A069894 -id:A069894 - cp's OEIS frontend

A002522 a(n) = n^2 + 1.

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

Offset: 0

N. J. A. Sloane

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 k-th 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(n-1) + 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^2-2n+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
1 + 2/2 + 2/5 + 2/10 + ... = Pi*coth Pi [Jolley], see A113319. - Gary W. Adamson, Dec 21 2006
For n >= 1, a(n-1) is the minimal number of choices from an n-set 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
{a(k): 0 <= k < 4} = divisors of 10. - Reinhard Zumkeller, Jun 17 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 m-th 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..k-1} (x-j) (falling factorials). See the T. Koshy reference, pp. 263-4 (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(n-1) counts configurations of non-attacking bishops on a 2 X n strip [Chaiken et al., Ann. Combin. 14 (2010) 419]. - R. J. Mathar, Jun 16 2011
Also numbers k such that 4*k-4 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
If h (5,17,37,65,101,...) is prime is relatively prime to 6, then h^2-1 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)^2 - a(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 2n-1 nodes. See A245904 for more information on increasing strict binary trees. - Manda Riehl, Aug 07 2014
a(n-1) is the maximum number of stages in the Gale-Shapley 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 (1-q^(n*k+1))/(1-q) after making the simplification k=n. The unimodal polynomial is from O'Hara's proof of unimodality of q-binomials 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 q-binomials. 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
From Klaus Purath, Apr 03 2025: (Start)
The odd prime factors of these terms are always of the form 4*k + 1.
All a(n) = D satisfy the Pell equation (k*x)^2 - D*y^2 = -1. The values for k and the solutions x, y can be calculated using the following algorithm: k = n, x(0) = 1, x(1) = 4*D - 1, y(0) = 1, y(1) = 4*D - 3. The two recurrences are of the form (4*D - 2, -1). The solutions x, y of the Pell equations for n = {1 ... 14} are in OEIS.
It follows from the above that this sequence is a subsequence of A031396. (End)

			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 + ...

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.
Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley and Sons, New York, 2001.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000. Format corrected by _Peter Kagey_, Jan 25 2016
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, On the Salient Regularities of Strings of Assembly Theory, Preprints (2024). See p. 19.
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints.org (2025). See p. 28.
R. P. Boas and N. J. A. Sloane, Correspondence, 1974
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 44, 56.
Giulio Cerbai and Luca Ferrari, Permutation patterns in genome rearrangement problems: the reversal model, arXiv:1903.08774 [math.CO], 2019. See p. 19.
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.
R. M. Green and Tianyuan Xu, 2-roots for simply laced Weyl groups, arXiv:2204.09765 [math.RT], 2022.
Guo-Niu Han, Enumeration of Standard Puzzles
Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]
Cheyne Homberger, Patterns in Permutations and Involutions: A Structural and Enumerative Approach, arXiv:1410.2657 [math.CO], 2014.
C. Homberger and V. Vatter, On the effective and automatic enumeration of polynomial permutation classes, arXiv:1308.4946 [math.CO], 2013.
L. B. W. Jolley, Summation of Series, Dover, 1961, p. 176.
S. J. Leon, Linear Algebra with Applications: the Perron-Frobenius Theorem [Cached copy at the Wayback Machine]
T. Mansour and J. West, Avoiding 2-letter signed patterns, arXiv:math/0207204 [math.CO], 2002.
Michelle Rudolph-Lilith, On the Product Representation of Number Sequences, with Application to the Fibonacci Family, arXiv:1508.07894 [math.NT], 2015.
Eric Weisstein's World of Mathematics, Number Picking
Eric Weisstein's World of Mathematics, Near-Square Prime
Helmut Wielandt, Unzerlegbare, nicht negative Matrizen, Math. Z. 52 (1950), 642-648. volume 52
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).

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, A132411, A132414, A028872, A005408, A000124, A016813, A086514, A000125, A058331, A080856, A000127, A161701-A161704, A161706, A161707, A161708, A161710-A161713, 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.
Subsequence of A031396.

a002522 = (+ 1) . (^ 2)
a002522_list = scanl (+) 1 [1,3..]
-- Reinhard Zumkeller, Apr 06 2012

[n^2 + 1: n in [0..50]]; // Vincenzo Librandi, May 01 2011

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

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

A002522(n):=n^2+1$ makelist(A002522(n),n,0,30); /* Martin Ettl, Nov 07 2012 */

a(n)=n^2+1 \\ Charles R Greathouse IV, Jun 10 2011

O.g.f.: (1-x+2*x^2)/((1-x)^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)/(1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a*(n-3). - R. J. Mathar, Apr 28 2008
For n > 0: a(n-1) = A143053(A000290(n)) - 1. - Reinhard Zumkeller, Jul 20 2008
A143053(a(n)) = A000290(n+1). - Reinhard Zumkeller, Jul 20 2008
a(n)*a(n-2) = (n-1)^4 + 4. - Reinhard Zumkeller, Feb 12 2009
a(n) = A156798(n)/A087475(n). - Reinhard Zumkeller, Feb 16 2009
From Reinhard Zumkeller, Mar 08 2010: (Start)
a(n) = A170949(A002061(n+1));
A170949(a(n)) = A132411(n+1);
A170950(a(n)) = A002061(n+1). (End)
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
From Eric Werley, Jun 27 2011: (Start)
a(n) = 2*a(n-1) - a(n-2) + 2.
a(n) = a(n-1) + 2*n - 1. (End)
a(n) = (n-1)^2 + 2(n-1) + 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
a(n) = A070216(n,1) for n > 0. - Reinhard Zumkeller, Nov 11 2012
E.g.f.: exp(x)*(1 + x + x^2). - Geoffrey Critzer, Aug 30 2013
a(n) = A254858(n-2,3) for n > 2. - Reinhard Zumkeller, Feb 09 2015
Sum_{n>=0} (-1)^n / a(n) = (1+Pi/sinh(Pi))/2 = 0.636014527491... = A367976 . - Vaclav Kotesovec, Feb 14 2015
Sum_{n>=0} 1/a(n) = (1 + Pi*coth(Pi))/2 = 2.076674... = A113319. - Vaclav Kotesovec, Apr 10 2016
4*a(n) = A001105(n-1) + A001105(n+1). - Bruno Berselli, Jul 03 2017
From Amiram Eldar, Jan 20 2021: (Start)
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)
Sum_{n>=0} a(n)/n! = 3*e. - Davide Rotondo, Feb 16 2025

Partially edited by Joerg Arndt, Mar 11 2010

A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

Original entry on oeis.org

1, 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513

Offset: 0

N. J. A. Sloane

These are Hogben's central polygonal numbers denoted by
...2...
....P..
...4.n.
Numbers of the form (k^2+1)/2 for k odd.
(y(2x+1))^2 + (y(2x^2+2x))^2 = (y(2x^2+2x+1))^2. E.g., let y = 2, x = 1; (2(2+1))^2 + (2(2+2))^2 = (2(2+2+1))^2, (2(3))^2 + (2(4))^2 = (2(5))^2, 6^2 + 8^2 = 10^2, 36 + 64 = 100. - Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 08 2002
a(n) is also the number of 3 X 3 magic squares with sum 3(n+1). - Sharon Sela (sharonsela(AT)hotmail.com), May 11 2002
For n > 0, a(n) is the smallest k such that zeta(2) - Sum_{i=1..k} 1/i^2 <= zeta(3) - Sum_{i=1..n} 1/i^3. - Benoit Cloitre, May 17 2002
Number of convex polyominoes with a 2 X (n+1) minimal bounding rectangle.
The prime terms are given by A027862. - Lekraj Beedassy, Jul 09 2004
First difference of a(n) is 4n = A008586(n). Any entry k of the sequence is followed by k + 2*(1 + sqrt(2k - 1)). - Lekraj Beedassy, Jun 04 2006
Integers of the form 1 + x + x^2/2 (generating polynomial is Schur's polynomial as in A127876). - Artur Jasinski, Feb 04 2007
If X is an n-set and Y and Z disjoint 2-subsets of X then a(n-4) is equal to the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Row sums of triangle A132778. - Gary W. Adamson, Sep 02 2007
Binomial transform of [1, 4, 4, 0, 0, 0, ...]; = inverse binomial transform of A001788: (1, 6, 24, 80, 240, ...). - Gary W. Adamson, Sep 02 2007
Narayana transform (A001263) of [1, 4, 0, 0, 0, ...]. Equals A128064 (unsigned) * [1, 2, 3, ...]. - Gary W. Adamson, Dec 29 2007
k such that the Diophantine equation x^3 - y^3 = x*y + k has a solution with y = x-1. If that solution is (x,y) = (m+1,m) then m^2 + (m+1)^2 = k. Note that this Diophantine equation is an elliptic curve and (m+1,m) is an integer point on it. - James R. Buddenhagen, Aug 12 2008
Numbers k such that (k, k, 2*k-2) are the sides of an isosceles triangle with integer area. Also, k such that 2*k-1 is a square. - James R. Buddenhagen, Oct 17 2008
a(n) is also the least weight of self-conjugate partitions having n+1 different odd parts. - Augustine O. Munagi, Dec 18 2008
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) = A153869 * (1, 2, 3, ...). - Gary W. Adamson, Jan 03 2009
Prefaced with a "1": (1, 1, 5, 13, 25, 41, ...) where a(n) = 2n*(n-1)+1, all tuples of square numbers (X-Y, X, X+Y) are produced by ((m*(a(n)-2n))^2, (m*a(n))^2, (m*(a(n)+2n-2))^2) where m is a whole number. - Doug Bell, Feb 27 2009
Equals (1, 2, 3, ...) convolved with (1, 3, 4, 4, 4, ...). E.g., a(3) = 25 = (1, 2, 3, 4) dot (4, 4, 3, 1) = (4 + 8 + 9 + 4). - Gary W. Adamson, May 01 2009
The running sum of squares taken two at a time. - Al Hakanson (hawkuu(AT)gmail.com), May 18 2009
Equals the odd integers convolved with (1, 2, 2, 2, ...). - Gary W. Adamson, May 25 2009
Equals the triangular numbers convolved with [1, 2, 1, 0, 0, 0, ...]. - Gary W. Adamson & Alexander R. Povolotsky, May 29 2009
When the positive integers are written in a square array by diagonals as in A038722, a(n) gives the numbers appearing on the main diagonal. - Joshua Zucker, Jul 07 2009
The finite continued fraction [n,1,1,n] = (2n+1)/(2n^2 + 2n + 1) = (2n+1)/a(n); and the squares of the first two denominators of the convergents = a(n). E.g., the convergents and value of [4,1,1,4] = 1/4, 1/5, 2/9, 9/41 where 4^2 + 5^2 = 41. - Gary W. Adamson, Jul 15 2010
From Keith Tyler, Aug 10 2010: (Start)
Running sum of A008574.
Square open pyramidal number; that is, the number of elements in a square pyramid of height (n) with only surface and no bottom nodes. (End)
For k>0, x^4 + x^2 + k factors over the integers iff sqrt(k) is in this sequence. - James R. Buddenhagen, Aug 15 2010
Create the simple continued fraction from Pythagorean triples to get [2n + 1; 2n^2 + 2n, 2n^2 + 2n + 1]; its value equals the rational number 2n + 1 + a(n) / (4n^4 + 8n^3 + 6n^2 + 2n + 1). - J. M. Bergot, Sep 30 2011
a(n), n >= 1, has in its prime number factorization only primes of the form 4*k+1, i.e., congruent to 1 (mod 4) (see A002144). This follows from the fact that a(n) is a primitive sum of two squares and odd. See Theorem 3.20, p. 164, in the given Niven-Zuckerman-Montgomery reference. E.g., a(3) = 25 = 5^2, a(6) = 85 = 5*17. - Wolfdieter Lang, Mar 08 2012
From Ant King, Jun 15 2012: (Start)
a(n) is congruent to 1 (mod 4) for all n.
The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 5, 4, 7, 5, 7, 4, 5, 1.
The units' digits of the a(n) form a purely periodic palindromic 5-cycle 1, 5, 3, 5, 1.
(End)
Number of integer solutions (x,y) of |x| + |y| <= n. Geometrically: number of lattice points inside a square with vertices (n,0), (0,-n), (-n,0), (0,n). - César Eliud Lozada, Sep 18 2012
(a(n)-1)/a(n) = 2*x / (1+x^2) where x = n/(n+1). Note that in this form, this is the velocity-addition formula according to the special theory of relativity (two objects traveling at 1/(n+1) slower than c relative to each other appear to travel at 1/a(n) less than c to a stationary observer). - Christian N. K. Anderson, May 20 2013 [Corrected by Rémi Guillaume, May 22 2025]
A geometric curiosity: the envelope of the circles x^2 + (y-a(n)/2)^2 = ((2n+1)/2)^2 is the parabola y = x^2, the n=0 circle being the osculating circle at the parabola vertex. - Jean-François Alcover, Dec 02 2013
Draw n ellipses in the plane (n>0), any 2 meeting in 4 points; a(n-1) gives the number of internal regions into which the plane is divided (cf. A051890, A046092); a(n-1) = A051890(n) - 1 = A046092(n-1) + 1. - Jaroslav Krizek, Dec 27 2013
a(n) is also, of course, the scalar product of the 2-vector (n, n+1) (or (n+1, n)) with itself. The unique inverse of (n, n+1) as vector in the Clifford algebra over the Euclidean 2-space is (1/a(n))(0, n, n+1, 0) (similarly for the other vector). In general the unique inverse of such a nonzero vector v (odd element in Cl_2) is v^(-1) = (1/|v|^2) v. Note that the inverse with respect to the scalar product is not unique for any nonzero vector. See the P. Lounesto reference, sects. 1.7 - 1.12, pp. 7-14. See also the Oct 15 2014 comment in A147973. - Wolfdieter Lang, Nov 06 2014
Subsequence of A004431, for n >= 1. - Bob Selcoe, Mar 23 2016
Numbers k such that 2k - 1 is a perfect square. - Juri-Stepan Gerasimov, Apr 06 2016
The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 574", based on the 5-celled von Neumann neighborhood. - Robert Price, May 13 2016
a(n) is the first integer in a sum of (2*n + 1)^2 consecutive integers that equals (2*n + 1)^4. - Patrick J. McNab, Dec 24 2016
Central elements of odd-length rows of the triangular array of positive integers. a(n) is the mean of the numbers in the (2*n + 1)-th row of this triangle. - David James Sycamore, Aug 01 2018
Intersection of A000982 and A080827. - David James Sycamore, Aug 07 2018
An off-diagonal of the array of Delannoy numbers, A008288, (or a row/column when the array is shown as a square). As such, this is one of the crystal ball sequences. - Jack W Grahl, Feb 15 2021 and Shel Kaphan, Jan 18 2023
a(n) appears as a solution to a "Riddler Express" puzzle on the FiveThirtyEight website. The Jan 21 2022 issue (problem) and the Jan 28 2022 issue (solution) present the following puzzle and include a proof. - Fold a square piece of paper in half, obtaining a rectangle. Fold again to obtain a square with 1/4 the size of the original square. Then make n cuts through the folded paper. a(n) is the greatest number of pieces of the unfolded paper after the cutting. - Manfred Boergens, Feb 22 2022
a(n) is (1/6) times the number of 2 X 2 triangles in the n-th order hexagram with 12*n^2 cells. - Donghwi Park, Feb 06 2024
If k is a centered square number, its index in this sequence is n = (sqrt(2k-1)-1)/2. - Rémi Guillaume, Mar 30 2025.
Row sums of the symmetric triangle of odd numbers [1]; [1, 3, 1]; [1, 3, 5, 3, 1]; [1, 3, 5, 7, 5, 3, 1]; .... - Marco Zárate, Jun 15 2025

			G.f.: 1 + 5*x + 13*x^2 + 25*x^3 + 41*x^4 + 61*x^5 + 85*x^6 + 113*x^7 + 145*x^8 + ...
The first few triples are (1,0,1), (3,4,5), (5,12,13), (7,24,25), ...
The first four such partitions, corresponding to n = 0,1,2,3, i.e., to a(n) = 1,5,13,25, are 1, 3+1+1, 5+3+3+1+1, 7+5+5+3+3+1+1. - _Augustine O. Munagi_, Dec 18 2008

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, p. 125, 1964.
L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
Pertti Lounesto, Clifford Algebras and Spinors, second edition, Cambridge University Press, 2001.
S. Mukai, An Introduction to Invariants and Moduli, Cambridge, 2003; see p. 483.
Ivan Niven, Herbert S. Zuckerman and Hugh L. Montgomery, An Introduction to the Theory Of Numbers, Fifth Edition, John Wiley and Sons, Inc., NY 1991.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Travers et al., The Mysterious Lost Proof, Using Advanced Algebra, (1976), pp. 27.

T. D. Noe, Table of n, a(n) for n = 0..1000
M. Ahmed, J. De Loera and R. Hemmecke, Polyhedral Cones of Magic Cubes and Squares, arXiv:math/0201108 [math.CO], 2002.
U. Alfred, n and n+1 consecutive integers with equal sums of squares, Math. Mag., 35 (1962), 155-164.
Bela Bajnok, Additive Combinatorics: A Menu of Research Problems, arXiv:1705.07444 [math.NT], May 2017. See Section 2.3.
Paul Barry, Centered polygon numbers, heptagons and nonagons, and the Robbins numbers, arXiv:2104.01644 [math.CO], 2021.
Matthias Beck, Moshe Cohen, Jessica Cuomo, and Paul Gribelyuk, The number of "magic" squares and hypercubes, arXiv:math/0201013 [math.CO], 2002-2005.
Arthur T. Benjamin and Doron Zeilberger, Pythagorean Primes and Palindromic Continued Fractions, Electronic Journal of Combinatorial Number Theory, 5(1) 2005, #A30.
J. A. De Loera, D. C. Haws and M. Koppe, Ehrhart Polynomials of Matroid Polytopes and Polymatroids, arXiv:0710.4346 [math.CO], 2007; Discrete Comput. Geom., 42 (2009), 670-702.
FiveThirtyEight, "Riddler Express" paper cutting problem and solution, Jan 28 2022.
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Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.
Clark Kimberling, Complementary Equations, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.4.
Clark Kimberling and John E. Brown, Partial Complements and Transposable Dispersions, J. Integer Seqs., Vol. 7, 2004.
Ron Knott, Pythagorean Triples and Online Calculators.
G. Kreweras, Sur les hiérarchies de segments, Cahiers Bureau Universitaire Recherche Opérationnelle, Cahier 20, Inst. Statistiques, Univ. Paris, 1973.
G. Kreweras, Sur les hiérarchies de segments, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, #20 (1973). (Annotated scanned copy)
A. O. Munagi, Pairing conjugate partitions by residue classes, Discrete Math., 308 (2008), 2492-2501.
Mitchell Paukner, Lucy Pepin, Manda Riehl, and Jarred Wieser, Pattern Avoidance in Task-Precedence Posets, arXiv:1511.00080 [math.CO], 2015-2016.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
John A. Jr. Rochowicz, Harmonic Numbers: Insights, Approximations and Applications, Spreadsheets in Education (eJSiE), 2015, Vol. 8: Iss. 2, Article 4.
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David James Sycamore, Triangular array
Leo Tavares, Illustration: Diamond Rows
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Index entries for sequences related to centered polygonal numbers
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
Nicolay Avilov, Graphical representation of the sequence members

X values are A005408; Y values are A046092.
Cf. A008586 (first differences), A005900 (partial sums), A254373 (digital roots).
Cf. A069894, A000217, A000290, A001263, A001788, A001845, A002061, A002144, A003215, A005448, A005891, A005917, A008844 (square terms), A027862 (prime terms), A048395, A051890, A056106, A101096, A127876, A128064, A132778, A147973, A153869, A240876, A251599, A000982, A080827.
Subsequence of A004431.
Right edge of A055096; main diagonal of A069480, A078475, A129312.
Row n=2 (or column k=2) of A008288.
Cf. A016754.

a001844 n = 2 * n * (n + 1) + 1
a001844_list = zipWith (+) a000290_list $ tail a000290_list
-- Reinhard Zumkeller, Dec 04 2012

[2*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Jan 19 2013

[n: n in [0..4400] | IsSquare(2*n-1)]; // Juri-Stepan Gerasimov, Apr 06 2016

A001844:=-(z+1)**2/(z-1)**3; # Simon Plouffe in his 1992 dissertation

Table[2n(n + 1) + 1, {n, 0, 50}]
FoldList[#1 + #2 &, 1, 4 Range@ 50] (* Robert G. Wilson v, Feb 02 2011 *)
maxn := 47; Flatten[Table[SeriesCoefficient[Series[(n + (n - 1)*x)/(1 - x)^2, {x, 0, maxn}], k], {n, maxn}, {k, n - 1, n - 1}]] (* L. Edson Jeffery, Aug 24 2014 *)
CoefficientList[ Series[-(x^2 + 2x + 1)/(x - 1)^3, {x, 0, 48}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {1, 5, 13}, 48] (* Robert G. Wilson v, Aug 01 2018 *)
Total/@Partition[Range[0,50]^2,2,1] (* Harvey P. Dale, Dec 05 2020 *)
Table[ j! Coefficient[Series[Exp[x]*(1 + 4*x + 2*x^2), {x, 0, 20}], x,
j], {j, 0, 20}] (* Nikolaos Pantelidis, Feb 07 2023 *)

PARI
```
{a(n) = 2*n*(n+1) + 1};
```

x='x+O('x^200); Vec((1+x)^2/(1-x)^3) \\ Altug Alkan, Mar 23 2016

print([2*n*(n+1)+1 for n in range(48)]) # Michael S. Branicky, Jan 05 2021

[i**2 + (i + 1)**2 for i in range(46)] # Zerinvary Lajos, Jun 27 2008

a(n) = 2*n^2 + 2*n + 1 = n^2 + (n+1)^2.
a(n) = 1 + 3 + 5 + ... + 2*n-1 + 2*n+1 + 2*n-1 + ... + 3 + 1. - Amarnath Murthy, May 28 2001
a(n) = 1/real(z(n+1)) where z(1)=i, (i^2=-1), z(k+1) = 1/(z(k)+2i). - Benoit Cloitre, Aug 06 2002
Nearest integer to 1/Sum_{k>n} 1/k^3. - Benoit Cloitre, Jun 12 2003
G.f.: (1+x)^2/(1-x)^3.
E.g.f.: exp(x)*(1+4x+2x^2).
a(n) = a(n-1) + 4n.
a(-n) = a(n-1).
a(n) = A064094(n+3, n) (fourth diagonal).
a(n) = 1 + Sum_{j=0..n} 4*j. - Xavier Acloque, Oct 08 2003
a(n) = A046092(n)+1 = (A016754(n)+1)/2. - Lekraj Beedassy, May 25 2004
a(n) = Sum_{k=0..n+1} (-1)^k*binomial(n, k)*Sum_{j=0..n-k+1} binomial(n-k+1, j)*j^2. - Paul Barry, Dec 22 2004
a(n) = ceiling((2n+1)^2/2). - Paul Barry, Jul 16 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=5, a(2)=13. - Jaume Oliver Lafont, Dec 02 2008
a(n)*a(n-1) = 4*n^4 + 1 for n > 0. - Reinhard Zumkeller, Feb 12 2009
Prefaced with a "1" (1, 1, 5, 13, 25, 41, ...): a(n) = 2*n*(n-1)+1. - Doug Bell, Feb 27 2009
a(n) = sqrt((A056220(n)^2 + A056220(n+1)^2) / 2). - Doug Bell, Mar 08 2009
a(n) = floor(2*(n+1)^3/(n+2)). - Gary Detlefs, May 20 2010
a(n) = A000330(n) - A000330(n-2). - Keith Tyler, Aug 10 2010
a(n) = A069894(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = 2*a(n-1) - a(n-2) + 4. - Ant King, Jun 12 2012
Sum_{n>=0} 1/a(n) = (Pi/2)*tanh(Pi/2) = 1.4406595199775... = A228048. - Ant King, Jun 15 2012
a(n) = A209297(2*n+1,n+1). - Reinhard Zumkeller, Jan 19 2013
a(n)^3 = A048395(n)^2 + A048395(-n-1)^2. - Vincenzo Librandi, Jan 19 2013
a(n) = A000217(2n+1) - n. - Ivan N. Ianakiev, Nov 08 2013
a(n) = A251599(3*n+1). - Reinhard Zumkeller, Dec 13 2014
a(n) = A101321(4,n). - R. J. Mathar, Jul 28 2016
From Ilya Gutkovskiy, Jul 30 2016: (Start)
a(n) = Sum_{k=0..n} A008574(k).
Sum_{n>=0} (-1)^(n+1)*a(n)/n! = exp(-1) = A068985. (End)
a(n) = 4 * A000217(n) + 1. - Bruce J. Nicholson, Jul 10 2017
a(n) = A002522(n) + A005563(n) = A002522(n+1) + A005563(n-1). - Bruce J. Nicholson, Aug 05 2017
Sum_{n>=0} a(n)/n! = 7*e. Sum_{n>=0} 1/a(n) = A228048. - Amiram Eldar, Jun 20 2020
a(n) = A000326(n+1) + A000217(n-1). - Charlie Marion, Nov 16 2020
a(n) = Integral_{x=0..2n+2} |1-x| dx. - Pedro Caceres, Dec 29 2020
From Amiram Eldar, Feb 17 2021: (Start)
Product_{n>=0} (1 + 1/a(n)) = cosh(sqrt(3)*Pi/2)*sech(Pi/2).
Product_{n>=1} (1 - 1/a(n)) = Pi*csch(Pi)*sinh(Pi/2). (End)
a(n) = A001651(n+1) + 1 - A028242(n). - Charlie Marion, Apr 05 2022
a(n) = A016754(n) - A046092(n). - Leo Tavares, Sep 16 2022
For n>0, a(n) = A101096(n+2) / 30. - Andy Nicol, Feb 06 2025
From Rémi Guillaume, Apr 21 2025: (Start)
a(n) = (2*A003215(n)+1)/3.
a(n) = (4*A005448(n+1)-1)/3.
a(n) + a(n-1) = A001845(n) - A001845(n-1), for n >= 1.
a(n) = (A005917(n+1))/(2n+1). (End)

Partially edited by Joerg Arndt, Mar 11 2010

A001653 Numbers k such that 2*k^2 - 1 is a square.

Original entry on oeis.org

1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925

Offset: 1

N. J. A. Sloane

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 (A069894).
(x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - Floor van Lamoen, Nov 29 2001
Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy, Jun 05 2002
Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - Benoit Cloitre, May 10 2003
Also, number of domino tilings in S_5 X P_2n. - Ralf Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.
If x is in the sequence then so is x*(8*x^2-3). - James R. Buddenhagen, Jan 13 2005
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, Jun 01 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n  >0, define C(n) to be the largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators = A002315 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008
Apparently Ljunggren shows that 169 is the last square term.
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p < r then r = 3p+2q+1 and s = 4p+3q+2. - Mohamed Bouhamida, Sep 02 2009
Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - Charlie Marion, May 02 2011
Apart from initial 1: subsequence of A198389, see also A198385. - Reinhard Zumkeller, Oct 25 2011
(a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, Mar 06 2012
Area of the Fibonacci snowflake of order n. - José Luis Ramírez Ramírez, Dec 13 2012
Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - José Luis Ramírez Ramírez, Dec 13 2012
For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014
Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014
Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014
The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - Stuart E Anderson, Feb 05 2015
Positive integers z such that z^2 is a centered square number (A001844). - Colin Barker, Feb 12 2015
The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - Peter Bala, Mar 25 2015
A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. A002315(n-1)/a(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - Bruno Berselli, Mar 19 2018
From Wolfdieter Lang, Jun 13 2018: (Start)
(a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture).
For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see (A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End)
Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k)  and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - Wolfdieter Lang, Jul 11 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=a(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
... (End)
G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - _Michael Somos_, Jun 26 2022

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Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
P. E. Trier, "Almost Isosceles" Right-Angled Triangles, Eureka, No. 4, May 1940, pp. 9 - 11.
Michel Waldschmidt, Continued fractions, Ecole de recherche CIMPA-Oujda, Théorie des Nombres et ses Applications, 18 - 29 mai 2015: Oujda (Maroc).
Eric Weisstein's World of Mathematics, NSW Number
Wikipedia, Tree of primitive Pythagorean triples.
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences, Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for two-way infinite sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Other two sides are A001652, A046090.
Cf. A001519, A001109, A005054, A122074, A056220, A056869 (subset of primes).
Cf. A000217, A000290, A002315, A002559, A005319, A069894.
Row 6 of array A094954.
Row 1 of array A188647.
Cf. similar sequences listed in A238379.

a:=[1,5];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

a001653 n = a001653_list !! n
a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
-- Reinhard Zumkeller, May 07 2013

I:=[1,5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 22 2014

a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
A001653:=-(-1+5*z)/(z**2-6*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

LinearRecurrence[{6,-1}, {1,5}, 40] (* Harvey P. Dale, Jul 12 2011 *)
a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* Michael Somos, Jul 22 2018 *)
Numerator[{1} ~Join~
Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* Greg Dresden, Sep 10 2019 *)

{a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* Michael Somos, Nov 02 2002 */

a(n)=([5,2;2,1]^(n-1))[1,1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de), corrected by Eric Chen, Jun 14 2018

{a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* Michael Somos, Jun 26 2022 */

G.f.: x*(1-x)/(1-6*x+x^2).
a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
4*a(n) = A077445(n).
Can be extended backwards by a(-n+1) = a(n).
a(n) = sqrt((A002315(n)^2 + 1)/2). [Inserted by N. J. A. Sloane, May 08 2000]
a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2), n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012]
a(n) = A000129(2n+1). - Ira M. Gessel, Sep 27 2002
a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 12 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 4) = a(n). - Benoit Cloitre, Nov 10 2002
For n and j >= 1, Sum_{k=0..j} a(k)*a(n) - Sum_{k=0..j-1} a(k)*a(n-1) = A001109(j+1)*a(n) - A001109(j)*a(n-1) = a(n+j); e.g., (1+5+29)*5 - (1+5)*1=169. - Charlie Marion, Jul 07 2003
From Charlie Marion, Jul 16 2003: (Start)
For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.
For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0..2*n-1} a(k) = 4*A001109(2n); e.g., 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416.
Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.
A001652(n) + A046090(n) - a(n) = A001542(n); e.g., 119 + 120 - 169 = 70.
(End)
Sum_{k=0...n} ((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g., 1*169 + 3*29 + 5*5 + 7*1 = 288 = 17^2 - 1; 1*29 + 3*5 + 5*1 = 49 = 7^2. - Charlie Marion, Jul 18 2003
Sum_{k=0...n} a(k)*a(n) = Sum_{k=0..n} a(2k) and Sum_{k=0..n} a(k)*a(n+1) = Sum_{k=0..n} a(2k+1); e.g., (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - Charlie Marion, Sep 22 2003
For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n > 0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n)) = Sum_{k=0..n} c(2*k+1); e.g., 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1)) = Sum_{k=0..n} c(2*k); e.g., 119*120*169/(20+21+29) = 1+29+985+33461 = 34476. - Charlie Marion, Dec 01 2003
Also solutions x > 0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2). - Benoit Cloitre, Feb 15 2004
a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
For n >= k, a(n)*a(n+2*k+1) - a(n+k)*a(n+k+1) = a(k)^2-1; e.g., 29*195025-985*5741 = 840 = 29^2-1; 1*169-5*29 = 24 = 5^2-1; a(n)*a(n+2*k)-a(n+k)^2 = A001542(k)^2; e.g., 169*195025-5741^2 = 144 = 12^2; 1*29-5^2 = 4 = 2^2. - Charlie Marion Jun 02 2004
For all k, a(n) is a factor of a((2n+1)*k+n). a((2*n+1)*k+n) = a(n)*(Sum_{j=0..k-1} (-1)^j*(a((2*n+1)*(k-j)) + a((2*n+1)*(k-j)-1))+(-1)^k); e.g., 195025 = 5*(33461+5741-169-29+1); 7645370045 = 169*(6625109+1136689-1).- Charlie Marion, Jun 04 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - Paul Barry, Aug 30 2004 [offset 0]
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - Paul Barry, Sep 30 2004 [offset 0]
For n < k, a(n)*A001541(k) = A011900(n+k)+A053141(k-n-1); e.g., 5*99 = 495 = 493+2. For n >= k, a(n)*A001541(k) = A011900(n+k)+A053141(n-k); e.g., 29*3 = 87 = 85+2. - Charlie Marion, Oct 18 2004
a(n) = (-1)^n*U(2*n, i*sqrt(4)/2) = (-1)^n*U(2*n, i), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - Paul Barry, Jul 18 2005 [offset 0]
a(n)*a(n+k) = A000129(k)^2 + A000129(2n+k+1)^2; e.g., 29*5741 = 12^2+169^2. - Charlie Marion, Aug 02 2005
Let a(n)*a(n+k) = x. Then 2*x^2-A001541(k)*x+A001109(k)^2 = A001109(2*n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2 = 40391^2; cf. A076218. - Charlie Marion, Aug 02 2005
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2*sqrt(2)). a(n) = A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - Lekraj Beedassy, Jul 19 2005
a(n) = Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*Pell(n-j+1), where Pell = A000129. - Paul Barry, May 19 2006 [offset 0]
a(n) = round(sqrt(A002315(n)^2/2)). - Lekraj Beedassy, Jul 15 2006
a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy, Aug 14 2006
a(n+1) = 3*a(n) + sqrt(8*a(n)^2-4), a(1)=1. - Richard Choulet, Sep 18 2007
6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29 = 29^2+5^2+4; 6*169*985 = 169^2+985^2+4. - Charlie Marion, Oct 07 2007
2*A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29 = 5^2+29^2+2^2; 2*99*29*5741 = 2*99*29*5741=29^2+5741^2+70^2. - Charlie Marion, Oct 12 2007
[a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
From Charlie Marion, Apr 10 2009: (Start)
In general, for n >= k, a(n+k) = 2*A001541(k)*a(n)-a(n-k);
e.g., a(n+0) = 2*1*a(n)-a(n); a(n+1) = 6*a(n)-a(n-1); a(6+0) = 33461 = 2*33461-33461; a(5+1) = 33461 = 6*5741-985; a(4+2) = 33461 = 34*985-29; a(3+3) = 33461 = 198*169-1.
(End)
G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - Johannes W. Meijer, Aug 01 2010
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(n) - a(n-1). - Charles L. Hohn, Apr 04 2011
Let T(n) be the n-th triangular number; then, for n > 0, T(a(n)) + A001109(n-1) = A046090(n)^2.  See also A046090. - Charlie Marion, Apr 25 2011
For k > 0, a(n+2*k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2*k) - a(n) = 4*A001109(k)*A002315(n+k-1). - Charlie Marion, Jan 06 2012
a(k+j+1) = (A001541(k)*A001541(j) + A002315(k)*A002315(j))/2. - Charlie Marion, Jun 25 2012
a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - Bruno Berselli, Oct 23 2012
a(n) = A143608(n-1)*A143608(n) + 1 = A182190(n-1)+1. - Charlie Marion, Dec 11 2012
G.f.: G(0)*(1-x)/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
a(n+1) = 4*A001652(n) + 3*a(n) + 2 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - Peter Bala, Mar 25 2015
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - David Pasino, Jul 09 2016
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + 2*cosh(2*sqrt(2)*x))*exp(3*x)/2. - Ilya Gutkovskiy, Jul 09 2016
a(n+2) = (a(n+1)^2 + 4)/a(n). - Vladimir M. Zarubin, Sep 06 2016
a(n) = 2*A053141(n)+1. - R. J. Mathar, Aug 16 2019
For n>1, a(n) is the numerator of the continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the denominators see A005319. - Greg Dresden, Sep 10 2019
a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - Paul Weisenhorn, May 23 2020
a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n+1) = 3*a(n) + A077444(n). - César Aguilera, Jul 13 2023

Additional comments from Wolfdieter Lang, Feb 10 2000
Better description from Harvey P. Dale, Jan 15 2002
Edited by N. J. A. Sloane, Nov 02 2002

A212959 Number of (w,x,y) such that w,x,y are all in {0,...,n} and |w-x| = |x-y|.

Original entry on oeis.org

1, 4, 11, 20, 33, 48, 67, 88, 113, 140, 171, 204, 241, 280, 323, 368, 417, 468, 523, 580, 641, 704, 771, 840, 913, 988, 1067, 1148, 1233, 1320, 1411, 1504, 1601, 1700, 1803, 1908, 2017, 2128, 2243, 2360, 2481, 2604, 2731, 2860, 2993, 3128, 3267

Offset: 0

Clark Kimberling, Jun 01 2012

In the following guide to related sequences: M=max(x,y,z), m=min(x,y,z), and R=range=M-m. In some cases, it is an offset of the listed sequence which fits the conditions shown for w,x,y. Each sequence satisfies a linear recurrence relation, some of which are identified in the list by the following code (signature):
  A: 2,  0, -2,  1, i.e., a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4);
  B: 3, -2, -2,  3, -1;
  C: 4, -6,  4, -1;
  D: 1,  2, -2, -1,  1;
  E: 2,  1, -4,  1,  2, -1;
  F: 2, -1,  1, -2,  1;
  G: 2, -1,  0,  1, -2,  1;
  H: 2, -1,  2, -4,  2, -1,  2, -1;
  I: 3, -3,  2, -3,  3, -1;
  J: 4, -7,  8, -7,  4, -1.
  ...
A212959 ... |w-x|=|x-y| ...... recurrence type A
A212960 ... |w-x| != |x-y| ................... B
A212683 ... |w-x| < |x-y| .................... B
A212684 ... |w-x| >= |x-y| ................... B
A212963 ... see entry for definition ......... B
A212964 ... |w-x| < |x-y| < |y-w| ............ B
A006331 ... |w-x| < y ........................ C
A005900 ... |w-x| <= y ....................... C
A212965 ... w = R ............................ D
A212966 ... 2*w = R
A212967 ... w < R ............................ E
A212968 ... w >= R ........................... E
A077043 ... w = x > R ........................ A
A212969 ... w != x and x > R ................. E
A212970 ... w != x and x < R ................. E
A055998 ... w = x + y - 1
A011934 ... w < floor((x+y)/2) ............... B
A182260 ... w > floor((x+y)/2) ............... B
A055232 ... w <= floor((x+y)/2) .............. B
A011934 ... w >= floor((x+y)/2) .............. B
A212971 ... w < floor((x+y)/3) ............... B
A212972 ... w >= floor((x+y)/3) .............. B
A212973 ... w <= floor((x+y)/3) .............. B
A212974 ... w > floor((x+y)/3) ............... B
A212975 ... R is even ........................ E
A212976 ... R is odd ......................... E
A212978 ... R = 2*n - w - x
A212979 ... R = average{w,x,y}
A212980 ... w < x + y and x < y .............. B
A212981 ... w <= x+y and x < y ............... B
A212982 ... w < x + y and x <= y ............. B
A212983 ... w <= x + y and x <= y ............ B
A002623 ... w >= x + y and x <= y ............ B
A087811 ... w = 2*x + y ...................... A
A008805 ... w = 2*x + 2*y .................... D
A000982 ... 2*w = x + y ...................... F
A001318 ... 2*w = 2*x + y .................... F
A001840 ... w = 3*x + y
A212984 ... 3*w = x + y
A212985 ... 3*w = 3*x + y
A001399 ... w = 2*x + 3*y
A212986 ... 2*w = 3*x + y
A008810 ... 3*x = 2*x + y .................... F
A212987 ... 3*w = 2*x + 2*y
A001972 ... w = 4*x + y ...................... G
A212988 ... 4*w = x + y ...................... G
A212989 ... 4*w = 4*x + y
A008812 ... 5*w = 2*x + 3*y
A016061 ... n < w + x + y <= 2*n ............. C
A000292 ... w + x + y <=n .................... C
A000292 ... 2*n < w + x + y <= 3*n ........... C
A212977 ... n/2 < w + x + y <= n
A143785 ... w < R < x ........................ E
A005996 ... w < R <= x ....................... E
A128624 ... w <= R <= x ...................... E
A213041 ... R = 2*|w - x| .................... A
A213045 ... R < 2*|w - x| .................... B
A087035 ... R >= 2*|w - x| ................... B
A213388 ... R <= 2*|w - x| ................... B
A171218 ... M < 2*m .......................... B
A213389 ... R < 2|w - x| ..................... E
A213390 ... M >= 2*m ......................... E
A213391 ... 2*M < 3*m ........................ H
A213392 ... 2*M >= 3*m ....................... H
A213393 ... 2*M > 3*m ........................ H
A213391 ... 2*M <= 3*m ....................... H
A047838 ... w = |x + y - w| .................. A
A213396 ... 2*w < |x + y - w| ................ I
A213397 ... 2*w >= |x + y - w| ............... I
A213400 ... w < R < 2*w
A069894 ... min(|w-x|,|x-y|) = 1
A000384 ... max(|w-x|,|x-y|) = |w-y|
A213395 ... max(|w-x|,|x-y|) = w
A213398 ... min(|w-x|,|x-y|) = x ............. A
A213399 ... max(|w-x|,|x-y|) = x ............. D
A213479 ... max(|w-x|,|x-y|) = w+x+y ......... D
A213480 ... max(|w-x|,|x-y|) != w+x+y ........ E
A006918 ... |w-x| + |x-y| > w+x+y ............ E
A213481 ... |w-x| + |x-y| <= w+x+y ........... E
A213482 ... |w-x| + |x-y| < w+x+y ............ E
A213483 ... |w-x| + |x-y| >= w+x+y ........... E
A213484 ... |w-x|+|x-y|+|y-w| = w+x+y
A213485 ... |w-x|+|x-y|+|y-w| != w+x+y ....... J
A213486 ... |w-x|+|x-y|+|y-w| > w+x+y ........ J
A213487 ... |w-x|+|x-y|+|y-w| >= w+x+y ....... J
A213488 ... |w-x|+|x-y|+|y-w| < w+x+y ........ J
A213489 ... |w-x|+|x-y|+|y-w| <= w+x+y ....... J
A213490 ... w,x,y,|w-x|,|x-y| distinct
A213491 ... w,x,y,|w-x|,|x-y| not distinct
A213493 ... w,x,y,|w-x|,|x-y|,|w-y| distinct
A213495 ... w = min(|w-x|,|x-y|,|w-y|)
A213492 ... w != min(|w-x|,|x-y|,|w-y|)
A213496 ... x != max(|w-x|,|x-y|)
A213498 ... w != max(|w-x|,|x-y|,|w-y|)
A213497 ... w = min(|w-x|,|x-y|)
A213499 ... w != min(|w-x|,|x-y|)
A213501 ... w != max(|w-x|,|x-y|)
A213502 ... x != min(|w-x|,|x-y|)
...
A211795 includes a guide for sequences that count 4-tuples (w,x,y,z) having all terms in {0,...,n} and satisfying selected properties.  Some of the sequences indexed at A211795 satisfy recurrences that are represented in the above list.
Partial sums of the numbers congruent to {1,3} mod 6 (see A047241). - Philippe Deléham, Mar 16 2014

			a(1)=4 counts these (x,y,z): (0,0,0), (1,1,1), (0,1,0), (1,0,1).
Numbers congruent to {1, 3} mod 6: 1, 3, 7, 9, 13, 15, 19, ...
a(0) = 1;
a(1) = 1 + 3 = 4;
a(2) = 1 + 3 + 7 = 11;
a(3) = 1 + 3 + 7 + 9 = 20;
a(4) = 1 + 3 + 7 + 9 + 13 = 33;
a(5) = 1 + 3 + 7 + 9 + 13 + 15 = 48; etc. - _Philippe Deléham_, Mar 16 2014

A. Barvinok, Lattice Points and Lattice Polytopes, Chapter 7 in Handbook of Discrete and Computational Geometry, CRC Press, 1997, 133-152.
P. Gritzmann and J. M. Wills, Lattice Points, Chapter 3.2 in Handbook of Convex Geometry, vol. B, North-Holland, 1993, 765-797.

Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A047241, A211795.

t = Compile[{{n, _Integer}}, Module[{s = 0},
(Do[If[Abs[w - x] == Abs[x - y], s = s + 1],
{w, 0, n}, {x, 0, n}, {y, 0, n}]; s)]];
m = Map[t[#] &, Range[0, 50]]   (* A212959 *)

a(n)=(6*n^2+8*n+3)\/4 \\ Charles R Greathouse IV, Jul 28 2015

a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4).
G.f.: (1+2*x+3*x^2)/((1+x)*(1-x)^3).
a(n) + A212960(n) = (n+1)^3.
a(n) = (6*n^2 + 8*n + 3 + (-1)^n)/4. - Luce ETIENNE, Apr 05 2014
a(n) = 2*A069905(3*(n+1)+2) - 3*(n+1). - Ayoub Saber Rguez, Aug 31 2021

A056109 Fifth spoke of a hexagonal spiral.

Original entry on oeis.org

1, 6, 17, 34, 57, 86, 121, 162, 209, 262, 321, 386, 457, 534, 617, 706, 801, 902, 1009, 1122, 1241, 1366, 1497, 1634, 1777, 1926, 2081, 2242, 2409, 2582, 2761, 2946, 3137, 3334, 3537, 3746, 3961, 4182, 4409, 4642, 4881, 5126, 5377, 5634, 5897, 6166, 6441

Offset: 0

Henry Bottomley, Jun 09 2000

Squared distance from (0,0,-1) to (n,n,n) in R^3. - James R. Buddenhagen, Jun 15 2013

			Illustration of initial terms:
.
.                                                o
.                           o                 o o o o
.            o           o o o o           o o o o o o o
.   o     o o o o     o o o o o o o     o o o o o o o o o o
.            o           o o o o           o o o o o o o
.                           o                 o o o o
.                                                o
.
.   1        6              17                   34
- _Aaron David Fairbanks_, Feb 16 2025

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Henry Bottomley, Illustration of initial terms
Tanya Khovanova, Recursive Sequences
G. Nebe and N. J. A. Sloane, Home page for hexagonal (or triangular) lattice A2
Leo Tavares, Triple Diamond Illustration
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A008810, A122430 (prime terms).
Other spokes: A003215, A056105, A056106, A056107, A056108.
Other spirals: A054552.
Cf. A000290.

List([0..50],n->3*n^2+2*n+1); # Muniru A Asiru, Oct 07 2018

[3*n^2 + 2*n + 1: n in [0..50]]; // Vincenzo Librandi, Mar 15 2013

seq(coeff(series(n!*(exp(x)*(3*x^2+5*x+1)),x,n+1), x, n), n = 0 .. 50); # Muniru A Asiru, Oct 07 2018

Table[3 n^2 + 2 n + 1, {n, 0, 100}] (* Vincenzo Librandi, Mar 15 2013 *)
LinearRecurrence[{3,-3,1},{1,6,17},60] (* Harvey P. Dale, Mar 28 2019 *)

{a(n) = 3*n^2 + 2*n + 1}; /* Michael Somos, Aug 03 2006 */

Vec((1+3*x+2*x^2)/(1-3*x+3*x^2-x^3)+O(x^100)) \\ Stefano Spezia, Oct 17 2018

a(n) = 3n^2+2n+1 = a(n-1)+6n-1 = 2a(n-1)-a(n-2)+6 = 3a(n-1)-3a(n-2)+a(n-3) = A056105(n)+4n = A056106(n)+3n = A056107(n)+2n = A056108(n)+n = A003215(n)-n.
G.f.: (1+3*x+2*x^2)/(1-3*x+3*x^2-x^3). - Colin Barker, Jan 04 2012
G.f.: (1 + x) * (1 + 2*x) / (1 - x)^3. - Michael Somos, Feb 04 2012
a(n) = A008810(3*n + 1) = A056105(-n). - Michael Somos, Aug 03 2006
E.g.f.: exp(x)*(1 + 5*x + 3*x^2). - Stefano Spezia, Oct 06 2018
a(n) = A000290(n+1) + 2*A000290(n). - Leo Tavares, May 29 2023
a(n) = A069894(n) - A000290(n+1). - Jarrod G. Sage, Jul 19 2024

A035597 Number of points of L1 norm 3 in cubic lattice Z^n.

Original entry on oeis.org

0, 2, 12, 38, 88, 170, 292, 462, 688, 978, 1340, 1782, 2312, 2938, 3668, 4510, 5472, 6562, 7788, 9158, 10680, 12362, 14212, 16238, 18448, 20850, 23452, 26262, 29288, 32538, 36020, 39742, 43712, 47938, 52428, 57190, 62232, 67562

Offset: 0

N. J. A. Sloane

Sums of the first n terms > 0 of A001105 in palindromic arrangement. a(n) = Sum_{i=1 .. n} A001105(i) + Sum_{i=1 .. n-1} A001105(i), e.g. a(3) = 38 = 2 + 8 + 18 + 8 + 2; a(4) = 88 = 2 + 8 + 18 + 32 + 18 + 8 + 2. - Klaus Purath, Jun 19 2020

Apart from multiples of 3, all divisors of n are also divisors of a(n), i.e. if n is not divisible by 3, a(n) is divisible by n. All divisors d of a(n) for d !== 0 (mod) 3 are also divisors of a(abs(n-d)) and a(n+d). For all n congruent to 0,2,7 (mod 9) a(n) is divisible by 3. If n is divisible by 3^k, a(n) is divisible by 3^(k-1). - Klaus Purath, Jul 24 2020

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
J. H. Conway and N. J. A. Sloane, Low-dimensional lattices. VII. Coordination sequences, Proc. Roy. Soc. Lond. A 458 (1996) 2369-2389.
Milzn Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.2.
Milan Janjic and B. Petkovic, A Counting Function, arXiv preprint arXiv:1301.4550 [math.CO], 2013. - _N. J. A. Sloane_, Feb 13 2013
Milan Janjic and B. Petkovic, A Counting Function Generalizing Binomial Coefficients and Some Other Classes of Integers, J. Int. Seq. 17 (2014) # 14.3.5.
R. J. Mathar, Bivariate generating functions for non-attacking Wazirs on rectangular boards, viXra:2404.0122 (2024) page 14
Joan Serra-Sagrista, Enumeration of lattice points in l_1 norm, Inf. Proc. Lett. 76 (1-2) (2000) 39-44.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Partial sums of A069894.
Cf. A001105, A005900, A084570, A097869.
Cf. A001844, A001845, A005899.
Column 3 of A035607, A266213, A343599.
Row 3 of A113413, A119800, A122542.

[(4*n^3 + 2*n)/3: n in [0..40]]; // Vincenzo Librandi, Sep 19 2011

f := proc(n,m) local i; sum( 2^i*binomial(n,i)*binomial(m-1,i-1),i=1..min(n,m)); end; # n=dimension, m=norm

Table[(4n^3+2n)/3,{n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,2,12,38},41] (* Harvey P. Dale, Sep 18 2011 *)

a(n) = (4*n^3 + 2*n)/3.
a(n) = 2*A005900(n). - R. J. Mathar, Dec 05 2009
a(0)=0, a(1)=2, a(2)=12, a(3)=38, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). G.f.: (2*x*(x+1)^2)/(x-1)^4. - Harvey P. Dale, Sep 18 2011
a(n) = -a(-n), a(n+1) = A097869(4n+3) = A084570(2n+1). - Bruno Berselli, Sep 20 2011
a(n) = 2*n*Hypergeometric2F1(1-n,1-k,2,2), where k=3. Also, a(n) = A001845(n) - A001844(n). - Shel Kaphan, Feb 26 2023
a(n) = A005899(n)*n/3. - Shel Kaphan, Feb 26 2023
a(n) = A006331(n)+A006331(n-1). - R. J. Mathar, Aug 12 2025

A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

Original entry on oeis.org

1, 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 845, 226, 965, 257, 1093, 290, 1229, 325, 1373, 362, 1525, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677

Offset: 0

Paul Curtz, Aug 15 2015

Using (n+sqrt(4+n^2))/2, after the integer 1 for n=0, the reduced metallic means are b(1) = (1+sqrt(5))/2, b(2) = 1+sqrt(2), b(3) = (3+sqrt(13))/2, b(4) = 2+sqrt(5), b(5) = (5+sqrt(29))/2, b(6) = 3+sqrt(10), b(7) = (7+sqrt(53))/2, b(8) = 4+sqrt(17), b(9) = (9+sqrt(85))/2, b(10) = 5+sqrt(26), b(11) = (11+sqrt(125))/2 = (11+5*sqrt(5))/2, ... . The last value yields the radicals in a(n) or A013946.
b(2) = 2.41, b(3) = 3.30, b(4) = 4.24, b(5) = 5.19 are "good" approximations of fractal dimensions corresponding to dimensions 3, 4, 5, 6: 2.48, 3.38, 4.33 and 5.45 based on models. See "Arbres DLA dans les espaces de dimension supérieure: la théorie des peaux entropiques" in Queiros-Condé et al. link. DLA: beginning of the title of the Witten et al. link.
Consider the symmetric array of the half extended Rydberg-Ritz spectrum of the hydrogen atom:
   0,    1/0,     1/0,     1/0,     1/0,      1/0,      1/0,     1/0, ...
-1/0,      0,     3/4,     8/9,   15/16,    24/25,    35/36,   48/49, ...
-1/0,   -3/4,       0,    5/36,    3/16,   21/100,      2/9,  45/196, ...
-1/0,   -8/9,   -5/36,       0,   7/144,   16/225,     1/12,  40/441, ...
-1/0, -15/16,   -3/16,  -7/144,       0,    9/400,    5/144,  33/784, ...
-1/0, -24/25, -21/100, -16/225,  -9/400,        0,   11/900, 24/1225, ...
-1/0, -35/36,    -2/9,   -1/12,  -5/144,  -11/900,        0, 13/1764, ...
-1/0, -48/49, -45/196, -40/441, -33/784, -24/1225, -13/1764,       0, ... .
The numerators are almost A165795(n).
Successive rows: A000007(n)/A057427(n), A005563(n-1)/A000290(n), A061037(n)/A061038(n), A061039(n)/A061040(n), A061041(n)/A061042(n), A061043(n)/A061044(n), A061045(n)/A061046(n), A061047(n)/A061048(n), A061049(n)/A061050(n).
A144433(n) or A195161(n+1) are the numerators of the second upper diagonal (denominators: A171522(n)).
c(n+1) = a(n) + a(n+1) = 6, 7, 15, 18, 34, 39, 63, 70, 102, 111, ... .
c(n+3) - c(n+1) = 9, 11, 19, 21, 29, 31, ... = A090771(n+2).
The final digit of a(n) is neither 4 nor 8. - Paul Curtz, Jan 30 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Diogo Queiros-Condé, Jean Chaline, and Jacques Dubois, Le monde des fractales La Nature trans-échelles, 478p., ellipses, Paris, 2015, page 220.
T. A. Witten, Jr. and L. M. Sander, Diffusion-Limited Aggregation, a Kinetic Critical Phenomenom, Phys. Rev. Lett., Vol. 47 (Nov 09 1981), pp. 1400-1403.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000007, A000290, A001622, A002522, A005563, A010685, A010698, A013946, A014176, A057427, A061035-A061050, A078370, A087475, A090771, A098316, A098317, A098318, A144433, A168077, A171621, A176398, A176439, A176458, A176522, A176537, A195161.

Cf. A069894.

[Numerator(1+n^2/4): n in [0..60]]; // Vincenzo Librandi, Aug 15 2015

A261327:=n->numer((4 + n^2)/4); seq(A261327(n), n=0..60); # Wesley Ivan Hurt, Aug 15 2015

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 60] (* Vincenzo Librandi, Aug 15 2015 *)
a[n_] := (n^2 + 4) / 4^Mod[n + 1, 2]; Table[a[n], {n, 0, 52}] (* Peter Luschny, Mar 18 2022 *)

vector(60, n, n--; numerator(1+n^2/4)) \\ Michel Marcus, Aug 15 2015

Vec((1+5*x-x^2-2*x^3+2*x^4+5*x^5)/(1-x^2)^3 + O(x^60)) \\ Colin Barker, Aug 15 2015

a(n)=if(n%2,n^2+4,(n/2)^2+1) \\ Charles R Greathouse IV, Oct 16 2015

[(n*n+4)//4**((n+1)%2) for n in range(60)] # Gennady Eremin, Mar 18 2022

[numerator(1+n^2/4) for n in (0..60)] # G. C. Greubel, Feb 09 2019

a(n) = numerator(1 + n^2/4). (Previous name.) See A010685 (denominators).
a(2*k) = 1 + k^2.
a(2*k+1) = 5 + 4*k*(k+1).
a(2*k+1) = 4*a(2*k) + 4*k + 1.
a(4*k+2) = A069894(k). - Paul Curtz, Jan 30 2019
a(-n) = a(n).
a(n+2) = a(n) + A144433(n) (or A195161(n+1)).
a(n) = A168077(n) + period 2: repeat 1, 4.
a(n) = A171621(n) + period 2: repeat 2, 8.
From Colin Barker, Aug 15 2015: (Start)
a(n) = (5 - 3*(-1)^n)*(4 + n^2)/8.
a(n) = n^2/4 + 1 for n even;
a(n) = n^2 + 4 for n odd.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.
G.f.: (1 + 5*x - x^2 - 2*x^3 + 2*x^4 + 5*x^5)/ (1 - x^2)^3. (End)
E.g.f.: (5/8)*(x^2 + x + 4)*exp(x) - (3/8)*(x^2 - x + 4)*exp(-x). - Robert Israel, Aug 18 2015
Sum_{n>=0} 1/a(n) = (4*coth(Pi)+tanh(Pi))*Pi/8 + 1/2. - Amiram Eldar, Mar 22 2022

New name by Peter Luschny, Mar 18 2022

cp's OEIS Frontend

A137224 Interleave 4*n^2, 1+4*n^2, 1+(2n+1)^2, (2n+1)^2 (or A016742, A053755, A069894, A016754).

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A002522 a(n) = n^2 + 1.

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A001844 Centered square numbers: a(n) = 2*n*(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

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A001653 Numbers k such that 2*k^2 - 1 is a square.

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A212959 Number of (w,x,y) such that w,x,y are all in {0,...,n} and |w-x| = |x-y|.

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A137224 Interleave 4n^2, 1+4n^2, 1+(2n+1)^2, (2n+1)^2 (or A016742, A053755, A069894, A016754).

A001844 Centered square numbers: a(n) = 2n(n+1)+1. Sums of two consecutive squares. Also, consider all Pythagorean triples (X, Y, Z=Y+1) ordered by increasing Z; then sequence gives Z values.

A108099 a(n) = 8n^2 + 8n + 4.

A164897 a(n) = 4n(n+1) + 3.