A350916 -id:A350916 - cp's OEIS frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

1, 2, 7, 26, 97, 362, 1351, 5042, 18817, 70226, 262087, 978122, 3650401, 13623482, 50843527, 189750626, 708158977, 2642885282, 9863382151, 36810643322, 137379191137, 512706121226, 1913445293767, 7141075053842, 26650854921601, 99462344632562, 371198523608647

Offset: 0

N. J. A. Sloane

Chebyshev's T(n,x) polynomials evaluated at x=2.
x = 2^n - 1 is prime if and only if x divides a(2^(n-2)).
Any k in the sequence is succeeded by 2*k + sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002
For all elements x of the sequence, 12*x^2 - 12 is a square. Lim_{n -> infinity} a(n)/a(n-1) = 2 + sqrt(3) = (4 + sqrt(12))/2 which preserves the kinship with the equation "12*x^2 - 12 is a square" where the initial "12" ends up appearing as a square root. - Gregory V. Richardson, Oct 10 2002
This sequence gives the values of x in solutions of the Diophantine equation x^2 - 3*y^2 = 1; the corresponding values of y are in A001353. The solution ratios a(n)/A001353(n) are obtained as convergents of the continued fraction expansion of sqrt(3): either as successive convergents of [2;-4] or as odd convergents of [1;1,2]. - Lekraj Beedassy, Sep 19 2003 [edited by Jon E. Schoenfield, May 04 2014]
a(n) is half the central value in a list of three consecutive integers, the lengths of the sides of a triangle with integer sides and area. - Eugene McDonnell (eemcd(AT)mac.com), Oct 19 2003
a(3+6*k) - 1 and a(3+6*k) + 1 are consecutive odd powerful numbers. See A076445. - T. D. Noe, May 04 2006
The intermediate convergents to 3^(1/2), beginning with 3/2, 12/7, 45/26, 168/97, comprise a strictly increasing sequence; essentially, numerators=A005320, denominators=A001075. - Clark Kimberling, Aug 27 2008
The upper principal convergents to 3^(1/2), beginning with 2/1, 7/4, 26/15, 97/56, comprise a strictly decreasing sequence; numerators=A001075, denominators=A001353. - Clark Kimberling, Aug 27 2008
a(n+1) is the Hankel transform of A000108(n) + A000984(n) = (n+2)*Catalan(n). - Paul Barry, Aug 11 2009
Also, numbers such that floor(a(n)^2/3) is a square: base 3 analog of A031149, A204502, A204514, A204516, A204518, A204520, A004275, A001541. - M. F. Hasler, Jan 15 2012
Pisano period lengths:  1, 2, 2, 4, 3, 2, 8, 4, 6, 6, 10, 4, 12, 8, 6, 8, 18, 6, 5, 12, ... - R. J. Mathar, Aug 10 2012
Except for the first term, positive values of x (or y) satisfying x^2 - 4*x*y + y^2 + 3 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 14*x*y + y^2 + 48 = 0. - Colin Barker, Feb 10 2014
From Gary W. Adamson, Jul 25 2016: (Start)
A triangle with row sums generating the sequence can be constructed by taking the production matrix M. Take powers of M, extracting the top rows.
M =
  1, 1, 0, 0, 0, 0, ...
  2, 0, 3, 0, 0, 0, ...
  2, 0, 0, 3, 0, 0, ...
  2, 0, 0, 0, 3, 0, ...
  2, 0, 0, 0, 0, 3, ...
  ...
The triangle generated from M is:
   1,
   1,  1,
   3,  1, 3,
  11,  3, 3, 9,
  41, 11, 9, 9, 27,
   ...
The left border is A001835 and row sums are (1, 2, 7, 26, 97, ...). (End)
Even-indexed terms are odd while odd-indexed terms are even. Indeed, a(2*n) = 2*(a(n))^2 - 1 and a(2*n+1) = 2*a(n)*a(n+1) - 2. - Timothy L. Tiffin, Oct 11 2016
For each n, a(0) divides a(n), a(1) divides a(2n+1), a(2) divides a(4*n+2), a(3) divides a(6*n+3), a(4) divides a(8*n+4), a(5) divides a(10n+5), and so on. Thus, a(k) divides a((2*n+1)*k) for each k > 0 and n >= 0. A proof of this can be found in Bhargava-Kedlaya-Ng's first solution to Problem A2 of the 76th Putnam Mathematical Competition. Links to the exam and its solutions can be found below. - Timothy L. Tiffin, Oct 12 2016
From Timothy L. Tiffin, Oct 21 2016: (Start)
If any term a(n) is a prime number, then its index n will be a power of 2. This is a consequence of the results given in the previous two comments. See A277434 for those prime terms.
a(2n) == 1 (mod 6) and a(2*n+1) == 2 (mod 6). Consequently, each odd prime factor of a(n) will be congruent to 1 modulo 6 and, thus, found in A002476.
a(n) == 1 (mod 10) if n == 0 (mod 6), a(n) == 2 (mod 10) if n == {1,-1} (mod 6), a(n) == 7 (mod 10) if n == {2,-2} (mod 6), and a(n) == 6 (mod 10) if n == 3 (mod 6). So, the rightmost digits of a(n) form a repeating cycle of length 6: 1, 2, 7, 6, 7, 2. (End)
a(A298211(n)) = A002350(3*n^2). - A.H.M. Smeets, Jan 25 2018
(2 + sqrt(3))^n = a(n) + A001353(n)*sqrt(3), n >= 0; integers in the quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 16 2018
Yong Hao Ng has shown that for any n, a(n) is coprime with any member of A001834 and with any member of A001835. - René Gy, Feb 26 2018
Positive numbers k such that 3*(k-1)*(k+1) is a square. - Davide Rotondo, Oct 25 2020
a(n)*a(n+1)-1 = a(2*n+1)/2 = A001570(n) divides both a(n)^6+1 and a(n+1)^6+1. In other words, for k = a(2*n+1)/2, (k+1)^6 has divisors congruent to -1 modulo k (cf. A350916). - Max Alekseyev, Jan 23 2022

			2^6 - 1 = 63 does not divide a(2^4) = 708158977, therefore 63 is composite. 2^5 - 1 = 31 divides a(2^3) = 18817, therefore 31 is prime.
G.f. = 1 + 2*x + 7*x^2 + 26*x^3 + 97*x^4 + 362*x^5 + 1351*x^6 + 5042*x^7 + ...

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
Eugene McDonnell, "Heron's Rule and Integer-Area Triangles", Vector 12.3 (January 1996) pp. 133-142.
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, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238.

Indranil Ghosh, Table of n, a(n) for n = 0..1745 (terms 0..200 from T. D. Noe)
Christian Aebi and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Christian Aebi and Grant Cairns, Less than Equable Triangles on the Eisenstein lattice, arXiv:2312.10866 [math.CO], 2023.
Krassimir T. Atanassov and Anthony G. Shannon, On intercalated Fibonacci sequences, Notes on Number Theory and Discrete Mathematics (2020) Vol. 26, No. 3, 218-223.
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
H. Brocard, Notes élémentaires sur le problème de Pell, Nouvelle Correspondance Mathématique, 4 (1878), 337-343.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 30, 43, 56.
Chris Caldwell, Primality Proving, Arndt's theorem.
J. B. Cosgrave and K. Dilcher, A role for generalized Fermat numbers, Math. Comp., 2016.
G. Dresden and Y. Li, Periodic Weighted Sums of Binomial Coefficients, arXiv:2210.04322 [math.NT], 2022.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
Margherita Maria Ferrari and Norma Zagaglia Salvi, Aperiodic Compositions and Classical Integer Sequences, Journal of Integer Sequences, Vol. 20 (2017), Article 17.8.8.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Kiran S. Kedlaya, The 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Kiran S. Kedlaya, Solutions to the 76th William Lowell Putnam Mathematical Competition, Dec 05 2015.
Tanya Khovanova, Recursive Sequences
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Pablo Lam-Estrada, Myriam Rosalía Maldonado-Ramírez, José Luis López-Bonilla, and Fausto Jarquín-Zárate, The sequences of Fibonacci and Lucas for each real quadratic fields Q(Sqrt(d)), arXiv:1904.13002 [math.NT], 2019.
Eugene McDonnell, Heron's Rule and Integer-Area Triangles, At Play With J, 2010.
Valcho Milchev and Tsvetelina Karamfilova, Domino tiling in grid - new dependence, arXiv:1707.09741 [math.HO], 2017.
Yong Hao Ng, A partition in three classes of the set of all prime numbers?, Mathematics Stack Exchange.
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
F. V. Waugh and M. W. Maxfield, Side-and-diagonal numbers, Math. Mag., 40 (1967), 74-83.
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-1).

Bisections are A011943 and A094347.

Cf. A001353, A065918, A071954, A001571, A001834, A003500, A016064, A082840, A026150, A277434.

a001075 n = a001075_list !! n
a001075_list =
   1 : 2 : zipWith (-) (map (4 *) $ tail a001075_list) a001075_list
-- Reinhard Zumkeller, Aug 11 2011

I:=[1, 2]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017

A001075 := proc(n)
    orthopoly[T](n,2) ;
end proc:
seq(A001075(n),n=0..30) ; # R. J. Mathar, Apr 14 2018

Table[ Ceiling[(1/2)*(2 + Sqrt[3])^n], {n, 0, 24}]
CoefficientList[Series[(1-2*x) / (1-4*x+x^2), {x, 0, 24}], x] (* Jean-François Alcover, Dec 21 2011, after Simon Plouffe *)
LinearRecurrence[{4,-1},{1,2},30] (* Harvey P. Dale, Aug 22 2015 *)
Round@Table[LucasL[2n, Sqrt[2]]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 15 2016 *)
ChebyshevT[Range[0, 20], 2] (* Eric W. Weisstein, May 26 2017 *)
a[ n_] := LucasL[2*n, x]/2 /. x->Sqrt[2]; (* Michael Somos, Sep 05 2022 *)

{a(n) = subst(poltchebi(abs(n)), x, 2)};

{a(n) = real((2 + quadgen(12))^abs(n))};

{a(n) = polsym(1 - 4*x + x^2, abs(n))[1 + abs(n)]/2};

a(n)=polchebyshev(n,1,2) \\ Charles R Greathouse IV, Nov 07 2016

my(x='x+O('x^30)); Vec((1-2*x)/(1-4*x+x^2)) \\ G. C. Greubel, Dec 19 2017

[lucas_number2(n,4,1)/2 for n in range(0, 25)] # Zerinvary Lajos, May 14 2009

def a(n):
    Q = QuadraticField(3, 't')
    u = Q.units()[0]
    return (u^n).lift().coeffs()[0]  # Ralf Stephan, Jun 19 2014

G.f.: (1 - 2*x)/(1 - 4*x + x^2). - Simon Plouffe in his 1992 dissertation
E.g.f.: exp(2*x)*cosh(sqrt(3)*x).
a(n) = 4*a(n-1) - a(n-2) = a(-n).
a(n) = (S(n, 4) - S(n-2, 4))/2 = T(n, 2), with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. U, resp. T, are Chebyshev's polynomials of the second, resp. first, kind. S(n-1, 4) = A001353(n), n >= 0. See A049310 and A053120.
a(n) = A001353(n+2) - 2*A001353(n+1).
a(n) = sqrt(1 + 3*A001353(n)) (cf. Richardson comment, Oct 10 2002).
a(n) = 2^(-n)*Sum_{k>=0} binomial(2*n, 2*k)*3^k = 2^(-n)*Sum_{k>=0} A086645(n, k)*3^k. - Philippe Deléham, Mar 01 2004
a(n) = ((2 + sqrt(3))^n + (2 - sqrt(3))^n)/2; a(n) = ceiling((1/2)*(2 + sqrt(3))^(n)).
a(n) = cosh(n * log(2 + sqrt(3))).
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*2^(n-2*k)*3^k. - Paul Barry, May 08 2003
a(n+2) = 2*a(n+1) + 3*Sum_{k>=0} a(n-k)*2^k. - Philippe Deléham, Mar 03 2004
a(n) = 2*a(n-1) + 3*A001353(n-1). - Lekraj Beedassy, Jul 21 2006
a(n) = left term of M^n * [1,0] where M = the 2 X 2 matrix [2,3; 1,2]. Right term = A001353(n). Example: a(4) = 97 since M^4 * [1,0] = [A001075(4), A001353(4)] = [97, 56]. - Gary W. Adamson, Dec 27 2006
Binomial transform of A026150: (1, 1, 4, 10, 28, 76, ...). - Gary W. Adamson, Nov 23 2007
First differences of A001571. - N. J. A. Sloane, Nov 03 2009
Sequence satisfies -3 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 4*u*v. - Michael Somos, Sep 19 2008
a(n) = Sum_{k=0..n} A201730(n,k)*2^k. - Philippe Deléham, Dec 06 2011
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x*(3*k - 4)/(x*(3*k - 1) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
a(n) = Sum_{k=0..n} A238731(n,k). - Philippe Deléham, Mar 05 2014
a(n) = (-1)^n*(A125905(n) + 2*A125905(n-1)), n > 0. - Franck Maminirina Ramaharo, Nov 11 2018
a(n) = (tan(Pi/12)^n + tan(5*Pi/12)^n)/2. - Greg Dresden, Oct 01 2020
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 4*x + sqrt(1 + 12*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - (3/2)/a(n)) = 1.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + (1/2)/a(n)) = 1/3.
Sum_{n >= 1} 1/(a(n)^2 - 3/2) = 1 - 1/sqrt(3). (End)
a(n) = binomial(2*n, n) + 2*Sum_{k > 0} binomial(2*n, n+2*k)*cos(k*Pi/3). - Greg Dresden, Oct 11 2022
2*a(n) + 2^n = 3*Sum_{k=-n..n} (-1)^k*binomial(2*n, n+6*k). - Greg Dresden, Feb 07 2023

More terms from James Sellers, Jul 10 2000

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A001570 Numbers k such that k^2 is centered hexagonal.

Original entry on oeis.org

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621

Offset: 1

N. J. A. Sloane

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
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).

G. C. Greubel, Table of n, a(n) for n = 1..870 (terms 1..101 from T. D. Noe)
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.
G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes, part C (in French).
Tanya Khovanova, Recursive Sequences
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.
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.
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics, Hex Number
Wikipedia, Beal's conjecture
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
Cf. A076139, A076140, A102871.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

[((2 + Sqrt(3))^(2*n - 1) + (2 - Sqrt(3))^(2*n - 1))/4: n in [1..50]]; // G. C. Greubel, Nov 04 2017

A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
  ] (* Complement of A041017 *)
a[12, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)

{a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

Original entry on oeis.org

1, 5, 65, 901, 12545, 174725, 2433601, 33895685, 472105985, 6575588101, 91586127425, 1275630195845, 17767236614401, 247465682405765, 3446752317066305, 48007066756522501, 668652182274248705

Offset: 1

Zak Seidov, Feb 23 2005

Corresponding areas are: 0, 12, 1848, 351780, 68149872, 13219419708, 2564481115560 (see A104009).
What is the next term? Is the sequence finite? The possible last two digits of "a" are (it may help in searching for more terms): {01, 05, 09, 15, 19, 25, 29, 33, 35, 39, 45, 49, 51, 55, 59, 65, 69, 75, 79, 83, 85, 89, 95, 99}.
Equivalently, positive integers a such that 3/16*a^4 + 1/4*a^3 - 1/8*a^2 - 1/4*a - 1/16 is a square (A000290), a direct result of Heron's formula. Conjecture: lim_{n->oo} a(n+1)/a(n) = 7 + 4*sqrt(3) (= 7 + A010502). - Rick L. Shepherd, Sep 04 2005
Values x^2 + y^2, where the pair (x, y) solves for x^2 - 3y^2=1, i.e., a(n)= (A001075(n))^2 + (A001353(n))^2 = A055793(n) + A098301(n). - Lekraj Beedassy, Jul 13 2006
Floretion Algebra Multiplication Program, FAMP Code: 1lestes[ 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e ]

Christian Aebi, and Grant Cairns, Lattice Equable Parallelograms, arXiv:2006.07566 [math.NT], 2020.
Project Euler, Problem 94: Almost Equilateral Triangles.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011922, A007655, A001353, A102341, A103975, A016064, A011945, A010502 (4*sqrt(3)), A000290 (square numbers), A350916.

A:=rsolve({-A(n+3)+15*A(n+2)-15*A(n+1)+A(n), A(0) = 1, A(1) = 5, A(2)=65}, A(n), makeproc); # Mihailovs

f[n_] := Simplify[((2 + Sqrt[3])^(2n) + (2 - Sqrt[3])^(2n) + 1)/3]; Table[ f[n], {n, 0, 16}] (* Or *)
a[1] = 1; a[2] = 5; a[3] = 65; a[n_] := a[n] = 15a[n - 1] - 15a[n - 2] + a[n - 3]; Table[ a[n], {n, 17}] (* Or *)
CoefficientList[ Series[(1 - 10x + 5x^2)/(1 - 15x + 15x^2 - x^3), {x, 0, 16}], x] (* Or *)
Range[0, 16]! CoefficientList[ Simplify[ Series[(E^x + E^((7 + 4Sqrt[3])x) + E^((7 - 4Sqrt[3])x))/3, {x, 0, 16}]], x] (* Robert G. Wilson v, Mar 24 2005 *)

for(a=1,10^6, b=a; c=a+1; s=(a+b+c)/2; if(issquare(s*(s-a)*(s-b)*(s-c)), print1(a,","))) /* Uses Heron's formula */ \\ Rick L. Shepherd, Sep 04 2005

Composite of comments from Alec Mihailovs (alec(AT)mihailovs.com) and David Terr, Mar 07 2005: (Start)
"a(n)^2 = A011922(n)^2 + (4*A007655(n))^2, so that A011922(n) = 1/2 base of triangles, A007655(n) = 1/4 height of triangles (conjectured by Paul Hanna).
Area is (a+1)/4*sqrt((3*a+1)*(a-1)). If a is even, the numerator is odd and the area is not an integer. That means a=2*k-1. In this case, Area=k*sqrt((3*k-1)*(k-1)).
Solving equation (3*k-1)*(k-1)=y^2, we get k=(2+sqrt(1+3*y^2))/3. That means that 1+3*y^2=x^2 with integer x and y. This is a Pell equation, all solutions of which have the form x=((2+sqrt(3))^n+(2-sqrt(3))^n)/2, y=((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3)). Therefore k=(x+2)/3 is an integer only for even n. Then a=2*k-1=(2*x+1)/3 with even n. Q.E.D.
a(n)=(1/3)*((2+sqrt(3))^(2*n-2)+(2-sqrt(3))^(2*n-2)+1).
Recurrence: a(n+3)=15*a(n+2)-15*a(n+1)+a(n), a(0)=1, a(1)=5, a(2)=65.
G.f.: x*(1-10*x+5*x^2)/(1-15*x+15*x^2-x^3).
E.g.f.: 1/3*(exp(x)+exp((7+4*sqrt(3))*x)+exp((7-4*sqrt(3))*x)).
a(n) = 4U(n)^2 + 1, where U(1) = 0, U(2)=1 and U(n+1) = 4U(n) - U(n-1) for n>1. (U(n), V(n)) is the n-th solution to Pell's equation 3U(n)^2 + 1 = V(n)^2. (U(n) is the sequence A001353.)" (End)
a(n+1) = A098301(n+1) + A055793(n+2) - Creighton Dement, Apr 18 2005
a(n) = floor((7+4*sqrt(3))*a(n-1))-4, n>=3. - Rick L. Shepherd, Sep 04 2005
a(n)= [1+14*A007655(n+2)-194*A007655(n+1)]/3. - R. J. Mathar, Nov 16 2007
For n>=3, a(n) = 14*a(n-1) - a(n-2) - 4. It is one of 10 second-order linear recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916. - Max Alekseyev, Jan 22 2022

More terms from Creighton Dement, Apr 18 2005

Edited by Max Alekseyev, Jan 22 2022

A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 2, 14, 134, 1322, 13082, 129494, 1281854, 12689042, 125608562, 1243396574, 12308357174, 121840175162, 1206093394442, 11939093769254, 118184844298094, 1169909349211682, 11580908647818722, 114639177128975534, 1134810862641936614, 11233469449290390602, 111199883630261969402

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Essentially the same as A157085. - R. J. Mathar, Feb 07 2022

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (11,-11,1).

Cf. A031138, A054318, A097784, A253175, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350924, A350925, A350926.

LinearRecurrence[{11, -11, 1}, {2, 2, 14}, 25] (* Paolo Xausa, May 30 2025 *)

G.f.: 2*(1 - 10*x + 7*x^2)/((1 - x)*(1 - 10*x + x^2)). - Stefano Spezia, Jan 22 2022
From Hugo Pfoertner, Jan 22 2022: (Start)
a(n) = A031138(n) + 1.
a(n) = 3*A054318(n) - 1.
a(n) = 12*A097784(n-2) + 2 for n >= 2. (End)
a(n) = 2 * A253175(n) for n>=1. - Alois P. Heinz, Jan 22 2022

A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

2, 5, 29, 194, 1325, 9077, 62210, 426389, 2922509, 20031170, 137295677, 941038565, 6449974274, 44208781349, 303011495165, 2076871684802, 14235090298445, 97568760404309, 668746232531714, 4583654867317685, 31416837838692077, 215334210003526850, 1475922632185995869

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.
From William P. Orrick, Dec 20 2023: (Start)
Every term is a Markov number (see A002559) and, for n > 1, corresponds to a node of the Markov tree A368546 whose sibling and ancestors are all odd-indexed Fibonacci numbers. For n > 1, a(n) is the label of the node obtained from the root by going left n - 2 times and then right. Its Farey index, described in the comments to A368546, is 2 / (2*n - 1).
For instance, a(3) = 194 comes from going left once from the root node of the Markov tree and then right, which corresponds to the sequence of Markov numbers 5, 13, 194.  The corresponding sequence of Farey indices is 1/2, 1/3, 2/5. The sibling of the final node corresponds to Markov number 34 and Farey index 1/4. (End)

Index entries for linear recurrences with constant coefficients, signature (8,-8,1).

Cf. A049684, A064170, A350916, A002559, A000045.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350923, A350924, A350925, A350926.

CoefficientList[Series[(2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)), {x, 0, 22}],x] (* James C. McMahon, Dec 22 2023 *)

G.f.: (2 - x)*(1 - 5*x)/((1 - x)*(1 - 7*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 3*A049684(n) + 2 = 3*A064170(n+2) - 1. - Hugo Pfoertner, Jan 22 2022
a(n) = 3*A000045(2*n - 1) * A000045(2*n + 1) - 1 = A000045(2*n - 1)^2 + A000045(2*n + 1)^2. - William P. Orrick, Jan 08 2023

A350917 a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

1, 2, 41, 937, 21506, 493697, 11333521, 260177282, 5972743961, 137112933817, 3147624733826, 72258255944177, 1658792261982241, 38079963769647362, 874180374439907081, 20068068648348215497, 460691398537569049346, 10575834097715739919457, 242783492848924449098161, 5573444501427546589338242, 127946440039984647105681401, 2937194676418219336841333977

Offset: 0

Max Alekseyev, Jan 21 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.
Other properties for all n:
(a(n)+1)*(a(n+2)+1) = (a(n+1)+1)*(a(n+1)+26);
((105*a(n) - 20)^2 - 50^2) / 21 is an integer square.

Sebastian et al., Are there infinitely many positive integer solutions to (3+3k+l)^2=m(kl-k^3-1)?, MathOverflow, 2022.
Index entries for linear recurrences with constant coefficients, signature (24,-24,1).

Cf. A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350919, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

a(n) = 17/42*A090731(n) - 15/2*A097778(n-1) + 4/21.

G.f.: ( -1+22*x-17*x^2 ) / ( (x-1)*(x^2-23*x+1) ). - R. J. Mathar, Feb 07 2022

A350919 a(0) = 9, a(1) = 9, and a(n) = 3*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

9, 9, 14, 29, 69, 174, 449, 1169, 3054, 7989, 20909, 54734, 143289, 375129, 982094, 2571149, 6731349, 17622894, 46137329, 120789089, 316229934, 827900709, 2167472189, 5674515854, 14856075369, 38893710249, 101825055374, 266581455869, 697919312229, 1827176480814, 4783610130209, 12523653909809, 32787351599214, 85838400887829, 224727851064269

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (4,-4,1).

Cf. A032908, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350920, A350921, A350922, A350923, A350924, A350925, A350926.

nxt[{a_,b_}]:={b,3b-a-4}; NestList[nxt,{9,9},40][[;;,1]] (* or *) LinearRecurrence[{4,-4,1},{9,9,14},40] (* Harvey P. Dale, Jul 19 2024 *)

a(n) = 5*A032908(n) - 1. - Hugo Pfoertner, Jan 22 2022
G.f.: (3 - 2*x)*(3 - 7*x)/((1 - x)*(1 - 3*x + x^2)). - Stefano Spezia, Jan 22 2022
a(n) = 5*A001519(n) +4. - R. J. Mathar, Feb 07 2022

A350920 a(0) = 5, a(1) = 5, and a(n) = 4*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

5, 5, 11, 35, 125, 461, 1715, 6395, 23861, 89045, 332315, 1240211, 4628525, 17273885, 64467011, 240594155, 897909605, 3351044261, 12506267435, 46674025475, 174189834461, 650085312365, 2426151414995, 9054520347611, 33791929975445, 126113199554165, 470660868241211, 1756530273410675, 6555460225401485, 24465310628195261, 91305782287379555

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001835, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350921, A350922, A350923, A350924, A350925, A350926.

a(n) = 3*A001835(n) + 2. - Hugo Pfoertner, Jan 22 2022

G.f.: (5 - 20*x + 11*x^2)/((1 - x)*(1 - 4*x + x^2)). - Stefano Spezia, Jan 22 2022

A350921 a(0) = 3, a(1) = 3, and a(n) = 6*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

3, 3, 11, 59, 339, 1971, 11483, 66923, 390051, 2273379, 13250219, 77227931, 450117363, 2623476243, 15290740091, 89120964299, 519435045699, 3027489309891, 17645500813643, 102845515571963, 599427592618131, 3493720040136819, 20362892648202779, 118683635849079851, 691738922446276323

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A001653, A011900, A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350922, A350923, A350924, A350925, A350926.

G.f.: (3 - 18*x + 11*x^2)/((1 - x)*(1 - 6*x + x^2)). - Stefano Spezia, Jan 22 2022

a(n) = 2*A001653(n) + 1 = 4*A011900(n-1) - 1 for n >= 1. - Hugo Pfoertner, Jan 22 2022

A350924 a(0) = 1, a(1) = 3, and a(n) = 16*a(n-1) - a(n-2) - 4 for n >= 2.

Original entry on oeis.org

1, 3, 43, 681, 10849, 172899, 2755531, 43915593, 699893953, 11154387651, 177770308459, 2833170547689, 45152958454561, 719614164725283, 11468673677149963, 182779164669674121, 2912997961037635969, 46425188211932501379, 739890013429882386091, 11791815026666185676073

Offset: 0

Max Alekseyev, Jan 22 2022

One of 10 linear second-order recurrence sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4 and together forming A350916.

Index entries for linear recurrences with constant coefficients, signature (17,-17,1).

Cf. A350916.

Other sequences satisfying (a(n)*a(n-1)-1) * (a(n)*a(n+1)-1) = (a(n)+1)^4: A103974, A350917, A350919, A350920, A350921, A350922, A350923, A350925, A350926.

nxt[{a_,b_}]:={b,16b-a-4}; NestList[nxt,{1,3},20][[All,1]] (* or *) LinearRecurrence[ {17,-17,1},{1,3,43},20] (* Harvey P. Dale, Jan 08 2023 *)

a350924 = [1, 3]
for k in range(2, 100): a350924.append(16*a350924[k-1]-a350924[k-2]-4)
print(a350924) # Karl-Heinz Hofmann, Jan 22 2022

G.f.: (1 - 14*x + 9*x^2)/((1 - x)*(1 - 16*x + x^2)). - Stefano Spezia, Jan 22 2022

7*a(n) = 2 +5*A077412(n) -61*A077412(n-1). - R. J. Mathar, Feb 07 2022

cp's OEIS Frontend

A001075 a(0) = 1, a(1) = 2, a(n) = 4*a(n-1) - a(n-2).

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A001570 Numbers k such that k^2 is centered hexagonal.

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A103974 Smaller sides (a) in (a,a,a+1)-integer triangle with integer area.

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A350923 a(0) = 2, a(1) = 2, and a(n) = 10*a(n-1) - a(n-2) - 4 for n >= 2.

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A350922 a(0) = 2, a(1) = 5, and a(n) = 7*a(n-1) - a(n-2) - 4 for n >= 2.

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A350917 a(0) = 1, a(1) = 2, and a(n) = 23*a(n-1) - a(n-2) - 4 for n >= 2.

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