A007654 -id:A007654 - cp's OEIS frontend

A007655 Standard deviation of A007654.

0, 1, 14, 195, 2716, 37829, 526890, 7338631, 102213944, 1423656585, 19828978246, 276182038859, 3846719565780, 53577891882061, 746243766783074, 10393834843080975, 144767444036350576, 2016350381665827089, 28084137899285228670, 391161580208327374291, 5448177985017298011404

Offset: 1

N. J. A. Sloane

a(n) corresponds also to one-sixth the area of Fleenor-Heronian triangle with middle side A003500(n). - Lekraj Beedassy, Jul 15 2002
a(n) give all (nontrivial, integer) solutions of Pell equation b(n+1)^2 - 48*a(n+1)^2 = +1 with b(n+1)=A011943(n), n>=0.
For n>=3, a(n) equals the permanent of the (n-2) X (n-2) tridiagonal matrix with 14's along the main diagonal, and i's along the superdiagonal and the subdiagonal (i is the imaginary unit). - John M. Campbell, Jul 08 2011
For n>1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,13}. - Milan Janjic, Jan 25 2015
6*a(n)^2 = 6*S(n-1, 14)^2 is the triangular number Tri((T(n, 7) - 1)/2) with Tri = A000217 and T = A053120. This is instance k = 3 of the general k-identity given in a comment to A001109. - Wolfdieter Lang, Feb 01 2016

			G.f. = x^2 + 14*x^3 + 195*x^4 + 2716*x^5 + 37829*x^6 + 526890*x^7 + ...

D. A. Benaron, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Indranil Ghosh, Table of n, a(n) for n = 1..874 (terms 1..100 from T. D. Noe)
R. Flórez, R. A. Higuita and A. Mukherjee, Alternating Sums in the Hosoya Polynomial Triangle, Article 14.9.5 Journal of Integer Sequences, Vol. 17 (2014).
D. S. Hale, 3165. Perfect Squares of the Form 48n^2+1, Math. Gaz., Oct. 1966, page 307.
Milan Janjic, On Linear Recurrence Equations Arising from Compositions of Positive Integers, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.7.
Tanya Khovanova, Recursive Sequences
Murray S. Klamkin, Perfect Squares of the Form (m^2 - 1)a_n^2 + t, Math. Mag., 1969, page 111.
E. K. Lloyd, The standard deviation of 1, 2, ..., n, Pell's equation and rational triangles, The Mathematical Gazette, Vol. 81, No. 491 (Jul., 1997), pp. 231-243.
Dino Lorenzini and Z. Xiang, Integral points on variable separated curves, Preprint 2016.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Cf. A000217, A001570, A003500, A011922, A011943, A011945, A028230, A046184, A049310, A053120, A055793, A067900, A098301, A101950, A103974.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), A001109 (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), this sequence (m=7), A077412 (m=8), A049660 (m=9), A075843 (m=10), A077421 (m=11), A077423 (m=12), A097309 (m=13), A097311 (m=14), A097313 (m=15), A029548 (m=16), A029547 (m=17), A144128 (m=18), A078987 (m=19), A097316 (m=33).
Cf. A323182.

m:=7;; a:=[0,1];; for n in [3..20] do a[n]:=2*m*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 23 2019

[n le 2 select n-1 else 14*Self(n-1)-Self(n-2): n in [1..70]]; // Vincenzo Librandi, Feb 02 2016

0,seq(orthopoly[U](n,7),n=0..30); # Robert Israel, Feb 04 2016

Table[GegenbauerC[n, 1, 7], {n,0,20}] (* Vladimir Joseph Stephan Orlovsky, Sep 11 2008 *)
LinearRecurrence[{14,-1}, {0,1}, 20] (* Vincenzo Librandi, Feb 02 2016 *)
ChebyshevU[Range[21] -2, 7] (* G. C. Greubel, Dec 23 2019 *)
Table[Sum[Binomial[n, 2 k - 1]*7^(n - 2 k + 1)*48^(k - 1), {k, 1, n}], {n, 0, 15}] (* Horst H. Manninger, Jan 16 2022 *)

concat(0, Vec((x^2/(1-14*x+x^2) + O(x^30)))) \\ Michel Marcus, Feb 02 2016

vector(21, n, polchebyshev(n-2, 2, 7) ) \\ G. C. Greubel, Dec 23 2019

[lucas_number1(n,14,1) for n in range(0,20)] # Zerinvary Lajos, Jun 25 2008

[chebyshev_U(n,7) for n in (-1..20)] # G. C. Greubel, Dec 23 2019

a(n) = 14*a(n-1) - a(n-2).
G.f.: x^2/(1-14*x+x^2).
a(n+1) ~ 1/24*sqrt(3)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n+1) = S(n-1, 14), n>=0, with S(n, x) := U(n, x/2) Chebyshev's polynomials of the second kind. S(-1, x) := 0. See A049310.
a(n+1) = ( (7+4*sqrt(3))^n - (7-4*sqrt(3))^n )/(8*sqrt(3)).
a(n+1) = sqrt((A011943(n)^2 - 1)/48), n>=0.
Chebyshev's polynomials U(n-2, x) evaluated at x=7.
a(n) = A001353(2n)/4. - Lekraj Beedassy, Jul 15 2002
4*a(n+1) + A046184(n) = A055793(n+2) + A098301(n+1) 4*a(n+1) + A098301(n+1) + A055793(n+2) = A046184(n+1) (4*a(n+1))^2 = A098301(2n+1) (conjectures). - Creighton Dement, Nov 02 2004
(4*a(n))^2 = A103974(n)^2 - A011922(n-1)^2. - Paul D. Hanna, Mar 06 2005
From Mohamed Bouhamida, May 26 2007: (Start)
a(n) = 13*( a(n-1) + a(n-2) ) - a(n-3).
a(n) = 15*( a(n-1) - a(n-2) ) + a(n-3). (End)
a(n) = b such that (-1)^n/4*Integral_{x=-Pi/2..Pi/2} (sin((2*n-2)*x))/(2-sin(x)) dx = c+b*log(3). - Francesco Daddi, Aug 02 2011
a(n+2) = Sum_{k=0..n} A101950(n,k)*13^k. - Philippe Deléham, Feb 10 2012
Product {n >= 1} (1 + 1/a(n)) = 1/3*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
Product {n >= 2} (1 - 1/a(n)) = 1/7*(3 + 2*sqrt(3)). - Peter Bala, Dec 23 2012
a(n) = (A028230(n) - A001570(n))/2. - Richard R. Forberg, Nov 14 2013
E.g.f.: 1 - exp(7*x)*(12*cosh(4*sqrt(3)*x) - 7*sqrt(3)*sinh(4*sqrt(3)*x))/12. - Stefano Spezia, Dec 11 2022

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

Original entry on oeis.org

1, -3, 13, -48, 181, -675, 2521, -9408, 35113, -131043, 489061, -1825200, 6811741, -25421763, 94875313, -354079488, 1321442641, -4931691075, 18405321661, -68689595568, 256353060613, -956722646883, 3570537526921, -13325427460800, 49731172316281

Offset: 0

Creighton Dement, Jul 21 2005

In reference to program code, 2baseiseq[X](n) = ((-1)^n)*A001353(n) (a(n)^2 + 1 is a perfect square.) 1tesseq[X](n) = (-1^(n+1))*A097948(n).

Floretion Algebra Multiplication Program, FAMP Code: 1ibaseiseq[X] with X = .5'i + .5i' + 'ii' - .5'jj' + 1.5'kk' - 1 (* Corrected by Creighton Dement, Dec 11 2009 *)

Robert Munafo, Sequences Related to Floretions
Index entries for linear recurrences with constant coefficients, signature (-4,0,4,1).

Cf. A007654, A001570, A076139. See also A117808, A122571 (same except for signs).

Cf. A001353, A097948.

/* By definition: */
m:=15; R:=PowerSeriesRing(Integers(), m);
A001570:=Coefficients(R!((1-x)/(1-14*x+x^2)));
A007654:=Coefficients(R!(-3*x^2*(1+x)/(-1+x)/(1-14*x+x^2)));
&cat[[A001570[i],-A007654[i]]: i in [1..m-2]]; // Bruno Berselli, Feb 05 2013

seriestolist(series((x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)), x=0,25));

LinearRecurrence[{-4,0,4,1},{1,-3,13,-48},30] (* Harvey P. Dale, Jun 15 2018 *)

G.f.: (x^2+x+1)/((1-x)*(x+1)*(x^2+4*x+1)).
Floor(((2 + sqrt(3))^n + (2 - sqrt(3))^n)/4) produces this sequence with a different offset and without signs. - James R. Buddenhagen, May 20 2010
Define c(n) = a(n) - 4*a(n+1) - a(n+2) and d(n) = -a(n) - 4*a(n+1) - a(n+2); Conjectures: I: c(2n) = 24*A076139(n); (Triangular numbers that are one-third of another triangular number) II: c(2n+1) = -A011943(n+1); (Numbers n such that any group of n consecutive integers has integral standard deviation) III: d(2n) = -2; IV: d(2n+1) = -1

A132592 X-values of solutions to the equation X(X + 1) - 8Y^2 = 0.

Original entry on oeis.org

0, 8, 288, 9800, 332928, 11309768, 384199200, 13051463048, 443365544448, 15061377048200, 511643454094368, 17380816062160328, 590436102659356800, 20057446674355970888, 681362750825443653408, 23146276081390728245000, 786292024016459316676608, 26710782540478226038759688

Offset: 0

Mohamed Bouhamida, Nov 14 2007

Equivalently, numbers k such that both k/2 and k+1 are squares. - Karl-Heinz Hofmann, Sep 20 2022

Equivalently, numbers k such that the k-dimensional volume and total (k-1)-dimensional volume are equal, with side length being a positive integer, for all regular polyhedra constructible in k dimensions. - Matt Moir, Jul 09 2024

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Intersection between A132411 and A001105.
Cf. A007654.
Cf. A001541, A058331, A001079, A037270, A055792, A071253, A108741, A132592, A146311, A146312, A146313, A173115, A173116, A173121.

I:=[0,8,288]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+ Self(n-3): n in [1..30]]; // Vincenzo Librandi, Dec 24 2018

Table[Round[N[Sinh[2 n ArcCosh[Sqrt[2]]]^2, 100]], {n, 0, 20}] (* Artur Jasinski, Feb 10 2010 *)
LinearRecurrence[{35, -35, 1}, {0, 8, 288}, 30] (* Vincenzo Librandi, Dec 24 2018 *)

A132592 = [0, 8]
for n in range(2, 18): A132592.append(34 * A132592[-1] - A132592[-2] + 16)
print(A132592) # Karl-Heinz Hofmann, Sep 20 2022

a(0)=0, a(1)=8 and a(n) = 34*a(n-1) - a(n-2) + 16.
a(n) = (A056771(n) - 1)/2. - Max Alekseyev, Nov 13 2009
a(n) = sinh(2*n*arccosh(sqrt(2))^2) (n=0,1,2,3,...). - Artur Jasinski, Feb 10 2010
G.f.: -8*x*(x+1)/((x-1)*(x^2-34*x+1)). - Colin Barker, Oct 24 2012
a(n) = A055792(n+1)-1 = A001541(n)^2 - 1. - Antti Karttunen, Oct 03 2016

More terms from Max Alekseyev, Nov 13 2009

A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

Original entry on oeis.org

1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

E. K. Lloyd

If k is in the sequence, then it has successor 7*k + 4*sqrt(3*(k^2 - 1)). - Lekraj Beedassy, Jun 28 2002
Chebyshev's polynomials T(n,x) evaluated at x=7.
a(n+1) give all (nontrivial) solutions of Pell equation a(n+1)^2 - 48*b(n+1)^2 = +1 with b(n+1)=A007655(n+2), n >= 0.
Also all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - Herbert Kociemba, Jun 05 2022
Also numbers x of the form 3j+1 such that x^2 = 3m^2+1. Also solutions of x in x^2 - 3*y^2 = 1 in A001075 if x = 3j+1, j=1,2,... - Cino Hilliard, Mar 05 2005
In addition to having integral standard deviation, these k consecutive integers also have integral mean. This question was posed by Jim Delany of Cal Poly in 1989. The solution appeared in the American Mathematical Monthly Vol. 97, No. 5, (May, 1990), pp. 432 as problem E3302. - Ronald S. Tiberio, Jun 23 2008
Lebl and Lichtblau give the formula a(d) = ((7+4*sqrt(3))^d + (7-4*sqrt(3))^d)/2 in Theorem 1.2(iii), p. 4. - Jonathan Vos Post, Aug 05 2008
In a (square pyramidal) pile of cannonballs, paint the ball at the top and the balls on 2 opposite sides of the base. The sequence, after its 1st term, gives the numbers of painted balls, k, in piles where the total number of balls is twice a square multiple of k. - Peter Munn, Feb 06 2025

P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Robert Israel, Table of n, a(n) for n = 1..788
Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, Problem E 3302: Averaging to Integers, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Tanya Khovanova, Recursive Sequences
Jiri Lebl and Daniel Lichtblau, Uniqueness of certain polynomials constant on a hyperplane, arXiv:0808.0284 [math.CV], 2008-2010.
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Ronald S. Tiberio, Solution to Problem E 3302 [Broken link]
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = A001075(2n).
Row 2 of array A188644
Cf. A000330, A007654, A007655, A011944, A102344.

I:=[1,7]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

seq(orthopoly[T](n,7), n = 0..50); # Robert Israel, Jun 02 2015
a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
seq(simplify(a(n)), n=1..20); # Peter Luschny, Jul 26 2020

LinearRecurrence[{14,-1},{1,7},30] (* Harvey P. Dale, Dec 16 2013 *)
a[n_]:=1/2((7-4 Sqrt[3])^n+(7+4 Sqrt[3])^n); Table[a[n] // Simplify,{n,0,20}] (* Gerry Martens, May 30 2015 *)

PARI
```
a(n)=if(n<0,0,subst(poltchebi(n),x,7))
```

g(n) = forstep(x=1,n,3,y=(x^2-1)/3;if(issquare(y),print1(x","))) \\ Cino Hilliard, Mar 05 2005

a(n) = 14*a(n-1) - a(n-2).
a(n) = sqrt(12*A011944(n)^2 + 1).
a(n) ~ (1/2)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = T(n, 7) = (S(n, 14)-S(n-2, 14))/2 = T(2*n, 2) with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 14)=A007655(n+2).
a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
a(n) = sqrt(48*A007655(n+1)^2 + 1).
G.f.: (1-7*x)/(1-14*x+x^2).
a(n) = cosh(2n*arcsinh(sqrt(3))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020
E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - Stefano Spezia, Dec 12 2022

Better description from Lekraj Beedassy, Jun 27 2002
Chebyshev comments from Wolfdieter Lang, Nov 08 2002
More terms from Vincenzo Librandi, Apr 19 2015

A132596 X-values of solutions to the equation X(X + 1) - 6Y^2 = 0.

Original entry on oeis.org

0, 2, 24, 242, 2400, 23762, 235224, 2328482, 23049600, 228167522, 2258625624, 22358088722, 221322261600, 2190864527282, 21687323011224, 214682365584962, 2125136332838400, 21036680962799042, 208241673295152024

Offset: 0

Mohamed Bouhamida, Nov 14 2007

Or, 3*A000217(X) is a square, (3*A004189(n))^2. - Zak Seidov, Apr 08 2009

"You can find an infinite number of [different] triangular numbers such that when multiplied together form a square number. For example, for every triangular number, T_n, there are an infinite number of other triangular numbers, T_m, such that T_n*T_m is a square. For example, T_2 * T_24 = 30^2." [Pickover] - Robert G. Wilson v, Apr 01 2010

Clifford A. Pickover, The Loom of God, Tapestries of Mathematics and Mysticism, Sterling, NY, 2009, page 33.

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

Cf. A007654, A001079, A000217, A098297, A108741 (Y^2), A004189 (Y).

LinearRecurrence[{11, -11, 1}, {0, 2, 24}, 19] (* Jean-François Alcover, Feb 26 2019 *)

a(n) = 10*a(n-1) - a(n-2) + 4, a(0)=0, a(1)=2.
a(n) = (A001079(n) - 1)/2. - Max Alekseyev, Nov 13 2009
From R. J. Mathar, Apr 20 2010: (Start)
a(n) = 11*a(n-1) - 11*a(n-2) + a(n-3) = 2*A098297(n).
G.f.: -2*x*(1+x) / ( (x-1)*(x^2-10*x+1) ). (End)
a(n) = 2*A098297(n) = (1/2)*(T(2*n,sqrt(3)) - 1), where T(n,x) is the n-th Chebyshev polynomial of the first kind. - Peter Bala, Dec 31 2012

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 8, 2, 0, 0, 49, 24, 3, 0, 0, 288, 242, 48, 4, 0, 0, 1681, 2400, 675, 80, 5, 0, 0, 9800, 23762, 9408, 1444, 120, 6, 0, 0, 57121, 235224, 131043, 25920, 2645, 168, 7, 0, 0, 332928, 2328482, 1825200, 465124, 58080, 4374, 224, 8, 0

Offset: 0

Seiichi Manyama, Dec 23 2018

			Square array begins:
   0, 0,   0,    0,      0,       0,        0, ...
   0, 1,   8,   49,    288,    1681,     9800, ...
   0, 2,  24,  242,   2400,   23762,   235224, ...
   0, 3,  48,  675,   9408,  131043,  1825200, ...
   0, 4,  80, 1444,  25920,  465124,  8346320, ...
   0, 5, 120, 2645,  58080, 1275125, 27994680, ...
   0, 6, 168, 4374, 113568, 2948406, 76545000, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

Columns 0-5 give A000004, A001477, A033996, A322675, A322677, A322745.
Rows 0-9 give A000004, A001108, A132596, A007654(n+1), A132584, A322707, A322708, A322709, A132592, A132593.
Main diagonal gives A322746.
Cf. A173175 (A(n,2*n)), A322790.

Unprotect[Power]; 0^0 := 1; Protect[Power]; Table[(-1 + Sum[Binomial[2 k, 2 j] (# + 1)^(k - j)*#^j, {j, 0, k}])/2 &[n - k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jan 01 2019 *)
nmax = 9; row[n_] := LinearRecurrence[{4n+3, -4n-3, 1}, {0, n, 4n(n+1)}, nmax+1]; T = Array[row, nmax+1, 0]; A[n_, k_] := T[[n+1, k+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 06 2019 *)

def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  (0..n).map{|i| (0..k).inject(-1){|s, j| s + ncr(2 * k, 2 * j) * (i + 1) ** (k - j) * i ** j} / 2}
end
def A322699(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A322699(10)

sqrt(A(n,k)+1) + sqrt(A(n,k)) = (sqrt(n+1) + sqrt(n))^k.
sqrt(A(n,k)+1) - sqrt(A(n,k)) = (sqrt(n+1) - sqrt(n))^k.
A(n,0) = 0, A(n,1) = n and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) + 2*n for k > 1.
A(n,k) = (T_{k}(2*n+1) - 1)/2 where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = (2*n+1-1)/2 = n.

A221075 Simple continued fraction expansion of an infinite product.

Original entry on oeis.org

2, 12, 1, 24, 1, 192, 1, 360, 1, 2700, 1, 5040, 1, 37632, 1, 70224, 1, 524172, 1, 978120, 1, 7300800, 1, 13623480, 1, 101687052, 1, 189750624, 1, 1416317952, 1, 2642885280, 1, 19726764300, 1, 36810643320, 1

Offset: 0

Peter Bala, Jan 06 2013

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 4. For other cases see A221073 (m = 2), A221074 (m = 3) and A221076 (m = 5).

If we denote the present sequence by [2; 12, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-2*{(2-sqrt(3))^(2*k+1)}^(4*n+3)]/[1 - 2*{(2-sqrt(3))^(2*k+1)}^(4*n+1)]. An example is given below

			Product {n >= 0} {1 - 2*(2 - sqrt(3))^(4*n+3)}/{1 - 2*(2 - sqrt(3))^(4*n+1)} = 2.07715 13807 08976 70415 ...
= 2 + 1/(12 + 1/(1 + 1/(24 + 1/(1 + 1/(192 + 1/(1 + 1/(360 + ...))))))).
Since (2 - sqrt(3))^3 = 26 - 15*sqrt(3) we have the following simple continued fraction expansion:
product {n >= 0} {1 - 2*(26 - 15*sqrt(3))^(4*n+3)}/{1 - 2*(26 - 15*sqrt(3))^(4*n+1)} = 1.04000 05921 62729 43797 ... = 1 + 1/(24 + 1/(1 + 1/(2700 + 1/(1 + 1/(70224 + 1/(1 + 1/(7300800 + ...))))))).

P. Bala, Some simple continued fraction expansions for an infinite product, Part 1
Index entries for linear recurrences with constant coefficients, signature (0,1,0,14,0,-14,0,-1,0,1).

Cf. A001353, A007654, A045899, A076139, A076140, A098301, A123480, A174500, A217855, A221073 (m = 2), A221074 (m = 3), A221076 (m = 5).

a(2*n) = 1 for n >= 1. For n >= 1 we have
a(4*n - 3) = (2 + sqrt(3))^(2*n) + (2 - sqrt(3))^(2*n) - 2;
a(4*n - 1) = 1/2*{(2 + sqrt(3) )^(2*n + 1) + (2 - sqrt(3))^(2*n + 1)} - 2.
a(4*n - 3) = 12*A098301(n) = 12*A001353(n)^2 = 4*A007654(n);
a(4*n - 1) = 24*A076139(n) = 12*A217855 = 8*A076140(n) = 6*A123480(n) = 3*A045899(n).
O.g.f.: 2 + x^2/(1 - x^2) + 12*x*(1 + x^2)^2/(1 - 15*x^4 + 15*x^8 - x^12) = 2 + 12*x + x^2 + 24*x^3 + x^4 + 192*x^5 + ....
O.g.f.: (x^10-2*x^8-14*x^6+28*x^4-12*x^3+x^2-12*x-2) / ((x-1)*(x+1)*(x^4-4*x^2+1)*(x^4+4*x^2+1)). - Colin Barker, Jan 10 2014

A132584 a(0)=0, a(1)=4; for n > 1, a(n) = 18*a(n-1) - a(n-2) + 8.

Original entry on oeis.org

0, 4, 80, 1444, 25920, 465124, 8346320, 149768644, 2687489280, 48225038404, 865363202000, 15528312597604, 278644263554880, 5000068431390244, 89722587501469520, 1610006506595061124, 28890394531209630720, 518417095055178291844, 9302617316461999622480

Offset: 0

Mohamed Bouhamida, Nov 14 2007

The old definition given for this sequence was "Sequence allows us to find X values of the equation: X(X + 1) - 5*Y^2 = 0".

With this old definition, if X = a(n), then Y = A207832(n). Now, with u = 2X+1, this Diophantine equation becomes the Pell-Fermat equation u^2 - 20*Y^2 = 1, and then, u = A023039(n) and Y = A207832(n). - Bernard Schott, Jan 25 2023

Seiichi Manyama, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

Cf. A000032, A000045, A000217, A007654, A023039, A049683, A292443, A207832.

I:=[0,4,80]; [n le 3 select I[n] else 18*Self(n-1)-Self(n-2)+8: n in [1..30]]; // Vincenzo Librandi, Dec 24 2018

LinearRecurrence[{19, -19, 1}, {0, 4, 80}, 40] (* Vincenzo Librandi, Dec 24 2018 *)
nxt[{a_,b_}]:={b,18b-a+8}; NestList[nxt,{0,4},20][[;;,1]] (* Harvey P. Dale, Aug 25 2024 *)

a(n) = (A023039(n) - 1)/2. - Max Alekseyev, Nov 13 2009
G.f.: -4*x*(x+1)/((x-1)*(x^2-18*x+1)). - Colin Barker, Oct 24 2012
From Amiram Eldar, Jan 11 2022: (Start)
a(n) = 5*Fibonacci(3*n)^2/4 - 1 if n is odd and 5*Fibonacci(3*n)^2/4 if n is even.
A000217(a(n)) = A292443(n). (End)
a(n) = (Lucas(6*n)-2)/4. - Jeffrey Shallit, Jan 20 2023
a(n) = 4 * A049683(n). - Alois P. Heinz, Jan 20 2023

More terms from Max Alekseyev, Nov 13 2009

New definition by Antti Karttunen, Oct 24 2012

A011944 a(n) = 14*a(n-1) - a(n-2) with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 28, 390, 5432, 75658, 1053780, 14677262, 204427888, 2847313170, 39657956492, 552364077718, 7693439131560, 107155783764122, 1492487533566148, 20787669686161950, 289534888072701152

Offset: 0

E. K. Lloyd

Standard deviation of A011943.
Product x*y, where the pair (x, y) solves for x^2 - 3y^2 = 1, i.e., a(n)=A001075(n)*A001353(n). - Lekraj Beedassy, Jul 13 2006
Solutions m to the Diophantine equation where square m^2 = k*(k+1)/3, corresponding solutions k are in A007654. - Bernard Schott, Apr 10 2021
All solutions for y in Pell equation x^2 - 12*y^2 = 1. Corresponding values for x are in A011943. - Herbert Kociemba, Jun 05 2022

Tanya Khovanova, Recursive Sequences
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = 2 * A007655 = {A001353(2n)}/2. Cf. A011943.

Cf. A007654.

LinearRecurrence[{14,-1},{0,2},20] (* Harvey P. Dale, Oct 17 2019 *)
Table[2 ChebyshevU[-1 + n, 7], {n, 0, 18}] (* Herbert Kociemba, Jun 05 2022 *)

For all members x of the sequence, 12*x^2 +1 is a square. Lim_{n->infinity} a(n)/a(n-1) = 7 + sqrt(12). - Gregory V. Richardson, Oct 13 2002
a(n) = ((7+2*sqrt(12))^(n-1) - (7-2*sqrt(12))^(n-1)) / (2*sqrt(12)). - Gregory V. Richardson, Oct 13 2002
a(n) = 13*(a(n-1) + a(n-2)) - a(n-3). a(n) = 15*(a(n-1) - a(n-2)) + a(n-3). - Mohamed Bouhamida, Sep 20 2006
a(n) = sinh(2n*arcsinh(sqrt(3)))/sqrt(12). - Herbert Kociemba, Apr 24 2008
G.f.: 2x/(1-14*x+x^2). - Philippe Deléham, Nov 17 2008

A132593 Nonnegative integer solutions X to the equation: X(X + 1) - 10*Y^2 = 0.

Original entry on oeis.org

0, 9, 360, 13689, 519840, 19740249, 749609640, 28465426089, 1080936581760, 41047124680809, 1558709801289000, 59189925324301209, 2247658452522156960, 85351831270517663289, 3241121929827149048040, 123077281502161146162249

Offset: 0

Mohamed Bouhamida, Nov 14 2007

Also, numbers n such that 5*A000217(n) is a square. [Bruno Berselli, Dec 16 2013]

Seiichi Manyama, Table of n, a(n) for n = 0..500
Kenneth M. Wilke, Problem 269, Crux Mathematicorum, Vol. 3, No. 7 (1977), p. 190; Solution to Problem 269 by Lindsay Reynolds, W. J. Blundon and M. S. Klamkin, ibid., Vol. 4, No. 3 (1978), pp. 79-82; Comment by the MaScoT Problems Group, ibid., Vol. 6, No. 2 (1980), pp. 44-46.
Index entries for linear recurrences with constant coefficients, signature (39,-39,1).

Cf. A007654, A078986.

Cf. A233474 (numbers n such that 5*A000217(n)-1 is a square).

LinearRecurrence[{39,-39,1},{0,9,360},30] (* Harvey P. Dale, Jun 01 2014 *)

a(0)=0, a(1)=9 and a(n) = 38*a(n-1) - a(n-2) + 18.
a(n) = (A078986(n) - 1)/2. - Max Alekseyev, Nov 13 2009
G.f.: -9*x*(x+1)/((x-1)*(x^2-38*x+1)). - Colin Barker, Oct 24 2012
From Amiram Eldar, Feb 15 2022: (Start)
sqrt(a(n)+1) - sqrt(n) = (sqrt(10)-3)^n (Wilke, 1977).
a(n) = ((Sum_{k=0..n} binomial(2*n, 2*k) * 10^(n-k) * 9*k)- 1)/2 (Klamkin, 1978).
a(n) = sinh(n*log(sqrt(10)+3))^2 (MaScoT Problems Group, 1980). (End)

More terms from Max Alekseyev, Nov 13 2009

cp's OEIS Frontend

A007655 Standard deviation of A007654.

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A108946 a(2n) = A001570(n), a(2n+1) = -A007654(n+1).

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A132592 X-values of solutions to the equation X*(X + 1) - 8*Y^2 = 0.

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A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

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A132596 X-values of solutions to the equation X*(X + 1) - 6*Y^2 = 0.

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A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j).

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A132592 X-values of solutions to the equation X(X + 1) - 8Y^2 = 0.

A132596 X-values of solutions to the equation X(X + 1) - 6Y^2 = 0.

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2k,2j)(n+1)^(k-j)n^j).