A200182 -id:A200182 - cp's OEIS frontend

A014206 a(n) = n^2 + n + 2.

2, 4, 8, 14, 22, 32, 44, 58, 74, 92, 112, 134, 158, 184, 212, 242, 274, 308, 344, 382, 422, 464, 508, 554, 602, 652, 704, 758, 814, 872, 932, 994, 1058, 1124, 1192, 1262, 1334, 1408, 1484, 1562, 1642, 1724, 1808, 1894, 1982, 2072, 2164, 2258, 2354, 2452, 2552

Offset: 0

N. J. A. Sloane

Draw n + 1 circles in the plane; sequence gives maximal number of regions into which the plane is divided. Cf. A051890, A386480.
Number of binary (zero-one) bitonic sequences of length n + 1. - Johan Gade (jgade(AT)diku.dk), Oct 15 2003
Also the number of permutations of n + 1 which avoid the patterns 213, 312, 13452 and 34521. Example: the permutations of 4 which avoid 213, 312 (and implicitly 13452 and 34521) are 1234, 1243, 1342, 1432, 2341, 2431, 3421, 4321. - Mike Zabrocki, Jul 09 2007
If Y is a 2-subset of an n-set X then, for n >= 3, a(n-3) is equal to the number of (n-3)-subsets and (n-1)-subsets of X having exactly one element in common with Y. - Milan Janjic, Dec 28 2007
With a different offset, competition number of the complete tripartite graph K_{n, n, n}. [Kim, Sano] - Jonathan Vos Post, May 14 2009. Cf. A160450, A160457.
A related sequence is A241119. - Avi Friedlich, Apr 28 2015
From Avi Friedlich, Apr 28 2015: (Start)
This sequence, which also represents the number of Hamiltonian paths in K_2 X P_n (A200182), may be represented by interlacing recursive polynomials in arithmetic progression (discriminant =-63). For example:
a(3*k-3) = 9*k^2 - 15*k + 8,
a(3*k-2) = 9*k^2 - 9*k  + 4,
a(3*k-1) = 9*k^2 - 3*k  + 2,
a(3*k)   = 3*(k+1)^2  - 1. (End)
a(n+1) is the area of a triangle with vertices at (n+3, n+4), ((n-1)*n/2, n*(n+1)/2),((n+1)^2, (n+2)^2) with n >= -1. - J. M. Bergot, Feb 02 2018
For prime p and any integer k, k^a(p-1) == k^2 (mod p^2). - Jianing Song, Apr 20 2019
From Bernard Schott, Jan 01 2021: (Start)
For n >= 1, a(n-1) is the number of solutions x in the interval 0 <= x <= n of the equation x^2 - [x^2] = (x - [x])^2, where [x] = floor(x). For n = 3, the a(2) = 8 solutions in the interval [0, 3] are 0, 1, 3/2, 2, 9/4, 5/2, 11/4 and 3.
This is a variant of the 4th problem proposed during the 20th British Mathematical Olympiad in 1984 (see A002061). The interval [1, n] of the Olympiad problem becomes here [0, n], and only the new solution x = 0 is added. (End)
See A386480 for the almost identical sequence 1, 2, 4, 8, 14, 22, 32, 44, 58, 74, 92, 112, 134, ... which is the maximum number of regions that can be formed in the plane by drawing n circles, and the maximum number of regions that can be formed on the sphere by drawing n great circles. - N. J. A. Sloane, Aug 01 2025

			a(0) = 0^2 + 0 + 2 = 2.
a(1) = 1^2 + 1 + 2 = 4.
a(2) = 2^2 + 2 + 2 = 8.
a(6) = 4*5/5 + 5*6/5 + 6*7/5 + 7*8/5 + 8*9/5 = 44. - _Bruno Berselli_, Oct 20 2016

K. E. Batcher, Sorting Networks and their Applications. Proc. AFIPS Spring Joint Comput. Conf., Vol. 32, pp. 307-314 (1968). [for bitonic sequences]
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 73, Problem 3.
T. H. Cormen, C. E. Leiserson and R. L. Rivest, Introduction to Algorithms. MIT Press / McGraw-Hill (1990) [for bitonic sequences]
Indiana School Mathematics Journal, vol. 14, no. 4, 1979, p. 4.
D. E. Knuth, The Art of Computer Programming, vol3: Sorting and Searching, Addison-Wesley (1973) [for bitonic sequences]
J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 177.
Derrick Niederman, Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed, A Perigee Book, NY, 2009, p. 83.
A. M. Yaglom and I. M. Yaglom, Challenging Mathematical Problems with Elementary Solutions. Vol. I. Combinatorial Analysis and Probability Theory. New York: Dover Publications, Inc., 1987, p. 13, #44 (First published: San Francisco: Holden-Day, Inc., 1964)

N. J. A. Sloane, Table of n, a(n) for n = 0..1000
A. Burstein, S. Kitaev, and T. Mansour, Partially ordered patterns and their combinatorial interpretations, PU. M. A. Vol. 19 (2008), No. 2-3, pp. 27-38.
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
S.-R. Kim and Y. Sano, The competition numbers of complete tripartite graphs, Discrete Appl. Math., 156 (2008) 3522-3524.
Hans Werner Lang, Bitonic sequences.
Daniel Q. Naiman and Edward R. Scheinerman, Arbitrage and Geometry, arXiv:1709.07446 [q-fin.MF], 2017.
Jean-Christoph Novelli and Anne Schilling, The Forgotten Monoid, arXiv 0706.2996 [math.CO], 2007.
Parabola, Problem #Q736, 24(1) (1988), p. 22.
Franck Ramaharo, Enumerating the states of the twist knot, arXiv:1712.06543 [math.CO], 2017.
Franck Ramaharo, Statistics on some classes of knot shadows, arXiv:1802.07701 [math.CO], 2018.
Franck Ramaharo, A generating polynomial for the two-bridge knot with Conway's notation C(n,r), arXiv:1902.08989 [math.CO], 2019.
Yoshio Sano, The competition numbers of regular polyhedra, arXiv:0905.1763 [math.CO], 2009.
Jeffrey Shallit, Recursivity: An Interesting but Little-Known Function, 2012. [Mentions this function in a blog post as the solution for small n to a problem involving Boolean matrices whose values for larger n are unknown.]
Eric Weisstein's World of Mathematics, Plane Division by Circles.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014206 (dim 2), A046127 (dim 3), A059173 (dim 4), A059174 (dim 5).
A row of A059250.
Cf. A000124, A051890, A002522, A241119, A033547 (partial sums).
Cf. A002061 (central polygonal numbers).
Cf. A003682, A160450, A160457, A200182, A386480.
Column 4 of A347570.

[n^2+n+2: n in [0..50]]; // Vincenzo Librandi, Apr 29 2015

Maple
```
A014206 := n->n^2+n+2;
```

Table[n^2 + n + 2, {n, 0, 50}] (* Stefan Steinerberger, Apr 08 2006 *)
LinearRecurrence[{3, -3, 1}, {2, 4, 8}, 50] (* Harvey P. Dale, May 14 2011 *)
CoefficientList[Series[2 (x^2 - x + 1)/(1 - x)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Apr 29 2015 *)

a(n)=n^2+n+2 \\ Charles R Greathouse IV, Jul 31 2011

x='x+O('x^100); Vec(2*x*(x^2-x+1)/(1-x)^3) \\ Altug Alkan, Nov 01 2015

G.f.: 2*(x^2 - x + 1)/(1 - x)^3.
n hyperspheres divide R^k into at most C(n-1, k) + Sum_{i = 0..k} C(n, i) regions.
a(n) = A002061(n+1) + 1 for n >= 0. - Rick L. Shepherd, May 30 2005
Equals binomial transform of [2, 2, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 18 2008
a(n) = A003682(n+1), n > 0. - R. J. Mathar, Oct 28 2008
a(n) = a(n-1) + 2*n (with a(0) = 2). - Vincenzo Librandi, Nov 20 2010
a(0) = 2, a(1) = 4, a(2) = 8, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. - Harvey P. Dale, May 14 2011
a(n + 1) = n^2 + 3*n + 4. - Alonso del Arte, Apr 12 2015
a(n) = Sum_{i=n-2..n+2} i*(i + 1)/5. - Bruno Berselli, Oct 20 2016
Sum_{n>=0} 1/a(n) = Pi*tanh(Pi*sqrt(7)/2)/sqrt(7). - Amiram Eldar, Jan 09 2021
From Amiram Eldar, Jan 29 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(11)*Pi/2)*sech(sqrt(7)*Pi/2).
Product_{n>=0} (1 - 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(sqrt(7)*Pi/2). (End)
a(n) = 2*A000124(n). - R. J. Mathar, Mar 14 2021
E.g.f.: exp(x)*(2 + 2*x + x^2). - Stefano Spezia, Apr 30 2022

More terms from Stefan Steinerberger, Apr 08 2006

A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

Original entry on oeis.org

14, 19, 31, 38, 54, 63, 83, 94, 118, 131, 159, 174, 206, 223, 259, 278, 318, 339, 383, 406, 454, 479, 531, 558, 614, 643, 703, 734, 798, 831, 899, 934, 1006, 1043, 1119, 1158, 1238, 1279, 1363, 1406, 1494, 1539, 1631, 1678, 1774, 1823, 1923, 1974, 2078, 2131

Offset: 1

T. D. Noe, Apr 17 2014

From Robert Israel, Apr 28 2015: (Start)
Numbers n >= 14 such that 3*n + 7 is a square.
This is because
C(n,i+3) - 3*C(n,i+2) + 3*C(n,i+1) - C(n,i) = n!/((n-i)!*(i+3)!) * g(n,i)
where g(n,i) = (n-3-2*i) * ((n-3-2*i)^2 - 3*n - 7). (End)

			Binomial(14,k) = (1, 14, 91, 364, 1001, 2002, 3003, 3432) for k = 0..7. The 4 terms beginning with 91 equal 182 - 273*x + 182*x^2 for x = 1..4.

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Sequence A241200 gives the position of the first of the 4 terms. Sequence A008865 gives the terms greater than 2 for which 3 consecutive terms satisfy a linear relation.
A014206 is a related sequence. - Avi Friedlich, Apr 28 2015
Cf. A062730 (3 terms in arithmetic progression in Pascal's triangle).

map(k -> (3*k^2+8*k+3,3*k^2+10*k+6),[$1..100]); # Robert Israel, Apr 28 2015

Select[Range[2500], MemberQ[Differences[Binomial[#, Range[0, #/2]], 3], 0] &]
LinearRecurrence[{1,2,-2,-1,1},{14,19,31,38,54},50] (* Harvey P. Dale, Oct 29 2017 *)

Vec(-x*(6*x^4-3*x^3-16*x^2+5*x+14)/((x-1)^3*(x+1)^2) + O(x^100)) \\ Colin Barker, Apr 29 2015

a(n)=(6*n^2+42*n+55-(-1)^n*(2*n+7))/8 \\ Charles R Greathouse IV, Apr 15 2016

a(n) = (55-7*(-1)^n-2*(-21+(-1)^n)*n+6*n^2)/8. G.f.: -x*(6*x^4-3*x^3-16*x^2+5*x+14) / ((x-1)^3*(x+1)^2). - Colin Barker, Apr 18 2014 and Apr 29 2015
The terms appear to satisfy a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5), with initial terms 14, 19, 31, 38, 54. - T. D. Noe, Apr 18 2014
Numbers are of the form A200182(3n+1) and A200182(3n-1). - Avi Friedlich, Apr 25 2015
a(2*k-1) = 3*k^2 + 8*k + 3, a(2*k) = 3*k^2 + 10*k + 6. - Robert Israel, Apr 28 2015

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A014206 a(n) = n^2 + n + 2.

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A241199 Numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2.

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Maple

Mathematica

PARI

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