A002593 -id:A002593 - cp's OEIS frontend

A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

504, 8721, 65472, 312375, 1119528, 3293829, 8388096, 19131147, 39999000, 77947353, 143325504, 250991871, 421651272, 683434125, 1073737728, 1641349779, 2448874296, 3575480097, 5119992000, 7204344903, 9977420904, 13619289621, 18345871872, 24414046875

Offset: 2

Vladimir Pletser, Jan 10 2015

Numbers c(n) such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers for M(n) being twice a squared integer (A001105) and b(n)=(A002593).
If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling to a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), and c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=2, M(n)=8, b(n)=28, c(n)=504.
See "File Triplets (M,b,c) for M=2n^2" link.

Vladimir Pletser, Table of n, a(n) for n = 2..50000
Vladimir Pletser, File Triplets (M,b,c) for M=2n^2
Vladimir Pletser, Number of terms, first term and square root of sums of consecutive cubed integers equal to integer squares, Research Gate, 2015.
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (8,-28,56,-70,56,-28,8,-1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253681, A253707, A253709, A002593, A253725.

[n^3*(4*n^4 - 1): n in [2..30]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 2 to 50000 do a:= n^3*(4*n^4 - 1): print (a); end do:

f[n_] := n^3 (4 n^4 - 1); Rest@Array[f, 32] (* Michael De Vlieger, Jan 28 2015 *)

Vec(-3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168)/(x-1)^8 + O(x^100)) \\ Colin Barker, Jan 14 2015

a(n) = n^3(4n^4 - 1).

G.f.: -3*x^2*(x^7-8*x^6+27*x^5-216*x^4-1521*x^3-3272*x^2-1563*x-168) / (x-1)^8. - Colin Barker, Jan 14 2015

A253725 Integer squares c^2 that are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Original entry on oeis.org

254016, 76055841, 4286582784, 97578140625, 1253342942784, 10849309481241, 70360154505216, 366000785535609, 1599920001000000, 6075789839706609, 20542200096854016, 62996919308080641, 177789795179217984, 467082203214515625, 1152912708530601984

Offset: 2

Vladimir Pletser, Jan 10 2015

Numbers a(n)=c^2 such that b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2 has nontrivial solutions over the integers where M(n) is twice a squared integer (A001105) and b(n)=(A002593).
If M is twice a squared integer, there always exists at least one nontrivial solution for the sum of M consecutive cubed integers starting at b^3 and equaling a squared integer c^2. For n>=1, M(n)= 2n^2 (A001105), b(n) = M(M-1)/2 = n^2(2n^2 - 1) (A002593), c(n)= sqrt(M/2) (M(M^2-1)/2)= n^3(4n^4 - 1) (A253724) and a(n)=c(n)^2 (this sequence).
The trivial solutions with M < 1 and b < 2 are not considered here.

			For n=2, M(n)=8, b(n)=28, c(n)=504, a(n)=c^2=254016.
See "File Triplets (M,b,c) for M=2n^2" link.

Vladimir Pletser, Table of n, a(n) for n = 2..50000
Vladimir Pletser, File Triplets (M,b,c) for M=2n^2
Vladimir Pletser, General solutions of sums of consecutive cubed integers equal to squared integers, arXiv:1501.06098 [math.NT], 2015.
R. J. Stroeker, On the sum of consecutive cubes being a perfect square, Compositio Mathematica, 97 no. 1-2 (1995), pp. 295-307.
Index entries for linear recurrences with constant coefficients, signature (15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1).

Cf. A116108, A116145, A126200, A126203, A163392, A163393, A253679, A253680, A253707, A253708, A002593, A253724.

[(n^3*(4*n^4-1))^2: n in [2..20]]; // Vincenzo Librandi, Feb 19 2015

restart: for n from 2 to 50000 do a:=(n^3*(4*n^4 - 1))^2: print (a); end do:

f[n_] := (n^3 (4 n^4 - 1))^2; Rest[f /@ Range@16] (* Michael De Vlieger, Jan 28 2015 *)
LinearRecurrence[{15,-105,455,-1365,3003,-5005,6435,-6435,5005,-3003,1365,-455,105,-15,1},{254016,76055841,4286582784,97578140625,1253342942784,10849309481241,70360154505216,366000785535609,1599920001000000,6075789839706609,20542200096854016,62996919308080641,177789795179217984,467082203214515625,1152912708530601984},20] (* Harvey P. Dale, Feb 18 2023 *)

a(n) = (n^3(4n^4 - 1))^2.

G.f.: -9*x^2*(x^14 -15*x^13 +106*x^12 +27754*x^11 +8028759*x^10 +352487303*x^9 +4572193580*x^8 +22833696108*x^7 +49725383807*x^6 +49725372367*x^5 +22833705546*x^4 +4572187210*x^3 +352490761*x^2 +8027289*x +28224) / (x -1)^15. - Colin Barker, Jan 14 2015

A000583 Fourth powers: a(n) = n^4.

Original entry on oeis.org

0, 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625, 65536, 83521, 104976, 130321, 160000, 194481, 234256, 279841, 331776, 390625, 456976, 531441, 614656, 707281, 810000, 923521, 1048576, 1185921

Offset: 0

N. J. A. Sloane

Figurate numbers based on 4-dimensional regular convex polytope called the 4-measure polytope, 4-hypercube or tesseract with Schlaefli symbol {4,3,3}. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004
Totally multiplicative sequence with a(p) = p^4 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A058649. The inverse binomial transforms yields the (finite) 0, 1, 14, 36, 24, the 4th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Generate Pythagorean triangles with parameters a and b to get sides of lengths x = b^2-a^2, y = 2*a*b, and z = a^2 + b^2. In particular use a=n-1 and b=n for a triangle with sides (x1,y1,z1) and a=n and b=n+1 for another triangle with sides (x2,y2,z2). Then x1*x2 + y1*y2 + z1*z2 = 8*a(n). - J. M. Bergot, Jul 22 2013
For n > 0, a(n) is the largest integer k such that k^4 + n is a multiple of k + n. Also, for n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n^2. - Derek Orr, Sep 04 2014
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
a(n+2)/2 is the area of a trapezoid with vertices at (T(n), T(n+1)), (T(n+1), T(n)), (T(n+1), T(n+2)), and (T(n+2), T(n+1)) with T(n)=A000292(n) for n >= 0. - J. M. Bergot, Feb 16 2018

John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
Dov Juzuk, Curiosa 56: An interesting observation, Scripta Mathematica 6 (1939), 218.
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).
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 47.

T. D. Noe, Table of n, a(n) for n = 0..1000
Henry Bottomley, Illustration of initial terms
Henry Bottomley, Some Smarandache-type multiplicative sequences
Ralph Greenberg, Math for Poets.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Sameen Ahmed Khan, Sums of the powers of reciprocals of polygonal numbers, Int'l J. of Appl. Math. (2020) Vol. 33, No. 2, 265-282.
Hyun Kwang Kim, On Regular Polytope Numbers, Proc. Amer. Math. Soc., Vol. 131, No. 1 (2002), pp. 65-75.
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
Kenneth A. Ross, First Digits of Squares and Cubes, Math. Mag. 85 (2012) 36-42.
Eric Weisstein's World of Mathematics, Biquadratic Number.
Index entries for "core" sequences
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Index entries for sequences related to Benford's law

Cf. A000538, A005917 (first differences), A000332, A014820, A092181, A092182, A092183.
Cf. A004831, A002646.
Cf. A002593, A260810. - Bruno Berselli, Jul 31 2015
Cf. A002415, A000290, A006008, A132366, A039623, A139584, A071270, A047928, A187756.
Cf. A062392, A231303, A267315.

a000583 = (^ 4)
a000583_list = scanl (+) 0 a005917_list
-- Reinhard Zumkeller, Nov 13 2014, Nov 11 2012

[n^4 : n in [0..50]]; // Wesley Ivan Hurt, Sep 05 2014

A000583 := n->n^4: seq(A000583(n), n=0..50);
A000583:=-(z+1)*(z**2+10*z+1)/(z-1)**5; # Simon Plouffe in his 1992 dissertation; gives sequence without initial zero
with (combinat):seq(fibonacci(3, n^2)-1, n=0..33); # Zerinvary Lajos, May 25 2008

Range[0,100]^4 (* Vladimir Joseph Stephan Orlovsky, Mar 14 2011 *)

makelist(n^4,n,0,30); /* Martin Ettl, Nov 12 2012 */

A000583(n) = n^4 \\ Michael B. Porter, Nov 09 2009

def a(n): return n**4
print([a(n) for n in range(34)]) # Michael S. Branicky, Nov 10 2022

a(n) = A123865(n)+1 = A002523(n)-1.
Multiplicative with a(p^e) = p^(4e). - David W. Wilson, Aug 01 2001
G.f.: x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5. More generally, g.f. for n^m is Euler(m, x)/(1-x)^(m+1), where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292).
Dirichlet generating function: zeta(s-4). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: (x + 7*x^2 + 6*x^3 + x^4)*e^x. More generally, the general form for the e.g.f. for n^m is phi_m(x)*e^x, where phi_m is the exponential polynomial of order n. - Franklin T. Adams-Watters, Sep 11 2005
Sum_{k>0} 1/a(k) = Pi^4/90 = A013662. - Jaume Oliver Lafont, Sep 20 2009
a(n) = C(n+3,4) + 11*C(n+2,4) + 11*C(n+1,4) + C(n,4). [Worpitzky's identity for powers of 4. See, e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n*A177342(n) - Sum_{i=1..n-1} A177342(i) - (n - 1), with n > 1. - Bruno Berselli, May 07 2010
a(n) + a(n+1) + 1 = 2*A002061(n+1)^2. - Charlie Marion, Jun 13 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 24. - Ant King, Sep 23 2013
From Amiram Eldar, Jan 20 2021: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/720 (A267315).
Product_{n>=2} (1 - 1/a(n)) = sinh(Pi)/(4*Pi). (End)

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

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 188.
W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
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).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.

T. D. Noe and Eric Chen, Table of n, a(n) for n = 1..1000 (terms 1..201 from T. D. Noe)
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
César Aguilera, Notes on Perfect Numbers, OSF Preprints, 2023, p 21.
R. C. Alperin, A family of nonlinear recurrences and their linear solutions, Fib. Q., 57:4 (2019), 318-321.
S. Barbero, U. Cerruti, and N. Murru, A Generalization of the Binomial Interpolated Operator and its Action on Linear Recurrent Sequences, J. Int. Seq. 13 (2010) # 10.9.7, proposition 16.
A. Blondin-Massé, S. Brlek, S. Labbé, and M. Mendès France, Fibonacci snowflakes, Special Issue dedicated to Paulo Ribenboim, Annales des Sciences Mathématiques du Québec 35, No 2 (2011).
A. J. C. Cunningham, Binomial Factorisations, Vols. 1-9, Hodgson, London, 1923-1929. See Vol. 1, page xxxv.
J.-P. Ehrmann et al., POLYA003: Integers of the form a/(bc) + b/(ca) + c/(ab).
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
Daniel C. Fielder, Errata:Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.
G. Frobenius, Uber die Markoff'schen Zahlen, Sitzungsber. Konig. Preuss. Akad. Wiss. (1913) p. 348-387, Parag. 9
L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.
Glass, Darren B. Critical groups of graphs with dihedral actions. II. Eur. J. Comb. 61, 25-46 (2017).
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.
R. J. Hetherington, Letter to N. J. A. Sloane, Oct 26 1974
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 403
Tanya Khovanova, Recursive Sequences
Ron Knott, Pythagorean Triples and Online Calculators
Giuseppe Lancia and Paolo Serafini, Polyhedra. Chapter 2 of Compact Extended Linear Programming Models (2018). EURO Advanced Tutorials on Operational Research. Springer, Cham., 11.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Hyperbola, Forum Geometricorum (2019) Vol. 19, 11-16.
A. Martin, Table of prime rational right-angled triangles, The Mathematical Magazine, 2 (1910), 297-324.
A. Martin, Table of prime rational right-angled triangles (annotated scans of a few pages).
Sam Northshield, Topographs; Conway and Otherwise, Fibonacci Quart. 58 (2020), no. 5, 172-189. See p. 16.
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
James M. Parks, Computing Pythagorean Triples, arXiv:2107.06891 [math.GM], 2021.
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
B. Polster and M. Ross, Marching in squares, arXiv preprint arXiv:1503.04658 [math.HO], 2015.
José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368 [cs.DM], 2012-2014.
Dan Romik, The dynamics of Pythagorean Triples, Trans. Amer. Math. Soc. 360 (2008), 6045-6064.
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

A056220 a(n) = 2*n^2 - 1.

Original entry on oeis.org

-1, 1, 7, 17, 31, 49, 71, 97, 127, 161, 199, 241, 287, 337, 391, 449, 511, 577, 647, 721, 799, 881, 967, 1057, 1151, 1249, 1351, 1457, 1567, 1681, 1799, 1921, 2047, 2177, 2311, 2449, 2591, 2737, 2887, 3041, 3199, 3361, 3527, 3697, 3871, 4049, 4231, 4417, 4607, 4801

Offset: 0

N. J. A. Sloane, Aug 06 2000

Image of squares (A000290) under "little Hankel" transform that sends [c_0, c_1, ...] to [d_0, d_1, ...] where d_n = c_n^2 - c_{n+1}*c_{n-1}. - Henry Bottomley, Dec 12 2000
Surround numbers of an n X n square. - Jason Earls, Apr 16 2001
Numbers n such that 2*n + 2 is a perfect square. - Cino Hilliard, Dec 18 2003, Juri-Stepan Gerasimov, Apr 09 2016
The sums of the consecutive integer sequences 2n^2 to 2(n+1)^2-1 are cubes, as 2n^2 + ... + 2(n+1)^2-1 = (1/2)(2(n+1)^2 - 1 - 2n^2 + 1)(2(n+1)^2 - 1 + 2n^2) = (2n+1)^3. E.g., 2+3+4+5+6+7 = 27 = 3^3, then 8+9+10+...+17 = 125 = 5^3. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 29 2005
X values (other than 0) of solutions to the equation 2*X^3 + 2*X^2 = Y^2. To find Y values: b(n) = 2n*(2*n^2 - 1). - Mohamed Bouhamida, Nov 06 2007
Average of the squares of two consecutive terms is also a square. In fact: (2*n^2 - 1)^2 + (2*(n+1)^2 - 1)^2 = 2*(2*n^2 + 2*n + 1)^2. - Matias Saucedo (solomatias(AT)yahoo.com.ar), Aug 18 2008
Equals row sums of triangle A143593 and binomial transform of [1, 6, 4, 0, 0, 0, ...] with n > 1. - Gary W. Adamson, Aug 26 2008
Start a spiral of square tiles. Trivially the first tile fits in a 1 X 1 square. 7 tiles fit in a 3 X 3 square, 17 tiles fit in a 5 X 5 square and so on. - Juhani Heino, Dec 13 2009
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=-2, A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 1, a(n) = coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jan 26 2010
For each n > 0, the recursive series, formula S(b) = 6*S(b-1) - S(b-2) - 2*a(n) with S(0) = 4n^2-4n+1 and S(1) = 2n^2, has the property that every even term is a perfect square and every odd term is twice a perfect square. - Kenneth J Ramsey, Jul 18 2010
Fourth diagonal of A154685 for n > 2. - Vincenzo Librandi, Aug 07 2010
First integer of (2*n)^2 consecutive integers, where the last integer is 3 times the first + 1. As example, n = 2: term = 7; (2*n)^2 = 16; 7, 8, 9, ..., 20, 21, 22: 7*3 + 1 = 22. - Denis Borris, Nov 18 2012
Chebyshev polynomial of the first kind T(2,n). - Vincenzo Librandi, May 30 2014
For n > 0, number of possible positions of a 1 X 2 rectangle in a (n+1) X (n+2) rectangular integer lattice. - Andres Cicuttin, Apr 07 2016
This sequence also represents the best solution for Ripà's n_1 X n_2 X n_3 dots problem, for any 0 < n_1 = n_2 < n_3 = floor((3/2)*(n_1 - 1)) + 1. - Marco Ripà, Jul 23 2018

			a(0) = 0^2-1*1 = -1, a(1) = 1^2 - 4*0 = 1, a(2) = 2^2 - 9*1 = 7, etc.
a(4) = 31 = (1, 3, 3, 1) dot (1, 6, 4, 0) = (1 + 18 + 12 + 0). - _Gary W. Adamson_, Aug 29 2008

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Jeremiah Bartz, Bruce Dearden, and Joel Iiams, Classes of Gap Balancing Numbers, arXiv:1810.07895 [math.NT], 2018.
Milan Janjić, Hessenberg Matrices and Integer Sequences, J. Int. Seq., Vol. 13 (2010), Article # 10.7.8.
Mitch Phillipson, Manda Riehl, and Tristan Williams, Enumeration of Wilf classes in Sn ~ Cr for two patterns of length 3, PU. M. A., Vol. 21, No. 2 (2010), pp. 321-338.
Marco Ripà, The rectangular spiral or the n1 X n2 X ... X nk Points Problem , Notes on Number Theory and Discrete Mathematics, Vol. 20, No. 1 (2014), pp. 59-71.
Leo Tavares, Illustration: Twin Squares
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000105, A000217, A000290, A000384, A001082, A001653, A001844, A002378, A002593, A003215, A005563, A028347, A036666, A046092, A047875, A062717, A069074, A077585, A087475, A119258, A143593, A154685, A162610, A188653, A225227.
Cf. A066049 (indices of prime terms)
Column 2 of array A188644 (starting at offset 1).

List([0..50], n-> 2*n^2-1); # Muniru A Asiru, Jul 24 2018

[2*n^2-1 : n in [0..50]]; // Vincenzo Librandi, May 30 2014

A056220:=n->2*n^2-1; seq(A056220(n), n=0..50); # Wesley Ivan Hurt, Jun 16 2014

Array[2 #^2 - 1 &, 50, 0] (* Robert G. Wilson v, Jul 23 2018 *)
CoefficientList[Series[(x^2 +4x -1)/(1-x)^3, {x, 0, 50}], x] (* or *)
LinearRecurrence[{3, -3, 1}, {-1, 1, 7}, 51] (* Robert G. Wilson v, Jul 24 2018 *)

PARI
```
a(n)=2*n^2-1;
```

[2*n^2-1 for n in (0..50)] # G. C. Greubel, Jul 07 2019

G.f.: (-1 + 4*x + x^2)/(1-x)^3. - Henry Bottomley, Dec 12 2000
a(n) = A119258(n+1,2) for n > 0. - Reinhard Zumkeller, May 11 2006
From Doug Bell, Mar 08 2009: (Start)
a(0) = -1,
a(n) = sqrt(A001844(n)^2 - A069074(n-1)),
a(n+1) = sqrt(A001844(n)^2 + A069074(n-1)) = sqrt(a(n)^2 + A069074(n-1)*2). (End)
a(n) + a(n+1) + 1 = (2n+1)^2. - Doug Bell, Mar 09 2009
a(n) = a(n-1) + 4*n - 2 (with a(0)=-1). - Vincenzo Librandi, Dec 25 2010
a(n) = A188653(2*n) for n > 0. - Reinhard Zumkeller, Apr 13 2011
a(n) = A162610(2*n-1,n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(n) = ( Sum_{k=0..2} (C(n+k,3)-C(n+k-1,3))*(C(n+k,3)+C(n+k+1,3)) ) - (C(n+2,3)-C(n-1,3))*(C(n,3)+C(n+3,3)), for n > 3. - J. M. Bergot, Jun 16 2014
a(n) = j^2 + k^2 - 2 or 2*j*k if n >= 2 and j = n + sqrt(2)/2 and k = n - sqrt(2)/2. - Avi Friedlich, Mar 30 2015
a(n) = A002593(n)/n^2. - Bruce J. Nicholson, Apr 03 2017
a(n) = A000384(n) + n - 1. - Bruce J. Nicholson, Nov 12 2017
a(n)*a(n+k) + 2k^2 = m^2 (a perfect square), m = a(n) + (2n*k), for n>=1. - Ezhilarasu Velayutham, May 13 2019
From Amiram Eldar, Aug 10 2020: (Start)
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(2)*Pi*cot(Pi/sqrt(2))/4.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(2)*Pi*csc(Pi/sqrt(2))/4 - 1/2. (End)
From Amiram Eldar, Feb 04 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = (Pi/sqrt(2))*csc(Pi/sqrt(2)).
Product_{n>=2} (1 - 1/a(n)) = (Pi/(4*sqrt(2)))*csc(Pi/sqrt(2)). (End)
a(n) = A003215(n) - A000217(n-2)*2. - Leo Tavares, Jun 29 2021
Let T(n) = n*(n+1)/2. Then a(n)^2 = T(2n-2)*T(2n+1) + n^2. - Charlie Marion, Feb 12 2023
E.g.f.: exp(x)*(2*x^2 + 2*x - 1). - Stefano Spezia, Jul 08 2023

A002309 Sum of fourth powers of first n odd numbers.

Original entry on oeis.org

1, 82, 707, 3108, 9669, 24310, 52871, 103496, 187017, 317338, 511819, 791660, 1182285, 1713726, 2421007, 3344528, 4530449, 6031074, 7905235, 10218676, 13044437, 16463238, 20563863, 25443544, 31208345, 37973546, 45864027, 55014652, 65570653, 77688014

Offset: 1

N. J. A. Sloane

			a(1) = 1^4 = 1.
a(2) = 1^4 + 3^4 = 82.
a(3) = 1^4 + 3^4 + 5^4 = 707.

F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
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).

T. D. Noe, Table of n, a(n) for n = 1..1000
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359. [Annotated scanned copy]
J. L. Bailey, Jr., A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., 2 (1931), 355-359.
F. E. Croxton and D. J. Cowden, Applied General Statistics, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955 [Annotated scans of just pages 742-743]
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
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000027, A000290, A000447, A002593, A005408.

A002309:=(1+76*z+230*z**2+76*z**3+z**4)/(z-1)**6; # conjectured by Simon Plouffe in his 1992 dissertation

Table[(48*n^5 - 40*n^3 + 7*n)/15, {n, 0, 40}] (* Stefan Steinerberger, Apr 10 2006 *)
s = 0; lst = {s}; Do[s += n^4; AppendTo[lst, s], {n, 1, 60, 2}]; lst (* Zerinvary Lajos, Jul 12 2009 *)
Accumulate[Range[1,63,2]^4] (* Harvey P. Dale, Oct 24 2011 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,82,707,3108,9669,24310},20] (* Harvey P. Dale, Sep 29 2015 *)

a(n)=(48*n^5-40*n^3+7*n)/15 \\ Charles R Greathouse IV, Apr 07 2016

def A002309(n): return n*(n**2*(6*n**2-5<<3)+7)//15 # Chai Wah Wu, Oct 02 2024

a(n) = (48*n^5 - 40*n^3 + 7*n)/15. - Ralf Stephan, Jan 29 2003
a(1)=1, a(2)=82, a(3)=707, a(4)=3108, a(5)=9669, a(6)=24310, a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Harvey P. Dale, Oct 24 2011
a(n) = v(n,n-2) - v(n,n-1)*V(n,n-1), where v(n,k) and V(n,k) are the central factorial numbers of the first kind and the second kind, respectively, with odd indices. - Mircea Merca, Jan 25 2014
From Wolfdieter Lang, Mar 11 2017: (Start)
G.f.: x*(1 + 76*x + 230*x^2 + 76*x^3 + 1*x^4)/(1-x)^6.
E.g.f. (with offset 0): exp(x)*(1 + 81*x + 544*x^2/2! + 1232*x^3/3! + 1152*x^4/4! + 384*x^5/5!). (End)

Definition changed by David A. Corneth, Mar 11 2017

Name clarified by Mohammed Yaseen, Jul 24 2023

A105636 Transform of n^3 by the Riordan array (1/(1-x^2), x).

Original entry on oeis.org

0, 1, 8, 28, 72, 153, 288, 496, 800, 1225, 1800, 2556, 3528, 4753, 6272, 8128, 10368, 13041, 16200, 19900, 24200, 29161, 34848, 41328, 48672, 56953, 66248, 76636, 88200, 101025, 115200, 130816, 147968, 166753, 187272, 209628, 233928, 260281, 288800, 319600

Offset: 0

Paul Barry, Apr 16 2005

Recurrence a(n) = a(n-2) + n^3, starting with a(0)=0, a(1)=1. Also, in physics, a(n)/4 is the trace of the spin operator |S_z|^3 for a particle with spin S=n/2. For example, when S=3/2, the S_z eigenvalues are -3/2, -1/2, +1/2, +3/2 and therefore the sum of the absolute values of their 3rd powers is 2*28/8 = a(3)/4. - Stanislav Sykora, Nov 07 2013
Also the number of 3-cycles in the (n+1)-triangular honeycomb queen graph. - Eric W. Weisstein, Jul 14 2017
With zero prepended and offset 1, the sequence starts 0,0,1,8,28,... for n=1,2,3,... Call this b(n). Consider the partitions of n into two parts (p,q). Then b(n) is the total volume of the family of cubes with side length |q - p|. - Wesley Ivan Hurt, Apr 14 2018

Stanislav Sykora, Table of n, a(n) for n = 0..9999
Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Eric Weisstein's World of Mathematics, Graph Cycle.
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A002620, A000292, A000578, A011934, A231303.

Cf. A289705 (4-cycles), A289706 (5-cycles), A289707 (6-cycles).

List([0..30], n -> (2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16); # G. C. Greubel, Dec 16 2018

[(2*n^4+8*n^3+8*n^2-1)/16+(-1)^n/16: n in [0..50]]; // Vincenzo Librandi, Oct 27 2014

LinearRecurrence[{4, -5, 0, 5, -4, 1}, {0, 1, 8, 28, 72, 153}, 60] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)
CoefficientList[Series[x (1 + 4 x + x^2)/((1 + x) (1 - x)^5), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 26 2012 *)
Table[((-1)^n + 2 n^2 (n + 2)^2 - 1)/16, {n, 0, 20}] (* Eric W. Weisstein, Jul 14 2017 *)

my(x='x+O('x^99)); concat(0, Vec(x*(1+4*x+x^2)/((1+x)*(1-x)^5))) \\ Altug Alkan, Apr 16 2018

[(2*n^4 +8*n^3 +8*n^2 -1+(-1)^n)/16 for n in range(30)] # G. C. Greubel, Dec 16 2018

G.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^5).
a(n) = 4*a(n-1) - 5*a(n-2) + 5*a(n-4) - 4*a(n-5) + a(n-6).
a(n) = (2*n^4 + 8*n^3 + 8*n^2 - 1 + (-1)^n)/16.
a(n) = Sum_{k=0..floor((n-1)/2)} (n-2*k)^3.
a(n+1) = Sum_{k=0..n} k^3*(1 - (-1)^(n+k-1))/2.
a(n) = ((((x^2 - (x mod 2) - 4)/4)^2 - (((x^2 - (x mod 2) - 4)/4) mod 2))/8) = floor(((floor(x^2/4) - 1)^2)/8) where x = 2*n + 2. Replace x with 2*n - 1 to obtain A050534(n) = 3*A000332(n+1). Note that a(2*n) = A060300(n)/2 and a(2*n + 1) = A002593(n+1). - Raphie Frank, Jan 30 2014
a(n) = floor(1/(exp(2/n^2) - 1)^2)/2. Also a(n) = A007590(n+1)*A074148(n-1)/2. - Richard R. Forberg, Oct 26 2014
Sum_{n>=1} 1/a(n) = -cot(Pi/sqrt(2))*Pi/sqrt(2) - 1/2. - Amiram Eldar, Aug 25 2022

A260810 a(n) = n^2(3n^2 - 1)/2.

Original entry on oeis.org

0, 1, 22, 117, 376, 925, 1926, 3577, 6112, 9801, 14950, 21901, 31032, 42757, 57526, 75825, 98176, 125137, 157302, 195301, 239800, 291501, 351142, 419497, 497376, 585625, 685126, 796797, 921592, 1060501, 1214550, 1384801, 1572352, 1778337, 2003926, 2250325, 2518776

Offset: 0

Bruno Berselli, Jul 31 2015

Pentagonal numbers with square indices.

After 0, a(k) is a square if k is in A072256.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Subsequence of A001318 and A245288 (see Formula field).
Cf. A000326, A193218 (first differences).
Cf. A000583 (squares with square indices), A002593 (hexagonal numbers with square indices).
Cf. A232713 (pentagonal numbers with pentagonal indices), A236770 (pentagonal numbers with triangular indices).

Magma
```
[n^2*(3*n^2-1)/2: n in [0..40]];
```

I:=[0,1,22,117,376]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Aug 23 2015

A260810:=n->n^2*(3*n^2 - 1)/2: seq(A260810(n), n=0..50); # Wesley Ivan Hurt, Apr 25 2017

Table[n^2 (3 n^2 - 1)/2, {n, 0, 40}]
LinearRecurrence[{5, -10, 10, -5, 1}, {0, 1, 22, 117, 376}, 40] (* Vincenzo Librandi, Aug 23 2015 *)

PARI
```
vector(40, n, n--; n^2*(3*n^2-1)/2)
```
Sage
```
[n^2*(3*n^2-1)/2 for n in (0..40)]
```

G.f.: x*(1 + x)*(1 + 16*x + x^2)/(1 - x)^5.
a(n) = a(-n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = A245288(2*n^2).
a(n) = A001318(2*n^2-1) with A001318(-1) = 0.
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 3 - Pi^2/3 - sqrt(3)*Pi*cot(Pi/sqrt(3)).
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi*cosec(Pi/sqrt(3)) - Pi^2/6 - 3. (End)

A254371 Sum of cubes of the first n even numbers (A016743).

Original entry on oeis.org

0, 8, 72, 288, 800, 1800, 3528, 6272, 10368, 16200, 24200, 34848, 48672, 66248, 88200, 115200, 147968, 187272, 233928, 288800, 352800, 426888, 512072, 609408, 720000, 845000, 985608, 1143072, 1318688, 1513800, 1729800, 1968128, 2230272, 2517768, 2832200, 3175200

Offset: 0

Luciano Ancora, Mar 16 2015

Property: for n >= 2, each (a(n), a(n)+1, a(n)+2) is a triple of consecutive terms that are the sum of two nonzero squares; precisely: a(n) = (n*(n + 1))^2 + (n*(n + 1))^2, a(n)+1 = (n^2+2n)^2 + (n^2-1)^2 and a(n)+2 = (n^2+n+1)^2 + (n^2+n-1)^2 (see Diophante link). - Bernard Schott, Oct 05 2021

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Luciano Ancora, The Square Pyramidal Number and other figurate numbers, ch. 3.
Diophante, Une miniature avec trois entiers consécutifs (in French).
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000537 (sum of first n cubes); A002593 (sum of first n odd cubes).
Cf. A060300 (2*a(n)).
First bisection of A105636; second bisection of A212892.
Cf. A000217, A000415, A002378, A035287, A163102.

List([0..35],n->2*(n*(n+1))^2); # Muniru A Asiru, Oct 24 2018

[2*n^2*(n+1)^2: n in [0..40]]; // Bruno Berselli, Mar 23 2015

A254371:=n->2*n^2*(n + 1)^2: seq(A254371(n), n=0..50); # Wesley Ivan Hurt, Apr 28 2017

Table[2 n^2 (n+1)^2, {n, 0, 40}] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {0, 8, 72, 288, 800}, 40]
Accumulate[Range[0,80,2]^3] (* Harvey P. Dale, Jun 26 2017 *)

a(n)=sum(i=0, n, 8*i^3); \\ Michael B. Porter, Mar 16 2015

G.f.: 8*x*(1 + 4*x + x^2)/(1 - x)^5.
a(n) = 2*n^2*(n + 1)^2.
a(n) = 2*A035287(n+1) = 2*A002378(n)^2 = 8*A000217(n)^2. - Bruce J. Nicholson, Apr 23 2017
a(n) = 8*A000537(n). - Michel Marcus, Apr 23 2017
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/6 - 3/2.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - 2*log(2). (End)
From Elmo R. Oliveira, Aug 14 2025: (Start)
E.g.f.: 2*x*(2 + x)*(2 + 6*x + x^2)*exp(x).
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = 4*A163102(n) = A060300(n)/2. (End)

A099764 a(n) = n^2 * (n+1)^2 * (n+2)^2 = 36*A001249(n-1).

Original entry on oeis.org

0, 36, 576, 3600, 14400, 44100, 112896, 254016, 518400, 980100, 1742400, 2944656, 4769856, 7452900, 11289600, 16646400, 23970816, 33802596, 46785600, 63680400, 85377600, 112911876, 147476736, 190440000, 243360000, 308002500, 386358336

Offset: 0

Kari Lajunen (Kari.Lajunen(AT)Welho.com), Nov 11 2004

			a(0) = 1^3 - 1^3 = 0;
a(1) = (1+3)^3 - (1^3+3^3) = 64 - 28 = 36;
a(2) = (1+3+5)^3 - (1^3+3^3+5^3) = 729 - 153 = 576;
a(3) = (1+3+5+7)^3 - (1^3+3^3+5^3+7^3) = 4096 - 496 = 3600;
a(4) = (1+3+5+7+9)^3 - (1^3+3^3+5^3+7^3+9^3) = 15625 - 1225 = 14400; etc. - _Philippe Deléham_, Mar 10 2014

Jolley, Summation of Series, Dover (1961).

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A001014, A002593.

List([1..30], n-> (n*(n^2-1))^2); # G. C. Greubel, Sep 03 2019

[n^2*(n+1)^2*(n+2)^2: n in [0..30]]; // Vincenzo Librandi, Oct 04 2011

A099764:=n->n^2*(n+1)^2*(n+2)^2; seq(A099764(n), n=0..30); # Wesley Ivan Hurt, Mar 09 2014

Table[n^2*(n+1)^2*(n+2)^2, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)
Times@@@Partition[Range[0,30]^2,3,1] (* Harvey P. Dale, Sep 02 2016 *)

vector(30, n, (n*(n^2-1))^2) \\ G. C. Greubel, Sep 03 2019

[(n*(n^2-1))^2 for n in (1..30)] # G. C. Greubel, Sep 03 2019

Sum_{n>=1} 1/a(n) = Pi^2/4-39/16 =  0.029901100272... [Jolley eq 241]
G.f.: 36*x*(1+x)*(1 +8*x +x^2)/(1-x)^7 . - R. J. Mathar, Oct 03 2011
a(n) = (Sum_{k=0..n} (2*k+1))^3 - Sum_{k=0..n} (2*k+1)^3. - Philippe Deléham, Mar 10 2014
a(n) = A001014(n+1) - A002593(n+1). - Philippe Deléham, Mar 10 2014
E.g.f.: exp(x)*x*(36+252*x+330*x^2+138*x^3+21*x^4+x^5). - Stefano Spezia, Sep 04 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/24 - 7/16. - Amiram Eldar, Jul 02 2020

cp's OEIS Frontend

A253724 Numbers c(n) whose squares are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A253725 Integer squares c^2 that are equal to the sums of a number M(n) of consecutive cubed integers b^3 + (b+1)^3 + ... + (b+M-1)^3 = c^2, starting at b(n) (A002593) for M(n) being twice a squared integer (A001105).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A000583 Fourth powers: a(n) = n^4.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Magma

Maple

Mathematica

Maxima

PARI

Python

Formula

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

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

GAP

Haskell

Magma

Maple

Mathematica

PARI

PARI

PARI

Formula

Extensions

A056220 a(n) = 2*n^2 - 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

PARI

Sage

A260810 a(n) = n^2(3n^2 - 1)/2.