

A002522


a(n) = n^2 + 1.


358



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



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
From Cino Hilliard, Feb 21 2006: (Start)
Also, except for the first term, numbers that cannot be expressed as a perfect power, i.e., x^2 + 1 != y^n for all x,y,n > 1. Proof: We assume the truth of the following theorem. Proofs can be found in elementary texts on number theory and online. Theorem I: A number N is a sum of two squares if and only if all prime factors of N of the form 4m+3 have even exponents.
We are now ready to prove x^2 + 1 != y^n for all x,y,n > 1. We assume equality and seek a contradiction for n even and n odd. If n is even = 2k, x^2 + 1 = y^2k = (y^k)^2 and (y^k  x)(y^k + x) = 1. This implies y^kx = y^k+x = 1 or 2x = 0 contrary to x > 1. So n must be odd for equality to hold.
Then x^2+1 = y^(2k+1) implies all prime factors of y, including those of the form 4m+3 are raised to an odd exponent contrary to Theorem I. So we have shown x^2+1 = y^n is false for n even or n odd. Therefore x^2 + 1 != y^n as was desired. (End)
Note that in the above proof, y doesn't necessarily have any prime factors of the form 4m+3.  Jon Perry, Aug 06 2012
Also, numbers m such m^3m^2 is a perfect square, (n*(1 + n^2))^2.  Zak Seidov
1 + 2/2 + 2/5 + 2/10 +...= Pi*coth Pi [Jolley], see A113319.  Gary W. Adamson, Dec 21 2006
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
Sequence allows us to find X values of 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 (bhmd95(AT)yahoo.fr), Nov 29 2007
For n>0: a(n1)=A143053(A000290(n))1.  Reinhard Zumkeller, Jul 20 2008
A143053(a(n))=A000290(n+1).  Reinhard Zumkeller, Jul 20 2008
a(n) = A156798(n)/A087475(n).  Reinhard Zumkeller, Feb 16 2009
{a(k): 0 <= k < 4} = divisors of 10.  Reinhard Zumkeller, Jun 17 2009
Number of units of a(n) belongs to a periodic sequence: 1, 2, 5, 0, 7, 6, 7, 0, 5, 2.  Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 04 2009
a(n)=A170949(A002061(n+1)); A170949(a(n))=A132411(n+1); A170950(a(n))=A002061(n+1).  Reinhard Zumkeller, Mar 08 2010
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(xj,j=0..k1)(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 length 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
Also numbers n such that 4*n4 is a square. Hence this sequence is the union of A053755 and A069894.  Arkadiusz Wesolowski, Aug 02 2011
a(n) is also the Moore lower bound on the order, A191595(n), of an (n,5)cage.  Jason Kimberley, Oct 17 2011
Left edge of the triangle in A195437: a(n+1) = A195437(n,0).  Reinhard Zumkeller, Nov 23 2011
a(n) = A070216(n,1) for n > 0.  Reinhard Zumkeller, Nov 11 2012
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
The identity (4*n^2+2)^2(n^2+1)*(4*n)^2=4 can be written as A005899(n)^2a(n)*A008586(n)^2=4.  Vincenzo Librandi, Jun 15 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
a(n) = A254858(n2,3) for n > 2.  Reinhard Zumkeller, Feb 09 2015
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
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674047468581174... .  Vaclav Kotesovec, Apr 10 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


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

Vincenzo Librandi, Table of n, a(n) for n = 0..1000. Format corrected by Peter Kagey, Jan 25 2016
R. P. Boas & N. J. A. Sloane, Correspondence, 1974
S. Chaiken et al., Nonattacking Queens in a Rectangular Strip, arXiv:1105.5087 [math.CO], 2011.
Bryan Ek, Unimodal Polynomials and Lattice Walk Enumeration with Experimental Mathematics, arXiv:1804.05933 [math.CO], 2018.
GuoNiu Han, Enumeration of Standard Puzzles
GuoNiu Han, Enumeration of Standard Puzzles [Cached copy]
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv preprint 1410.2657 [math.CO], 2014.
C. Homberger, V. Vatter, On the effective and automatic enumeration of polynomial permutation classes, arXiv preprint arXiv:1308.4946 [math.CO], 2013.
S. J. Leon, Linear Algebra with Applications: the PerronFrobenius heorem
T. Mansour and J. West, Avoiding 2letter signed patterns, arXiv:math/0207204 [math.CO], 2002.
Michelle RudolphLilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv preprint arXiv:1508.07894 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Number Picking
Eric Weisstein's World of Mathematics, NearSquare Prime
Helmut Wielandt, Unzerlegbare nicht negativen Matrizen, Math. Z. 52 (1950), 642648.
Reinhard Zumkeller, Enumerations of Divisors
Index to values of cyclotomic polynomials of integer argument
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


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
a(n)*a(n2) = (n1)^4 + 4.  Reinhard Zumkeller, Feb 12 2009
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.  Eric Werley, Jun 27 2011
a(n) = (n1)^2 + 2(n1) + 2 = 122 read in base n  1 (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
E.g.f.: exp(x)*(1 + x + x^2).  Geoffrey Critzer, Aug 30 2013
4*a(n) = A001105(n1) + A001105(n+1).  Bruno Berselli, Jul 03 2017


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

A002522 := proc(n)
numtheory[cyclotomic](4, n) ;
end proc:
seq(A002522(n), n=0..20) ; # R. J. Mathar, Feb 07 2014


MATHEMATICA

Table[n^2 + 1, {n 0, 50}]; (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)


PROG

(MAGMA) [n^2 + 1: n in [0..50]]; // Vincenzo Librandi, May 01 2011
(PARI) a(n)=n^2+1 \\ Charles R Greathouse IV, Jun 10 2011
(Haskell)
a002522 = (+ 1) . (^ 2)
a002522_list = scanl (+) 1 [1, 3..]
 Reinhard Zumkeller, Apr 06 2012
(Maxima) A002522(n):=n^2+1$ makelist(A002522(n), n, 0, 30); /* Martin Ettl, Nov 07 2012 */


CROSSREFS

Left edge of A055096.
Cf. A059100, A117950, A087475, A117951, A114949, A117619 (sequences of form n^2 + K).
a(n+1) = A101220(n, n+1, 3).
Cf. A059592, A124808, A117950, 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
Cf. A002496 (primes).
Cf. A254858.
Cf. A302612, A302644, A302645, A302646.
Sequence in context: A082607 A303372 A159547 * A217990 A300164 A248193
Adjacent sequences: A002519 A002520 A002521 * A002523 A002524 A002525


KEYWORD

nonn,easy,changed


AUTHOR

N. J. A. Sloane


EXTENSIONS

More terms from Zerinvary Lajos, May 28 2006
Partially edited by Joerg Arndt, Mar 11 2010


STATUS

approved



