

A002522


a(n) = n^2 + 1.


427



1, 2, 5, 10, 17, 26, 37, 50, 65, 82, 101, 122, 145, 170, 197, 226, 257, 290, 325, 362, 401, 442, 485, 530, 577, 626, 677, 730, 785, 842, 901, 962, 1025, 1090, 1157, 1226, 1297, 1370, 1445, 1522, 1601, 1682, 1765, 1850, 1937, 2026, 2117, 2210, 2305, 2402, 2501
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OFFSET

0,2


COMMENTS

An n X n nonnegative matrix A is primitive (see A070322) iff every element of A^k is > 0 for some power k. If A is primitive then the power which should have all positive entries is <= n^2  2n + 2 (Wielandt).
a(n) = Phi_4(n), where Phi_k is the kth cyclotomic polynomial.
As the positive solution to x=2n+1/x is x=n+sqrt(a(n)), the continued fraction expansion of sqrt(a(n)) is {n; 2n, 2n, 2n, 2n, ...}.  Benoit Cloitre, Dec 07 2001
a(n) is one less than the arithmetic mean of its neighbors: a(n) = (a(n1) + a(n+1))/2  1. E.g., 2 = (1+5)/2  1, 5 = (2+10)/2  1.  Amarnath Murthy, Jul 29 2003
Equivalently, the continued fraction expansion of sqrt(a(n)) is (n;2n,2n,2n,....).  Franz Vrabec, Jan 23 2006
Number of {12,1*2*,21}avoiding signed permutations in the hyperoctahedral group.
The number of squares of side 1 which can be drawn without lifting the pencil, starting at one corner of an n X n grid and never visiting an edge twice is n^22n+2.  Sébastien Dumortier, Jun 16 2005
Also, numbers m such that m^3  m^2 is a square, (n*(1 + n^2))^2.  Zak Seidov
For n >= 1, a(n1) is the minimal number of choices from an nset such that at least one particular element has been chosen at least n times or each of the n elements has been chosen at least once. Some games define "matches" this way; e.g., in the classic Parker Brothers, now Hasbro, board game Risk, a(2)=5 is the number of cards of three available types (suits) required to guarantee at least one match of three different types or of three of the same type (ignoring any jokers or wildcards).  Rick L. Shepherd, Nov 18 2007
Positive X values of solutions to the equation X^3 + (X  1)^2 + X  2 = Y^2. To prove that X = n^2 + 1: Y^2 = X^3 + (X  1)^2 + X  2 = X^3 + X^2  X  1 = (X  1)(X^2 + 2X + 1) = (X  1)*(X + 1)^2 it means: (X  1) must be a perfect square, so X = n^2 + 1 and Y = n(n^2 + 2).  Mohamed Bouhamida, Nov 29 2007
Number of units of a(n) belongs to a periodic sequence: 1, 2, 5, 0, 7, 6, 7, 0, 5, 2.  Mohamed Bouhamida, Sep 04 2009
Appears in A054413 and A086902 in relation to sequences related to the numerators and denominators of continued fractions convergents to sqrt((2*n)^2/4 + 1), n=1, 2, 3, ... .  Johannes W. Meijer, Jun 12 2010
For n>0, continued fraction [n,n] = n/a(n); e.g., [5,5] = 5/26.  Gary W. Adamson, Jul 15 2010
The only real solution of the form f(x) = A*x^p with negative p which satisfies f^(m)(x) = f^[1](x), x >= 0, m >= 1, with f^(m) the mth derivative and f^[1] the compositional inverse of f, is obtained for m=2*n, p=p(n)= (sqrt(a(n))n) and A=A(n)=(fallfac(p(n),2*n))^(p(n)/(p(n)+1)), with fallfac(x,k):=Product_{j=0..k1} (xj) (falling factorials). See the T. Koshy reference, pp. 2634 (there are also two solutions for positive p, see the corresponding comment in A087475).  Wolfdieter Lang, Oct 21 2010
n + sqrt(a(n)) = [2*n;2*n,2*n,...] with the regular continued fraction with period 1. This is the even case. For the general case see A087475 with the Schroeder reference and comments. For the odd case see A078370.
a(n1) counts configurations of nonattacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419].  R. J. Mathar, Jun 16 2011
If h (5,17,37,65,101,..) is prime is relatively prime to 6, then h^21 is divisible by 24.  Vincenzo Librandi, Apr 14 2014
a(n) is also the number of permutations simultaneously avoiding 213 and 321 in the classical sense which can be realized as labels on an increasing strict binary tree with 2n1 nodes. See A245904 for more information on increasing strict binary trees.  Manda Riehl, Aug 07 2014
Sum_{n>=0} (1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... .  Vaclav Kotesovec, Feb 14 2015
a(n1) is the maximum number of stages in the GaleShapley algorithm for finding a stable matching between two sets of n elements given an ordering of preferences for each element (see Gura et al.).  Melvin Peralta, Feb 07 2016
Because of Fermat's little theorem, a(n) is never divisible by 3.  Altug Alkan, Apr 08 2016
For n > 0, if a(n) points are placed inside an n X n square, it will always be the case that at least two of the points will be a distance of sqrt(2) units apart or less.  Melvin Peralta, Jan 21 2017
Also the limit as q>1^ of the unimodal polynomial (1q^(n*k+1))/(1q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of qbinomials after making the restriction to partitions of size <=1. See G_1(n,k) from arXiv:1711.11252. As the size restriction s increases, G_s>G_infinity=G: the qbinomials. Then substituting k=n and q=1 yields the central binomial coefficients: A000984.  Bryan T. Ek, Apr 11 2018
a(n) is the smallest number congruent to both 1 (mod n), and 2 (mod n+1).  David James Sycamore, Apr 04 2019
a(n) is the number of permutations of 1,2,...,n+1 with exactly one reduced decomposition.  Richard Stanley, Dec 22 2022


REFERENCES

S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (see p. 120).
E. Gura and M. Maschler, Insights into Game Theory: An Alternative Mathematical Experience, Cambridge, 2008; p. 26.
L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 176.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.


LINKS



FORMULA

O.g.f.: (1x+2*x^2)/((1x)^3).  Eric Werley, Jun 27 2011
Sequences of the form a(n) = n^2 + K with offset 0 have o.g.f. (K  2*K*x + K*x^2 + x + x^2)/(1x)^3 and recurrence a(n) = 3*a(n1)  3*a(n2) + a*(n3).  R. J. Mathar, Apr 28 2008
For n > 1, a(n)^2 + (a(n) + 1)^2 + ... + (a(n) + n  2)^2 + (a(n) + n  1 + a(n) + n)^2 = (n+1) *(6*n^4 + 18*n^3 + 26*n^2 + 19*n + 6) / 6 = (a(n) + n)^2 + ... + (a(n) + 2*n)^2.  Charlie Marion, Jan 10 2011
a(n) = 2*a(n1)  a(n2) + 2.
a(n) = a(n1) + 2*n  1. (End)
a(n) = (n1)^2 + 2(n1) + 2 = 122 read in base n1 (for n > 3).  Jason Kimberley, Oct 20 2011
a(n)*a(n+1) = a(n*(n+1) + 1) so a(1)*a(2) = a(3). More generally, a(n)*a(n+k) = a(n*(n+k) + 1) + k^2  1.  Jon Perry, Aug 01 2012
a(n) = (n!)^2* [x^n] BesselI(0, 2*sqrt(x))*(1+x).  Peter Luschny, Aug 25 2012
Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi)*sinh(sqrt(2)*Pi).
Product_{n>=1} (1  1/a(n)) = Pi*csch(Pi). (End)


EXAMPLE

G.f. = 1 + 2*x + 5*x^2 + 10*x^3 + 17*x^4 + 26*x^5 + 37*x^6 + 50*x^7 + 65*x^8 + ...


MAPLE

numtheory[cyclotomic](4, n) ;
end proc:


MATHEMATICA



PROG

(Haskell)
a002522 = (+ 1) . (^ 2)
a002522_list = scanl (+) 1 [1, 3..]


CROSSREFS

Cf. A059592, A124808, A132411, A132414, A028872, A005408, A000124, A016813, A086514, A000125, A058331, A080856, A000127, A161701A161704, A161706, A161707, A161708, A161710A161713, A161715, A006261.
Moore lower bound on the order of a (k,g) cage: A198300 (square); rows: A000027 (k=2), A027383 (k=3), A062318 (k=4), A061547 (k=5), A198306 (k=6), A198307 (k=7), A198308 (k=8), A198309 (k=9), A198310 (k=10), A094626 (k=11); columns: A020725 (g=3), A005843 (g=4), this sequence (g=5), A051890 (g=6), A188377 (g=7).  Jason Kimberley, Oct 30 2011


KEYWORD

nonn,easy


AUTHOR



EXTENSIONS



STATUS

approved



