = -1, and A[i, j] = 0 otherwise. Then, for n >= 3, a(n - 1) = coeff(charpoly(A, x), x^(n - 2)). - Milan Janjic, Jan 26 2010
Except for the first term of [A002522] and [A058331] if X = [A058331], Y = [A087113], A = [A002522], we have, for all other terms, Pell's equation: [A058331]^2 - [A002522]*[A087113]^2 = 1; (X^2 - A*Y^2 = 1); e.g., 3^2 -2*2^2 = 1; 9^2 - 5*4^2 = 1; 129^2 - 65*16^2 = 1, and so on. - Vincenzo Librandi, Aug 07 2010
Niven (1961) gives this formula as an example of a formula that does not contain all odd integers, in contrast to 2n + 1 and 2n - 1. - Alonso del Arte, Dec 05 2012
Numbers m such that 2*m-2 is a square. - Vincenzo Librandi, Apr 10 2015
Number of n-tuples from the set {1,0,-1} where at most two elements are nonzero. - Michael Somos, Oct 19 2022
a(n) gives the x-value of the integral solution (x,y) of the Pellian equation x^2 - (n^2 + 1)*y^2 = 1. The y-value is given by 2*n (see Tattersall). - Stefano Spezia, Jul 23 2025

Examples

			a(1) = 3 since (0 0 / 0 0), (1 0 / 0 1) and (0 1 / 1 0) have different determinants.
G.f. = 1 + 3*x + 9*x^2 + 19*x^3 + 33*x^4 + 51*x^5 + 73*x^6 + ... - _Michael Somos_, Oct 19 2022
		

References

  • Ivan Niven, Numbers: Rational and Irrational, New York: Random House for Yale University (1961): 17.
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 256.

Crossrefs

Cf. A000124.
Second row of array A099597.
See A120062 for sequences related to integer-sided triangles with integer inradius n.
Cf. A112295.
Column 2 of array A188645.
Cf. A001105 and A247375. - Bruno Berselli, Sep 16 2014

Programs

  • Haskell
    a058331 = (+ 1) . a001105  -- Reinhard Zumkeller, Dec 13 2014
    
  • Magma
    [2*n^2 + 1 : n in [0..100]]; // Wesley Ivan Hurt, Feb 02 2017
  • Mathematica
    b[g_] := Length[Union[Map[Det, Flatten[ Table[{{i, j}, {k, l}}, {i, 0, g}, {j, 0, g}, {k, 0, g}, {l, 0, g}], 3]]]] Table[b[g], {g, 0, 20}]
    2*Range[0, 49]^2 + 1 (* Alonso del Arte, Dec 05 2012 *)
  • PARI
    a(n)=2*n^2+1 \\ Charles R Greathouse IV, Jun 16 2011
    

Formula

G.f.: (1 + 3x^2)/(1 - x)^3. - Paul Barry, Apr 06 2003
a(n) = M^n * [1 1 1], leftmost term, where M = the 3 X 3 matrix [1 1 1 / 0 1 4 / 0 0 1]. a(0) = 1, a(1) = 3; a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). E.g., a(4) = 33 since M^4 *[1 1 1] = [33 17 1]. - Gary W. Adamson, Nov 11 2004
a(n) = cosh(2*arccosh(n)). - Artur Jasinski, Feb 10 2010
a(n) = 4*n + a(n-1) - 2 for n > 0, a(0) = 1. - Vincenzo Librandi, Aug 07 2010
a(n) = (((n-1)^2 + n^2))/2 + (n^2 + (n+1)^2)/2. - J. M. Bergot, May 31 2012
a(n) = A251599(3*n) for n > 0. - Reinhard Zumkeller, Dec 13 2014
a(n) = sqrt(8*(A000217(n-1)^2 + A000217(n)^2) + 1). - J. M. Bergot, Sep 03 2015
E.g.f.: (2*x^2 + 2*x + 1)*exp(x). - G. C. Greubel, Jul 14 2017
a(n) = A002378(n) + A002061(n). - Bruce J. Nicholson, Aug 06 2017
From Amiram Eldar, Jul 15 2020: (Start)
Sum_{n>=0} 1/a(n) = (1 + (Pi/sqrt(2))*coth(Pi/sqrt(2)))/2.
Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(2))*csch(Pi/sqrt(2)))/2. (End)
From Amiram Eldar, Feb 05 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(2))*sinh(Pi).
Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(2))*csch(Pi/sqrt(2)). (End)
From Leo Tavares, May 23 2022: (Start)
a(n) = A000384(n+1) - 3*n.
a(n) = 3*A000217(n) + A000217(n-2). (End)
a(n) = a(-n) for all n in Z and A037235(n) = Sum_{k=0..n-1} a(k). - Michael Somos, Oct 19 2022

Extensions

Revised description from Noam Katz (noamkj(AT)hotmail.com), Jan 28 2001

A122458 "Dropping time" of the reduced Collatz iteration starting with 2n+1.

Original entry on oeis.org

0, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 37, 1, 35, 1, 2, 1, 5, 1, 3, 1, 34, 1, 2, 1, 3, 1, 4, 1, 34, 1, 2, 1, 32, 1, 3, 1, 5, 1, 2, 1, 3, 1, 28, 1, 5, 1, 2, 1, 26, 1, 3, 1, 19, 1, 2, 1, 3, 1, 5, 1, 9, 1, 2, 1, 4, 1, 3, 1, 4, 1, 2, 1, 3, 1, 25, 1, 13, 1, 2, 1, 18, 1, 3, 1, 5, 1, 2, 1, 3, 1, 4, 1, 8, 1, 2, 1, 5
Offset: 0

Views

Author

T. D. Noe, Sep 08 2006

Keywords

Comments

We count only the 3x+1 steps of the usual Collatz iteration. We stop counting when the iteration produces a number less than the initial 2n+1. For a fixed dropping time k, let N(k)=A100982(k) and P(k)=2^(A020914(k)-1). There are exactly N(k) odd numbers less than P(k) with dropping time k. Moreover, the sequence is periodic: if d is one of the N(k) odd numbers, then k=a(d)=a(d+i*P(k)) for all i>=0. This periodicity makes it easy to compute the average dropping time of the reduced Collatz iteration: Sum_{k>0} k*N(k)/P(k) = 3.492651852186... (A122791).

Examples

			a(3)=4 because, starting with 7, the iteration produces 11,17,13,5 and the last term is less than 7.
n = 13: the fr trajectory for 2*13+1 = 27 is 41, 31, 47, 71, 107, 161, 121, 91, 137, 103, 155, 233, 175, 263, 395, 593, 445, 167, 251, 377, 283, 425, 319, 479, 719, 1079, 1619, 2429, 911, 1367, 2051, 3077, 577, 433, 325, 61, 23, 35, 53, 5, 1 with 41 terms (without 27), hence fr^[37] = 23 < 27  and  a(13) = 37. - _Wolfdieter Lang_, Feb 20 2019
		

References

  • Victor Klee and Stan Wagon, Old and New Unsolved Problems in Plane Geometry and Number Theory, Mathematical Association of America (1991) pp. 225-229, 308-309. [called on p. 225 stopping time for 2n+1 and the function C(2*n+1) = A075677(n+1), n >= 0. - Wolfdieter Lang, Feb 20 2019]

Crossrefs

Cf. A000265, A060445, A075677 (one step of the reduced Collatz iteration), A075680.
Cf. A087113 (indices of 1's), A017077 (indices of 2's), A122791 (limit mean).

Programs

  • Mathematica
    nextOddK[n_]:=Module[{m=3n+1}, While[EvenQ[m], m=m/2]; m]; dt[n_]:=Module[{m=n, cnt=0}, If[n>1, While[m=nextOddK[m]; cnt++; m>n]]; cnt]; Table[dt[n],{n,1,301,2}]

Formula

a(n) is the least k for which fr^[k](n) < 2*n + 1, for n >= 1 and k >= 1, where fr(n) = A075677(n+1) = A000265(3*n+2). No k satisfies this for n = 0: a(0) := 0 by convention. The dropping time a(n) is finite, for n >= 1, if the Collatz conjecture is true. - Wolfdieter Lang, Feb 20 2019
a(1+i*8) = 2, for i>=0, because A100982(2) = 1 is odd, and A020914(2) = 4 gives P(2) = 2^(4-1) = 8. - Ruud H.G. van Tol, Dec 19 2021
Showing 1-2 of 2 results.