A008844 -id:A008844 - cp's OEIS frontend

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.

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.
D. C. Haws, Matroids [Broken link, Oct 30 2017]
D. C. Haws, Matroids [Copy on website of Matthias Koeppe]
D. C. Haws, Matroids [Cached copy, pdf file only]
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, pp. 22 and 36.
Milan Janjic, Two Enumerative Functions. [Broken link; WayBackMachine archive.]
Milan Janjic, On a class of polynomials with integer coefficients, JIS 11 (2008) 08.5.
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.
Amelia Carolina Sparavigna, Groupoid of OEIS A001844 Numbers (centered square numbers), Politecnico di Torino, Italy (2019).
R. G. Stanton and D. D. Cowan, Note on a "square" functional equation, SIAM Rev., 12 (1970), 277-279.
David James Sycamore, Triangular array
Leo Tavares, Illustration: Diamond Rows
B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
Eric Weisstein's World of Mathematics, Centered Polygonal Number, Centered Square Number, Diamond, Pythagorean Triple, and von Neumann Neighborhood.
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

A046176 Indices of square numbers that are also hexagonal.

Original entry on oeis.org

1, 35, 1189, 40391, 1372105, 46611179, 1583407981, 53789260175, 1827251437969, 62072759630771, 2108646576008245, 71631910824649559, 2433376321462076761, 82663163018885960315, 2808114166320660573949

Offset: 1

Eric W. Weisstein

Bisection (even part) of Chebyshev sequence with Diophantine property.
(3*b(n))^2 - 2*(2*a(n+1))^2 = 1 with companion sequence b(n) = A077420(n), n >= 0.
Sequence also refers to inradius of primitive Pythagorean triangles with consecutive legs, odd followed by even. - Lekraj Beedassy, Apr 23 2003
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> oo} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12*sqrt(2). - Ant King, Nov 08 2011
Integers of the form sqrt((m+1)*(2*m+1)). The corresponding values of m form A078522. Subsequence of A284876. - Jonathan Sondow, Apr 07 2017

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Vincenzo Librandi, Table of n, a(n) for n = 1..200
Tanya Khovanova, Recursive Sequences.
E. Kilic, Y. T. Ulutas, and N. Omur, A Formula for the Generating Functions of Powers of Horadam's Sequence with Two Additional Parameters, J. Int. Seq. 14 (2011) #11.5.6, table 4, k=1, t=3.
Serge Perrine, About the diophantine equation z^2 = 32y^2 - 16, SCIREA Journal of Mathematics (2019) Vol. 4, Issue 5, 126-139.
Maciej Ulas, On certain diophantine equations related to triangular and tetrahedral numbers, arXiv:0811.2477 [math.NT] (2008) , v_n in Theorem 5.6.
P. H. van der Kamp, Global classification of two-component..., Found. Comput. Math. 9 (5) (2009) 559-597 near Eq. (4.7).
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A000129, A008844, A046177, A078522, A284876.

Cf. A001109, A001110 (partial sums).

List([0..20], n-> Lucas(2,-1, 4*n-2)[1]/2 ); # G. C. Greubel, Jan 13 2020

I:=[1,35]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 22 2011

seq( simplify(ChebyshevU(2*(n-1), 3)), n = 1..20); # G. C. Greubel, Jan 13 2020

LinearRecurrence[{34, -1}, {1, 35}, 15] (* Ant King, Nov 08 2011 *)
Fibonacci[4*Range[20] -2, 2]/2 (* G. C. Greubel, Jan 13 2020 *)

vector(21, n, polchebyshev(2*(n-1), 2, 3) ) \\ G. C. Greubel, Jan 13 2020

[lucas_number1(4*n-2, 2,-1)/2 for n in (1..20)] # G. C. Greubel, Jan 13 2020

a(n) = 34*a(n-1) - a(n-2); a(0)=-1, a(1)=1.
a(n+1) = S(2*n, 6) = S(n, 34) + S(n-1, 34), n >= 1, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. See A049310. S(n, 34) = A029547(n).
G.f.: x*(1+x)/(1-34*x+x^2).
a(n+1) = Sum_{k=0..n} (-1)^k*binomial(2*n-k, k)*6^(2*(n-k)), n >= 0.
a(n) = A001109(2n+1). - Lekraj Beedassy, Apr 23 2003
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(f(a(n-1),3),3). - Marcos Carreira, Dec 27 2006
From Antonio Alberto Olivares, Mar 22 2008: (Start)
a(n) = (sqrt(2)/8)*(3 + 2*sqrt(2))*(17 + 12*sqrt(2))^(n-1) - (sqrt(2)/8)*(3 - 2*sqrt(2))*(17 - 12*sqrt(2))^(n-1).
a(n) = (sqrt(2)/8)*( (17+12*sqrt(2))^(n-1/2) - (17-12*sqrt(2))^(n-1/2) ).
a(n) = (sqrt(2)/8)*( (3+2*sqrt(2))^(2n-1) - (3-2*sqrt(2))^(2n-1) ).
a(n) = (sqrt(2)/8)*( (1+sqrt(2))^(4n-2) - (1-sqrt(2))^(4n-2) ).
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3). (End)
a(n+1) = 17*a(n) + 6*sqrt(8*a(n)^2+1) for n >= 0. - Richard Choulet, May 01 2009
a(n) = b such that (-1)^(n+1) * Integral_{x=-Pi/2..Pi/2} cos((2*n-1)*x)/(3-sin(x)) dx = c + b*log(2). - Francesco Daddi, Aug 01 2011
a(n) are the nonzero integer square roots of A227970. - Richard R. Forberg, Aug 01 2013
a(n) = y/5, where y are solutions to: y^2 = 2x^2 - x - 3. - Richard R. Forberg, Nov 24 2013
a(n) = sqrt((A078522(n)+1)*(2*A078522(n)+1)). - Jonathan Sondow, Apr 07 2017
a(n) = Pell(4*n-2)/2. - G. C. Greubel, Jan 13 2020
a(n) = A001653(n)*A002315(n). - Gerry Martens, Mar 23 2024

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

Original entry on oeis.org

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Benoit Cloitre, May 10 2003

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
  a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
  a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 1..2608
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

A078522 Numbers k such that (k+1)(2k+1) is a perfect square.

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024

Offset: 1

Joseph L. Pe, Jan 07 2003

Equivalently, both k+1 and 2*k+1 are perfect squares.
The square roots of (k+1)*(2*k+1) are in A046176.
Also numbers k such that 3*A000217(k) + A000217(k+1) is a perfect square. - Bruno Berselli, Nov 17 2016
From Sergey Pavlov, Mar 14 2017: (Start)
The sequence of areas k(n)*q(n)/2, of the ordered Pythagorean triples (k(n), q(n) = k(n) + 2, c(n)) with k(1)=0, q(1)=2, c(1)=0, a(1)=0, and k(2)=6, q(2)=8, c(2)=10, a(2)=24 (conjectured).
Conjecture: let f(n) be a sequence of form x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n) = x(n) + v, z(n)) with x(1)=0, y(1)=v, z(1)=0, f(1)=0, where v is an even number. Then there exists such subset p(i) that p(1) = 0, p(2) = 24*(v/2)^2, for any i > 2, p(i) = 34*p(i-1) - p(i-2) + 24*(v/2)^2, and any p(i) is a term of the above sequence f(n) (see also the first formula by Benoit Cloitre in the Formula section).
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.

Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020

I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019

seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020

RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)

concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016

def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

From Benoit Cloitre, Jan 19 2003: (Start)
a(1) = 0, a(2) = 24; for n > 2, a(n) = 34*a(n-1) - a(n-2) + 24.
a(n) = floor(A*B^n), where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2).
a(n) = A008844(n) - 1. (End)
From R. J. Mathar, Sep 21 2011: (Start)
G.f.: 24*x^2/( (1-x)*(1-34*x+x^2) ).
a(n) = 24*A029546(n-2). (End)
a(n) = (A001653(n)^2 - 1)/2 = A002315(n-1)^2 - 1. - Tomohiro Yamada, Sep 15 2019
a(n) = (3/4)*(ChebyshevT(n, 17) - 16*Chebyshev(n-1, 17) - 1). - G. C. Greubel, Jan 13 2020
From Amiram Eldar, Dec 02 2024: (Start)
a(n) = A001542(n-1)*A001542(n).
Sum_{n>=2} 1/a(n) = (3 - 2*sqrt(2))/4. (End)

Edited by Bruno Berselli, Nov 17 2016

A008843 Squares of NSW numbers (A002315): x^2 such that x^2 - 2y^2 = -1 for some y.

Original entry on oeis.org

1, 49, 1681, 57121, 1940449, 65918161, 2239277041, 76069501249, 2584123765441, 87784138523761, 2982076586042449, 101302819786919521, 3441313796169221281, 116903366249966604049, 3971273138702695316401, 134906383349641674153601, 4582845760749114225906049, 155681849482120242006652081

Offset: 0

N. J. A. Sloane

Also indices of triangular numbers (A000217) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 16 2015

a(n)-th triangular number is a square; subsequence of A001108. - Jaroslav Krizek, Aug 05 2016

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.
P. Ribenboim, The Book of Prime Number Records. Springer-Verlag, NY, 2nd ed., 1989, p. 288.
P. F. Teilhet, Note #2094, L'Intermédiaire des Mathématiciens, 10 (1903), pp. 235-238.

M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
D. H. Lehmer, Lacunary recurrence formulas for the numbers of Bernoulli and Euler, Annals Math., 36 (1935), 637-649.
Index entries for sequences related to Bernoulli numbers.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000217, A002315, A016754, A146313, A253826.

LinearRecurrence[{35,-35,1},{1,49,1681},17] (* Stefano Spezia, Aug 17 2024 *)

a(n) = 34*a(n-1) - a(n-2) + 16 = A002315(n)^2 = 2*A001653(n)^2 - 1 = 2*A008844(n) - 1 = floor(A046176(n)*sqrt(2)) = 6*A055792(n+1) - a(n-1) + 4 = (6*A055792(n+2) + a(n-1) - 20)/35. - Henry Bottomley, Nov 13 2001
a(n) = A001108(2n+1). - Ira M. Gessel, Nov 05 2014
a(n) = Sum_{k=1..2*n+1} 2^(k-1)*binomial(4*n+2, 2*k). - Zoltan Zachar (zachar(AT)fellner.sulinet.hu), Oct 03 2003
O.g.f.: -(1+14*x+x^2)/((-1+x)*(1-34*x+x^2)). - R. J. Mathar, Nov 23 2007
a(n) = -(cosh((2*n - 1)*arctanh(sqrt(2))))^2 = -1 - (sinh((2*n - 1)*arctanh(sqrt(2))))^2. - Artur Jasinski, Oct 30 2008
a(n) = Sum_{k=0..4n+1} A000129(k), see Santana and Diaz-Barrero link at A002315. - Ivan N. Ianakiev, Jul 15 2022

a(14)-a(17) from Stefano Spezia, Aug 17 2024

A124124 Nonnegative integers n such that 2n^2 + 2n - 3 is square.

Original entry on oeis.org

1, 2, 6, 13, 37, 78, 218, 457, 1273, 2666, 7422, 15541, 43261, 90582, 252146, 527953, 1469617, 3077138, 8565558, 17934877, 49923733, 104532126, 290976842, 609257881, 1695937321, 3551015162, 9884647086, 20696833093, 57611945197, 120629983398, 335787024098

Offset: 1

John W. Layman, Nov 29 2006

First differences are apparently in A143608. [R. J. Mathar, Jul 17 2009]

Alternative definition: T_n and (T_n - 1)/2 are triangular numbers. - Raphie Frank, Sep 06 2012

Michael De Vlieger, Table of n, a(n) for n = 1..2613
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Counting families of generalized balancing numbers, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.
Hermann Stamm-Wilbrandt, 4 interlaced bisections
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001108, A001652, A001653, A006452, A008844, A046090, A046172, A077442, A098790, A216134.

A124124 := proc(n)
coeftayl(x*(1+x-2*x^2+x^3+x^4)/((1-x)*(x^2-2*x-1)*(x^2+2*x-1)), x=0, n);
end proc:
seq(A124124(n), n=1..20); # Wesley Ivan Hurt, Aug 04 2014
# Alternative:
a[1]:= 1: a[2]:= 2: a[3]:= 6:
for n from 4 to 1000 do
a[n]:= (3 + 2*(n mod 2))*(a[n-1]-a[n-2])+a[n-3]
od:
seq(a[n],n=1..100); # Robert Israel, Aug 13 2014

LinearRecurrence[{1,6,-6,-1,1},{1,2,6,13,37},40] (* Harvey P. Dale, Nov 05 2011 *)
CoefficientList[Series[(1 + x - 2*x^2 + x^3 + x^4)/((1 - x)*(x^2 - 2*x - 1)*(x^2 + 2*x - 1)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Aug 04 2014 *)

for(n=1,10^10,if(issquare(2*n^2+2*n-3),print1(n,", "))) \\ Derek Orr, Aug 13 2014

It appears that a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) if n is even, a(n) = 5*a(n-1)-5*a(n-2)+a(n-3) if n is odd. Can anyone confirm this?
Corrected and confirmed (using the g.f.) by Robert Israel, Aug 27 2014
2*a(n) = sqrt(7+2*A077442(n-1)^2)-1. - R. J. Mathar, Dec 03 2006
a(n) = a(n-1)+6*a(n-2)-6*a(n-3)-a(n-4)+a(n-5). G.f.: -x*(1+x-2*x^2+x^3+x^4)/((x-1)*(x^2-2*x-1)*(x^2+2*x-1)). [R. J. Mathar, Jul 17 2009]
For n>0, a(2n-1) = 2*A001653(n) - A046090(n-1) and a(2n) = 2*A001653(n) + A001652(n-1). - Charlie Marion, Jan 03 2012
From Raphie Frank, Sep 06 2012: (Start)
If y = A006452(n), then a(n) = 2y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n))).
Also see A216134 [a(n) = y + ((sqrt(8y^2 - 7) - 1)/2 - (1 - sgn(n)))].
(End)
From Hermann Stamm-Wilbrandt, Aug 27 2014: (Start)
a(2*n+2) = A098586(2*n).
a(2*n+1) = A098790(2*n).
a(n) = 7*a(n-2) - 7*a(n-4) + a(n-6), for n>6. (End)
a(2*n+1)^2 + (a(2*n+1)+1)^2 = A038761(n)^2 + 2^2. - Hermann Stamm-Wilbrandt, Aug 31 2014

More terms from Harvey P. Dale, Feb 07 2011

More terms from Wesley Ivan Hurt, Aug 04 2014

A046177 Squares (A000290) which are also hexagonal numbers (A000384).

Original entry on oeis.org

1, 1225, 1413721, 1631432881, 1882672131025, 2172602007770041, 2507180834294496361, 2893284510173841030625, 3338847817559778254844961, 3853027488179473932250054441, 4446390382511295358038307980025, 5131130648390546663702275158894481

Offset: 1

Eric W. Weisstein

Also, odd square-triangular numbers (or bisection of A001110 = {0, 1, 36, 1225, 41616, 1413721, 48024900, 1631432881, ...} = Numbers that are both triangular and square: a(n) = 34a(n-1) - a(n-2) + 2). - Alexander Adamchuk, Nov 06 2007
Let y^2 = x*(2*x-1) = H_x (x>=1). The least both hexagonal and square number which is greater than y^2 is given by the relation (24*x+17*y-6)^2 = H_{17*x+12*y-4}. - Richard Choulet, May 01 2009
As n increases, this sequence is approximately geometric with common ratio r = lim(n -> Infinity, a(n)/a(n-1)) = ( 1+ sqrt(2))^8 = 577 + 408 * sqrt(2). - Ant King Nov 08 2011
Also centered octagonal numbers (A016754) which are also triangular numbers (A000217). - Colin Barker, Jan 16 2015
Also hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015

M. Rignaux, Query 2175, L'Intermédiaire des Mathématiciens, 24 (1917), 80.

Colin Barker, Table of n, a(n) for n = 1..327
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Eric Weisstein's World of Mathematics, Square Triangular Number.
Index entries for linear recurrences with constant coefficients, signature (1155,-1155,1).

Cf. A008844, A046176, A253826.
Cf. A001110 (Numbers that are both triangular and square).
Cf. A000290, A000384, A016754, A253826.

LinearRecurrence[{1155, -1155, 1}, {1, 1225, 1413721}, 11] (* Ant King, Nov 08 2011 *)

Vec(x*(1+70*x+x^2)/((1-x)*(1-1154*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 16 2015

a(n) = A001110(2n-1). - Alexander Adamchuk, Nov 06 2007
a(n+1) = 577*a(n)+36+204*(8*a(n)^2+a(n))^0.5 for n>=1 (a(0)=1). - Richard Choulet, May 01 2009
a(n+2) = 1154*a(n+1) - a(n) + 72 for n>=0. - Richard Choulet, May 01 2009
From Ant King, Nov 07 2011: (Start)
a(n) = 1155*a(n-1) - 1155*a(n-2) + a(n-3).
a(n) = 1/32*((1 + sqrt(2))^(8*n - 4) + (1 - sqrt(2))^(8*n-4) - 2).
a(n) = floor(1/32*(1 + sqrt(2))^(8*n - 4)).
a(n) = 1/32*((tan(3*Pi/8))^(8*n-4) + (tan(Pi/8))^(8*n-4) - 2).
a(n) = floor(1/32*(tan(3*Pi/8))^(8*n-4)).
G.f.: x*(1 + 70*x + x^2)/((1 - x)*(1 - 1154*x + x^2)).
(End)
a(n) = A096979(4*n - 3). - Ivan N. Ianakiev, Sep 05 2016
a(n) = (1/2) * (A002315(n))^2 * ((A002315(n))^2 + 1) = ((2*x + 1)*sqrt(x^2 + (x+1)^2))^2, where x = (1/2)*(A002315(n)-1). - Ivan N. Ianakiev, Sep 05 2016

A377016 Semiperimeter of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

1, 28, 861, 28680, 970921, 32963140, 1119662181, 38034888528, 1292062686481, 43892073946540, 1491038320325421, 50651410052600280, 1720656899012149561, 58451683130389395028, 1985636569382856677301, 67453191675004485098400, 2291422880375627492063521, 77840924741066359629967420

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

a(0) = 1 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

			For n=2, the short leg is A002315(2) = 41 and the hypotenuse is A008844(n) = 841 so the semiperimeter is then a(2) = (41 + 840 + 841)/2 = 861.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (41,-246,246,-41,1).

Cf. A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b},a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2;b=(a^2-1)/2;{(a+2b+1)/2}];semis={};Do[semis=Join[semis,FullSimplify[s[n]]],{n,0,17}];semis

Vec((1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)) + O(x^20)) \\ Andrew Howroyd, Oct 14 2024

a(n) = (A002315(n) + 2*A008844(n) - 1)/2.

G.f.: (1 - 13*x - 41*x^2 + 21*x^3)/((1 - 34*x + x^2)*(1 - 6*x + x^2)*(1 - x)). - Andrew Howroyd, Oct 14 2024

A377017 Area of the unique primitive Pythagorean triple whose short leg is A002315(n) and such that its long leg and its hypotenuse are consecutive natural numbers.

Original entry on oeis.org

0, 84, 17220, 3412920, 675761016, 133797385260, 26491207202460, 5245125232676784, 1038508304885968560, 205619399242324129860, 40711602541676078766516, 8060691683852625858745320, 1595976241800278270688414120, 315995235184771245126273789084, 62565460590342906257639745449100

Offset: 0

Miguel-Ángel Pérez García-Ortega, Oct 13 2024

nonn

a(0)=0 is included by convention. This corresponds to the Pythagorean triple 1^2 + 0^2 = 1^2.

All terms in this sequence are divisible by 84.

			For n=2, the short leg is A002315(2) = 41 and the long leg is A008844(2)-1 = 840 so the area is then a(2) = 41*840/2 = 17220.

Miguel Ángel Pérez García-Ortega, José Manuel Sánchez Muñoz and José Miguel Blanco Casado, El Libro de las Ternas Pitagóricas, Preprint 2024.

Index entries for linear recurrences with constant coefficients, signature (204,-1190,204,-1).

Cf. A377016, A002315, A078522, A008844.

s[n_]:=s[n]=Module[{a, b}, a=((1+Sqrt[2])^(2n+1)-(Sqrt[2]-1)^(2n+1))/2; b=(a^2-1)/2; {(a*b)/2}]; areas={}; Do[areas=Join[areas, FullSimplify[s[n]]], {n, 0, 17}]; areas

a(n) = A002315(n)*(A008844(n)-1)/2.

G.f.: 84*x*(1 + x)/((1 - 198*x + x^2)*(1 - 6*x + x^2)). - Andrew Howroyd, Oct 14 2024

A156573 a(n) = 34*a(n-1) - a(n-2) - 4232 for n > 2; a(1)=529, a(2)=13225.

Original entry on oeis.org

529, 13225, 444889, 15108769, 513249025, 17435353849, 592288777609, 20120383080625, 683500735959409, 23218904639535049, 788759257008228025, 26794595833640213569, 910227499086759029089, 30920940373116166771225

Offset: 1

Klaus Brockhaus, Feb 11 2009

			a(3) = 34*a(2) - a(1) - 4232 = 34*13225 - 529 - 4232 = 444889.

G. C. Greubel, Table of n, a(n) for n = 1..600
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Second trisection of A156572.

Cf. A008844, A156164 (decimal expansion of 17+12*sqrt(2)), A156574, A156575.

LinearRecurrence[{35,-35,1}, {529,13225,444889}, 30] (* G. C. Greubel, Jan 04 2022 *)

{m=14; v=concat([529, 13225], vector(m-2)); for(n=3, m, v[n]=34*v[n-1]-v[n-2]-4232); v}

def a(n): return -529*bool(n==0) + (529/4) + (3/4)*(chebyshev_U(n, 17) - 33*chebyshev_U(n-1, 17))
[a(n) for n in (1..30)] # G. C. Greubel, Jan 04 2022

a(n) = 529*(2 + (3 - 2*sqrt(2))*(17 + 12*sqrt(2))^n + (3 + 2*sqrt(2))*(17 - 12*sqrt(2))^n)/8.
a(n) = 529*A008844(n).
G.f.: 529*x*(1 -10*x +x^2)/((1-x)*(1-34*x+x^2)). [corrected by Klaus Brockhaus, Sep 22 2009]
Limit_{n -> infinity} a(n)/a(n-1) = 17+12*sqrt(2).
a(n) = -529*[n=0] + (529/4) + (1587/4)*(ChebyshevU(n, 17) - 33*ChebyshevU(n-1, 17)). - G. C. Greubel, Jan 04 2022

Revised by Klaus Brockhaus, Feb 16 2009

cp's OEIS Frontend

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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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.

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