cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001109 a(n)^2 is a triangular number: a(n) = 6*a(n-1) - a(n-2) with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 35, 204, 1189, 6930, 40391, 235416, 1372105, 7997214, 46611179, 271669860, 1583407981, 9228778026, 53789260175, 313506783024, 1827251437969, 10650001844790, 62072759630771, 361786555939836, 2108646576008245, 12290092900109634, 71631910824649559, 417501372047787720
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

8*a(n)^2 + 1 = 8*A001110(n) + 1 = A055792(n+1) is a perfect square. - Gregory V. Richardson, Oct 05 2002
For n >= 2, A001108(n) gives exactly the positive integers m such that 1,2,...,m has a perfect median. The sequence of associated perfect medians is the present sequence. Let a_1,...,a_m be an (ordered) sequence of real numbers, then a term a_k is a perfect median if Sum_{j=1..k-1} a_j = Sum_{j=k+1..m} a_j. See Puzzle 1 in MSRI Emissary, Fall 2005. - Asher Auel, Jan 12 2006
(a(n), b(n)) where b(n) = A082291(n) are the integer solutions of the equation 2*binomial(b,a) = binomial(b+2,a). - Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de); comment revised by Michael Somos, Apr 07 2003
This sequence gives the values of y in solutions of the Diophantine equation x^2 - 8y^2 = 1. It also gives the values of the product xy where (x,y) satisfies x^2 - 2y^2 = +-1, i.e., a(n) = A001333(n)*A000129(n). a(n) also gives the inradius r of primitive Pythagorean triangles having legs whose lengths are consecutive integers, with corresponding semiperimeter s = a(n+1) = {A001652(n) + A046090(n) + A001653(n)}/2 and area rs = A029549(n) = 6*A029546(n). - Lekraj Beedassy, Apr 23 2003 [edited by Jon E. Schoenfield, May 04 2014]
n such that 8*n^2 = floor(sqrt(8)*n*ceiling(sqrt(8)*n)). - Benoit Cloitre, May 10 2003
For n > 0, ratios a(n+1)/a(n) may be obtained as convergents to continued fraction expansion of 3+sqrt(8): either successive convergents of [6;-6] or odd convergents of [5;1, 4]. - Lekraj Beedassy, Sep 09 2003
a(n+1) + A053141(n) = A001108(n+1). Generating floretion: - 2'i + 2'j - 'k + i' + j' - k' + 2'ii' - 'jj' - 2'kk' + 'ij' + 'ik' + 'ji' + 'jk' - 2'kj' + 2e ("jes" series). - Creighton Dement, Dec 16 2004
Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference). - Emeric Deutsch, Jun 19 2005
Number of D steps on the line y=x in all Delannoy paths of length n (a Delannoy path of length n is a path from (0,0) to (n,n), consisting of steps E=(1,0), N=(0,1) and D=(1,1)). Example: a(2)=6 because in the 13 (=A001850(2)) Delannoy paths of length 2, namely (DD), (D)NE, (D)EN, NE(D), NENE, NEEN, NDE, NNEE, EN(D), ENNE, ENEN, EDN and EENN, we have altogether six D steps on the line y=x (shown between parentheses). - Emeric Deutsch, Jul 07 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 smallest T-circle that does not intersect C(n-1). C(n) has radius a(n+1). Cf. A001653. - Charlie Marion, Sep 14 2005
Numbers such that there is an m with t(n+m)=2t(m), where t(n) are the triangular numbers A000217. For instance, t(20)=2*t(14)=210, so 6 is in the sequence. - Floor van Lamoen, Oct 13 2005
One half the bisection of the Pell numbers (A000129). - Franklin T. Adams-Watters, Jan 08 2006
Pell trapezoids: for n > 0, a(n) = (A000129(n-1)+A000129(n+1))*A000129(n)/2; see also A084158. - Charlie Marion, Apr 01 2006
Tested for 2 < p < 27: If and only if 2^p - 1 (the Mersenne number M(p)) is prime then M(p) divides a(2^(p-1)). - Kenneth J Ramsey, May 16 2006
If 2^p - 1 is prime then M(p) divides a(2^(p-1)-1). - Kenneth J Ramsey, Jun 08 2006; comment corrected by Robert Israel, Mar 18 2007
If 8*n+5 and 8*n+7 are twin primes then their product divides a(4*n+3). - Kenneth J Ramsey, Jun 08 2006
If p is an odd prime, then if p == 1 or 7 (mod 8), then a((p-1)/2) == 0 (mod p) and a((p+1)/2) == 1 (mod p); if p == 3 or 5 (mod 8), then a((p-1)/2) == 1 (mod p) and a((p+1)/2) == 0 (mod p). Kenneth J Ramsey's comment about twin primes follows from this. - Robert Israel, Mar 18 2007
a(n)*(a(n+b) - a(b-2)) = (a(n+1)+1)*(a(n+b-1) - a(b-1)). This identity also applies to any series a(0) = 0 a(1) = 1 a(n) = b*a(n-1) - a(n-2). - Kenneth J Ramsey, Oct 17 2007
For n < 0, let a(n) = -a(-n). Then (a(n+j) + a(k+j)) * (a(n+b+k+j) - a(b-j-2)) = (a(n+j+1) + a(k+j+1)) * (a(n+b+k+j-1) - a(b-j-1)). - Charlie Marion, Mar 04 2011
Sequence gives y values of the Diophantine equation: 0+1+2+...+x = y^2. If (a,b) and (c,d) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with aMohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 0+1+2+...+x = y^2 with p < r then r = 3*p+4*q+1 and s = 2*p+3*q+1. - Mohamed Bouhamida, Sep 02 2009
a(n)/A002315(n) converges to cos^2(Pi/8) (see A201488). - Gary Detlefs, Nov 25 2009
Binomial transform of A086347. - Johannes W. Meijer, Aug 01 2010
If x=a(n), y=A055997(n+1) and z = x^2+y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
In general, if b(0)=1, b(1)=k and for n > 1, b(n) = 6*b(n-1) - b(n-2), then
for n > 0, b(n) = a(n)*k-a(n-1); e.g.,
for k=2, when b(n) = A038725(n), 2 = 1*2 - 0, 11 = 6*2 - 1, 64 = 35*2 - 6, 373 = 204*2 - 35;
for k=3, when b(n) = A001541(n), 3 = 1*3 - 0, 17 = 6*3 - 1; 99 = 35*3 - 6; 577 = 204*3 - 35;
for k=4, when b(n) = A038723(n), 4 = 1*4 - 0, 23 = 6*4 - 1; 134 = 35*4 - 6; 781 = 204*4 - 35;
for k=5, when b(n) = A001653(n), 5 = 1*5 - 0, 29 = 6*5 - 1; 169 = 35*5 - 6; 985 = 204*5 - 35.
See also A002315, A054488, A038761, A054489, A054490.
- Charlie Marion, Dec 08 2010
See a Wolfdieter Lang comment on A001653 on a sequence of (u,v) values for Pythagorean triples (x,y,z) with x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, generated from (u(0)=1,v(0)=2), the triple (3,4,5), by a substitution rule given there. The present a(n) appears there as b(n). The corresponding generated triangles have catheti differing by one length unit. - Wolfdieter Lang, Mar 06 2012
a(n)*a(n+2k) + a(k)^2 and a(n)*a(n+2k+1) + a(k)*a(k+1) are triangular numbers. Generalizes description of sequence. - Charlie Marion, Dec 03 2012
a(n)*a(n+2k) + a(k)^2 is the triangular square A001110(n+k). a(n)*a(n+2k+1) + a(k)*a(k+1) is the triangular oblong A029549(n+k). - Charlie Marion, Dec 05 2012
From Richard R. Forberg, Aug 30 2013: (Start)
The squares of a(n) are the result of applying triangular arithmetic to the squares, using A001333 as the "guide" on what integers to square, as follows:
a(2n)^2 = A001333(2n)^2 * (A001333(2n)^2 - 1)/2;
a(2n+1)^2 = A001333(2n+1)^2 * (A001333(2n+1)^2 + 1)/2. (End)
For n >= 1, a(n) equals the number of 01-avoiding words of length n-1 on alphabet {0,1,...,5}. - Milan Janjic, Jan 25 2015
Panda and Rout call these "balancing numbers" and note that the period of the sequence modulo a prime p is the same as that modulo p^2 when p = 13, 31, 1546463. But these are precisely the p in A238736 such that p^2 divides A000129(p - (2/p)), where (2/p) is a Jacobi symbol. In light of the above observation by Franklin T. Adams-Watters that the present sequence is one half the bisection of the Pell numbers, i.e., a(n) = A000129(2*n)/2, it follows immediately that modulo a fixed prime p, or any power thereof, the period of a(n) is half that of A000129(n). - John Blythe Dobson, Mar 06 2015
The triangular number = square number identity Tri((T(n, 3) - 1)/2) = S(n-1, 6)^2 with Tri, T, and S given in A000217, A053120 and A049310, is the special case k = 1 of the k-family of identities Tri((T(n, 2*k+1) - 1)/2) = Tri(k)*S(n-1, 2*(2*k+1))^2, k >= 0, n >= 0, with S(-1, x) = 0. For k=2 see A108741(n) for S(n-1, 10)^2. This identity boils down to the identities S(n-1, 2*x)^2 = (T(2*n, x) - 1)/(2*(x^2-1)) and 2*T(n, x)^2 - 1 = T(2*n, x) with x = 2*k+1. - Wolfdieter Lang, Feb 01 2016
a(2)=6 is perfect. For n=2*k, k > 0, k not equal to 1, a(n) is a multiple of a(2) and since every multiple (beyond 1) of a perfect number is abundant, then a(n) is abundant. sigma(a(4)) = 504 > 408 = 2*a(4). For n=2*k+1, k > 0, a(n) mod 10 = A000012(n), so a(n) is odd. If a(n) is a prime number, it is deficient; otherwise a(n) has one or two distinct prime factors and is therefore deficient again. So for n=2k+1, k > 0, a(n) is deficient. sigma(a(5)) = 1260 < 2378 = 2*a(5). - Muniru A Asiru, Apr 14 2016
Behera & Panda call these the balancing numbers, and A001541 are the balancers. - Michel Marcus, Nov 07 2017
In general, a second-order linear recurrence with constant coefficients having a signature of (c,d) will be duplicated by a third-order recurrence having a signature of (x,c^2-c*x+d,-d*x+c*d). The formulas of Olivares and Bouhamida in the formula section which have signatures of (7,-7,1) and (5,5,-1), respectively, are specific instances of this general rule for x = 7 and x = 5. - Gary Detlefs, Jan 29 2021
Note that 6 is the largest triangular number in the sequence, because it is proved that 8 and 9 are the largest perfect powers which are consecutive (Catalan's conjecture). 0 and 1 are also in the sequence because they are also perfect powers and 0*1/2 = 0^2 and 8*9/2 = (2*3)^2. - Metin Sariyar, Jul 15 2021

Examples

			G.f. = x + 6*x^2 + 35*x^3 + 204*x^4 + 1189*x^5 + 6930*x^6 + 40391*x^7 + ...
6 is in the sequence since 6^2 = 36 is a triangular number: 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8. - _Michael B. Porter_, Jul 02 2016

References

  • Julio R. Bastida, Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163--166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 193, 197.
  • D. M. Burton, The History of Mathematics, McGraw Hill, (1991), p. 213.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 10.
  • P. Franklin, E. F. Beckenbach, H. S. M Coxeter, N. H. McCoy, K. Menger, and J. L. Synge, Rings And Ideals, No 8, The Carus Mathematical Monographs, The Mathematical Association of America, (1967), pp. 144-146.
  • A. Patra, G. K. Panda, and T. Khemaratchatakumthorn. "Exact divisibility by powers of the balancing and Lucas-balancing numbers." Fibonacci Quart., 59:1 (2021), 57-64; see B(n).
  • 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, pages 257-258.
  • P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022

Links

Crossrefs

Cf. A000217, A000290, A001108, A001542, A001653, A001850, A002315, A002965, A278310.
Chebyshev sequence U(n, m): A000027 (m=1), A001353 (m=2), this sequence (m=3), A001090 (m=4), A004189 (m=5), A004191 (m=6), A007655 (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.

Programs

  • GAP
    a:=[0,1];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Dec 18 2018
  • Haskell
    a001109 n = a001109_list !! n :: Integer
a001109_list = 0 : 1 : zipWith (-)
   (map (* 6) $ tail a001109_list) a001109_list
-- Reinhard Zumkeller, Dec 17 2011
  • Magma
    [n le 2 select n-1 else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jul 25 2015
  • Maple
    a[0]:=1: a[1]:=6: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n],n=0..26); # Emeric Deutsch
with (combinat):seq(fibonacci(2*n,2)/2, n=0..20); # Zerinvary Lajos, Apr 20 2008
  • Mathematica
    Transpose[NestList[Flatten[{Rest[#],ListCorrelate[{-1,6},#]}]&, {0,1}, 30]][[1]]  (* Harvey P. Dale, Mar 23 2011 *)
CoefficientList[Series[x/(1-6x+x^2),{x,0,30}],x]  (* Harvey P. Dale, Mar 23 2011 *)
LinearRecurrence[{6, -1}, {0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, Feb 12 2012 *)
a[ n_]:= ChebyshevU[n-1, 3]; (* Michael Somos, Sep 02 2012 *)
Table[Fibonacci[2n, 2]/2, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
TrigExpand@Table[Sinh[2 n ArcCsch[1]]/(2 Sqrt[2]), {n, 0, 10}] (* Federico Provvedi, Feb 01 2021 *)
  • PARI
    {a(n) = imag((3 + quadgen(32))^n)}; /* Michael Somos, Apr 07 2003 */
  • PARI
    {a(n) = subst( poltchebi( abs(n+1)) - 3 * poltchebi( abs(n)), x, 3) / 8}; /* Michael Somos, Apr 07 2003 */
  • PARI
    {a(n) = polchebyshev( n-1, 2, 3)}; /* Michael Somos, Sep 02 2012 */
  • PARI
    is(n)=ispolygonal(n^2,3) \\ Charles R Greathouse IV, Nov 03 2016
  • Sage
    [lucas_number1(n,6,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
  • Sage
    [chebyshev_U(n-1,3) for n in (0..20)] # G. C. Greubel, Dec 23 2019

Formula

G.f.: x / (1 - 6*x + x^2). - Simon Plouffe in his 1992 dissertation.
a(n) = S(n-1, 6) = U(n-1, 3) with U(n, x) Chebyshev's polynomials of the second kind. S(-1, x) := 0. Cf. triangle A049310 for S(n, x).
a(n) = sqrt(A001110(n)).
a(n) = A001542(n)/2.
a(n) = sqrt((A001541(n)^2-1)/8) (cf. Richardson comment).
a(n) = 3*a(n-1) + sqrt(8*a(n-1)^2+1). - R. J. Mathar, Oct 09 2000
a(n) = A000129(n)*A001333(n) = A000129(n)*(A000129(n)+A000129(n-1)) = ceiling(A001108(n)/sqrt(2)). - Henry Bottomley, Apr 19 2000
a(n) ~ (1/8)*sqrt(2)*(sqrt(2) + 1)^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
Limit_{n->oo} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 05 2002
a(n) = ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) / (4*sqrt(2)). - Gregory V. Richardson, Oct 13 2002. Corrected for offset 0, and rewritten. - Wolfdieter Lang, Feb 10 2015
a(2*n) = a(n)*A003499(n). 4*a(n) = A005319(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 21 2003
a(n) = floor((3+2*sqrt(2))^n/(4*sqrt(2))). - Lekraj Beedassy, Apr 23 2003
a(-n) = -a(n). - Michael Somos, Apr 07 2003
For n >= 1, a(n) = Sum_{k=0..n-1} A001653(k). - Charlie Marion, Jul 01 2003
For n > 0, 4*a(2*n) = A001653(n)^2 - A001653(n-1)^2. - Charlie Marion, Jul 16 2003
For n > 0, a(n) = Sum_{k = 0..n-1}((2*k+1)*A001652(n-1-k)) + A000217(n). - Charlie Marion, Jul 18 2003
a(2*n+1) = a(n+1)^2 - a(n)^2. - Charlie Marion, Jan 12 2004
a(k)*a(2*n+k) = a(n+k)^2 - a(n)^2; e.g., 204*7997214 = 40391^2 - 35^2. - Charlie Marion, Jan 15 2004
For j < n+1, a(k+j)*a(2*n+k-j) - Sum_{i = 0..j-1} a(2*n-(2*i+1)) = a(n+k)^2 - a(n)^2. - Charlie Marion, Jan 18 2004
From Paul Barry, Feb 06 2004: (Start)
a(n) = A000129(2*n)/2;
a(n) = ((1+sqrt(2))^(2*n) - (1-sqrt(2))^(2*n))*sqrt(2)/8;
a(n) = Sum_{i=0..n} Sum_{j=0..n} A000129(i+j)*n!/(i!*j!*(n-i-j)!)/2. (End)
E.g.f.: exp(3*x)*sinh(2*sqrt(2)*x)/(2*sqrt(2)). - Paul Barry, Apr 21 2004
A053141(n+1) + A055997(n+1) = A001541(n+1) + a(n+1). - Creighton Dement, Sep 16 2004
a(n) = Sum_{k=0..n} binomial(2*n, 2*k+1)*2^(k-1). - Paul Barry, Oct 01 2004
a(n) = A001653(n+1) - A038723(n); (a(n)) = chuseq[J]( 'ii' + 'jj' + .5'kk' + 'ij' - 'ji' + 2.5e ), apart from initial term. - Creighton Dement, Nov 19 2004, modified by Davide Colazingari, Jun 24 2016
a(n+1) = Sum_{k=0..n} A001850(k)*A001850(n-k), self convolution of central Delannoy numbers. - Benoit Cloitre, Sep 28 2005
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3), a(1) = 0, a(2) = 1, a(3) = 6, n > 3. Also a(n) = ( (1 + sqrt(2) )^(2*n) - (1 - sqrt(2) )^(2*n) ) / (4*sqrt(2)). - Antonio Alberto Olivares, Oct 23 2003
a(n) = 5*(a(n-1) + a(n-2)) - a(n-3). - Mohamed Bouhamida, Sep 20 2006
Define f(x,s) = s*x + sqrt((s^2-1)*x^2+1); f(0,s)=0. a(n) = f(a(n-1),3), see second formula. - Marcos Carreira, Dec 27 2006
The perfect median m(n) can be expressed in terms of the Pell numbers P() = A000129() by m(n) = P(n + 2) * (P(n + 2) + P(n + 1)) for n >= 0. - Winston A. Richards (ugu(AT)psu.edu), Jun 11 2007
For k = 0..n, a(2*n-k) - a(k) = 2*a(n-k)*A001541(n). Also, a(2*n+1-k) - a(k) = A002315(n-k)*A001653(n). - Charlie Marion, Jul 18 2007
[A001653(n), a(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
a(n) = Sum_{k=0..n-1} 4^k*binomial(n+k,2*k+1). - Paul Barry, Apr 20 2009
a(n+1)^2 - 6*a(n+1)*a(n) + a(n)^2 = 1. - Charlie Marion, Dec 14 2010
a(n) = A002315(m)*A011900(n-m-1) + A001653(m)*A001652(n-m-1) - a(m) = A002315(m)*A053141(n-m-1) + A001653(m)*A046090(n-m-1) + a(m) with m < n; otherwise a(n) = A002315(m)*A053141(m-n) - A001653(m)*A011900(m-n) + a(m) = A002315(m)*A053141(m-n) - A001653(m)*A046090(m-n) - a(m) = (A002315(n) - A001653(n))/2. - Kenneth J Ramsey, Oct 12 2011
16*a(n)^2 + 1 = A056771(n). - James R. Buddenhagen, Dec 09 2011
A010054(A000290(a(n))) = 1. - Reinhard Zumkeller, Dec 17 2011
In general, a(n+k)^2 - A003499(k)*a(n+k)*a(n) + a(n)^2 = a(k)^2. - Charlie Marion, Jan 11 2012
a(n+1) = Sum_{k=0..n} A101950(n,k)*5^k. - Philippe Deléham, Feb 10 2012
PSUM transform of a(n+1) is A053142. PSUMSIGN transform of a(n+1) is A084158. BINOMIAL transform of a(n+1) is A164591. BINOMIAL transform of A086347 is a(n+1). BINOMIAL transform of A057087(n-1). - Michael Somos, May 11 2012
a(n+k) = A001541(k)*a(n) + sqrt(A132592(k)*a(n)^2 + a(k)^2). Generalizes formula dated Oct 09 2000. - Charlie Marion, Nov 27 2012
a(n) + a(n+2*k) = A003499(k)*a(n+k); a(n) + a(n+2*k+1) = A001653(k+1)*A002315(n+k). - Charlie Marion, Nov 29 2012
From Peter Bala, Dec 23 2012: (Start)
Product_{n >= 1} (1 + 1/a(n)) = 1 + sqrt(2).
Product_{n >= 2} (1 - 1/a(n)) = (1/3)*(1 + sqrt(2)). (End)
G.f.: G(0)*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
G.f.: H(0)*x/2, where H(k) = 1 + 1/( 1 - x*(6-x)/(x*(6-x) + 1/H(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Feb 18 2014
a(n) = (a(n-1)^2 - a(n-3)^2)/a(n-2) + a(n-4) for n > 3. - Patrick J. McNab, Jul 24 2015
a(n-k)*a(n+k) + a(k)^2 = a(n)^2, a(n+k) + a(n-k) = A003499(k)*a(n), for n >= k >= 0. - Alexander Samokrutov, Sep 30 2015
Dirichlet g.f.: (PolyLog(s,3+2*sqrt(2)) - PolyLog(s,3-2*sqrt(2)))/(4*sqrt(2)). - Ilya Gutkovskiy, Jun 27 2016
4*a(n)^2 - 1 = A278310(n) for n > 0. - Bruno Berselli, Nov 24 2016
From Klaus Purath, Jan 18 2020: (Start)
a(n) = (a(n-3) + a(n+3))/198.
a(n) = Sum_{i=1..n} A001653(i), n>=1.
a(n) = sinh( 2 * n * arccsch(1) ) / ( 2 * sqrt(2) ). - Federico Provvedi, Feb 01 2021
(End)
a(n) = A002965(2*n)*A002965(2*n+1). - Jon E. Schoenfield, Jan 08 2022
a(n) = A002965(4*n)/2. - Gerry Martens, Jul 14 2023
a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n+k, 2*k+1)*8^k. - Peter Bala, Jul 17 2023

Extensions

Additional comments from Wolfdieter Lang, Feb 10 2000
Duplication of a formula removed by Wolfdieter Lang, Feb 10 2015

A156164 Decimal expansion of 17 + 12*sqrt(2).

Original entry on oeis.org

3, 3, 9, 7, 0, 5, 6, 2, 7, 4, 8, 4, 7, 7, 1, 4, 0, 5, 8, 5, 6, 2, 0, 2, 6, 4, 6, 9, 0, 5, 1, 6, 3, 7, 6, 9, 4, 2, 8, 3, 6, 0, 6, 2, 5, 0, 4, 5, 2, 3, 3, 7, 6, 8, 7, 8, 1, 2, 0, 1, 5, 6, 8, 5, 5, 8, 8, 8, 7, 8, 9, 7, 4, 1, 5, 4, 5, 2, 8, 4, 4, 6, 6, 2, 0, 4, 6, 5, 0, 4, 1, 1, 9, 3, 1, 6, 9, 8, 8, 7, 2, 8, 2, 0, 1
Offset: 2

Views

Author

Klaus Brockhaus, Feb 09 2009

Keywords

Comments

Lim_{n -> infinity} b(n)/b(n-1) = 17+12*sqrt(2) for b = A156160, A156161, A156162, A278310.
Conjecturally, the fractional part 0.97056 27484 ... of this constant equals ( (1 + 2 * Sum_{n >= 1} (-1)^n*exp(-2*Pi*n^2))/(1 + 2 * Sum_{n >= 1} exp(-2*Pi*n^2)) )^4. The series are rapidly converging. For example, summing both series from n = 1 to n = 2 approximates the fractional part of the constant as ( (1 - 2*exp(-2*Pi) + 2*exp(-8*Pi))/(1 + 2*exp(-2*Pi) + 2*exp(-8*Pi)) )^4 = 0.97056 27484 77140 58562 026(89) ..., correct to 23 decimal places. - Peter Bala, Jun 05 2019

Examples

			17 + 12*sqrt(2) = 33.97056274847714058562026469051637694283606250452337687...

Links

Crossrefs

Cf. A002193: decimal expansion of sqrt(2); A156035: decimal expansion of 3+2*sqrt(2); A156163: decimal expansion of (19+6*sqrt(2))/17.
Cf. A005259, A059415/A059416.

Programs

Formula

17+12*sqrt(2) = (3+2*sqrt(2))^2 = (1+sqrt(2))^4. - Klaus Brockhaus, Feb 14 2009. (corrected by Bruno Berselli, Feb 19 2013)

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

Original entry on oeis.org

0, 24, 840, 28560, 970224, 32959080, 1119638520, 38034750624, 1292061882720, 43892069261880, 1491038293021224, 50651409893459760, 1720656898084610640, 58451683124983302024, 1985636569351347658200, 67453191674820837076800, 2291422880374557112953024
Offset: 1

Views

Author

Joseph L. Pe, Jan 07 2003

Keywords

Comments

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

Links

Crossrefs

Cf. A000217, A001542, A001653, A002315, A008844, A009111, A029546, A029549, A046176.
Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

Programs

  • GAP
    a:=[0,24];; for n in [3..20] do a[n]:=34*a[n-1]-a[n-2]+24; od; a; # G. C. Greubel, Jan 13 2020
  • Magma
    I:=[0,24]; [n le 2 select I[n] else 34*Self(n-1) - Self(n-2) + 24: n in [1..20]]; // Marius A. Burtea, Sep 15 2019
  • Maple
    seq(coeff(series(24*x^2/((1-x)*(1-34*x+x^2)), x, n+1), x, n), n = 1..20); # G. C. Greubel, Jan 13 2020
  • Mathematica
    RecurrenceTable[{a[1]==0, a[2]==24, a[n]==34a[n-1] -a[n-2] +24}, a[n], {n,20}]
Drop[CoefficientList[Series[24*x^2/((1-x)*(1-34*x+x^2)), {x,0,20}], x], 1] (* Indranil Ghosh, Mar 15 2017 *)
Table[3*(ChebyshevT[n, 17] -16*ChebyshevU[n-1, 17] -1)/4, {n,20}] (* G. C. Greubel, Jan 13 2020 *)
  • PARI
    concat(0, Vec(24*x^2/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Nov 21 2016
  • Sage
    def A078522_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( 24*x^2/((1-x)*(1-34*x+x^2)) ).list()
a=A078522_list(20); a[1:] # G. C. Greubel, Jan 13 2020

Formula

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

Extensions

Edited by Bruno Berselli, Nov 17 2016

A084703 Squares k such that 2*k+1 is also a square.

Original entry on oeis.org

0, 4, 144, 4900, 166464, 5654884, 192099600, 6525731524, 221682772224, 7530688524100, 255821727047184, 8690408031080164, 295218051329678400, 10028723337177985444, 340681375412721826704, 11573138040695364122500, 393146012008229658338304, 13355391270239113019379844
Offset: 0

Views

Author

Amarnath Murthy, Jun 08 2003

Keywords

Comments

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008
Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011
From M. F. Hasler, Jan 17 2012: (Start)
Bisection of A079291. The squares 2*k+1 are given in A055792.
A204576 is this sequence written in binary. (End)
a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

Links

Crossrefs

Cf. A000129, A001109, A001110, A001333, A001541, A001542, A001652, A001653.
Cf. A014105, A029549, A046090, A046176, A055792, A075114, A079291, A084702.
Cf. A089928, A204576.
Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: this sequence (k=-1), A076218 (k=3), A278310 (k=-5).

Programs

  • Magma
    [4*Evaluate(ChebyshevU(n), 3)^2: n in [0..30]]; // G. C. Greubel, Aug 18 2022
  • Mathematica
    b[n_]:= b[n]= If[n<2, n, 34*b[n-1] -b[n-2] +2]; (* b=A001110 *)
a[n_]:= 4*b[n]; Table[a[n], {n, 0, 30}]
4*ChebyshevU[Range[-1,30], 3]^2 (* G. C. Greubel, Aug 18 2022 *)
  • SageMath
    [4*chebyshev_U(n-1, 3)^2 for n in (0..30)] # G. C. Greubel, Aug 18 2022

Formula

a(n) = 4*A001110(n) = A001542(n)^2.
a(n+1) = A001652(n)*A001652(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2. - Charlie Marion, Jul 01 2003
a(n) = A001653(k+n)*A001653(k-n) - A001653(k)^2, for k >= n >= 0; e.g. 144 = 5741*5 - 169^2. - Charlie Marion, Jul 16 2003
G.f.: 4*x*(1+x)/((1-x)*(1-34*x+x^2)). - R. J. Mathar, Dec 15 2008
a(n) = A079291(2n). - M. F. Hasler, Jan 16 2012
From George F. Johnson, Nov 02 2012: (Start)
a(n) = ((17+12*sqrt(2))^n + (17-12*sqrt(2))^n - 2)/8.
a(n+1) = 17*a(n) + 4 + 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1) = 17*a(n) + 4 - 12*sqrt(a(n)*(2*a(n) + 1)).
a(n-1)*a(n+1) = (a(n) - 4)^2.
2*a(n) + 1 = (A001541(n))^2.
a(n+1) = 34*a(n) - a(n-1) + 8 for n>1, a(0)=0, a(1)=4.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2) for n>0, a(0)=0, a(1)=4, a(2)=144.
a(n)*a(n+1) = (4*A029549(n))^2.
a(n+1) - a(n) = 4*A046176(n).
a(n) + a(n+1) = 4*(6*A029549(n) + 1).
a(n) = (2*A001333(n)*A000129(n))^2.
Limit_{n -> infinity} a(n)/a(n-r) = (17+12*sqrt(2))^r. (End)
Empirical: a(n) = A089928(4*n-2), for n > 0. - Alex Ratushnyak, Apr 12 2013
a(n) = 4*A001109(n)^2. - G. C. Greubel, Aug 18 2022
Product_{n>=2} (1 - 4/a(n)) = sqrt(2)/3 + 1/2 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Extensions

Edited and extended by Robert G. Wilson v, Jun 15 2003

A076218 Numbers n such that 2*n^2 - 3*n + 1 is a square.

Original entry on oeis.org

0, 1, 5, 145, 4901, 166465, 5654885, 192099601, 6525731525, 221682772225, 7530688524101, 255821727047185, 8690408031080165, 295218051329678401, 10028723337177985445, 340681375412721826705, 11573138040695364122501, 393146012008229658338305
Offset: 1

Views

Author

Gregory V. Richardson, Nov 03 2002

Keywords

Comments

Limit_{n -> infinity} a(n)/a(n-1) = 33.970562748477140585620264690516... = 17 + 12*sqrt(2).
Conjecture: a nonzero number occurs twice in A055524 if and only if it's in this sequence. - J. Lowell, Jul 23 2016
Equivalently, n=0 or both n-1 and 2*n-1 are perfect squares. - Sture Sjöstedt, Feb 22 2017

Examples

			5 is in the sequence since 2*5^2 - 3*5 + 1 = 50 - 15 + 1 = 36 is a square. - _Michael B. Porter_, Jul 24 2016

Links

Crossrefs

Cf. similar sequences with closed form ((1 + sqrt(2))^(4*r) + (1 - sqrt(2))^(4*r))/8 + k/4: A084703 (k=-1), this sequence (k=3), A278310 (k=-5).

Programs

  • Mathematica
    Join[{0},LinearRecurrence[{35,-35,1},{1,5,145},20]] (* Harvey P. Dale, Nov 27 2012 *)
  • PARI
    a(n)=if(n>1,([0,1,0;0,0,1;1,-35,35]^n*[145;5;1])[1,1],0) \\ Charles R Greathouse IV, Jul 24 2016
  • PARI
    concat(0, Vec(x^2*(1-30*x+5*x^2) / ((1-x)*(1-34*x+x^2)) + O(x^30))) \\ Colin Barker, Nov 21 2016

Formula

From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Nov 04 2002: (Start)
a(n) = ( (3+(17+12*sqrt(2))^(n-1)) + (3+(17-12*sqrt(2))^(n-1)) )/8 for n>=1.
a(n) = 35 * a(n-1) - 35 * a(n-2) + a(n-3).
G.f.: (x-30*x^2+5*x^3)/(1-35*x+35*x^2-x^3). (End)
Product of adjacent odd-indexed Pell numbers (A000129). - Gary W. Adamson, Jun 07 2003
sqrt(2) - 1 = 0.414213562... = 2/5 + 2/145 + 2/4901 + 2/166465 + ... = Sum_{n>=2} 2/a(n). - Gary W. Adamson, Jun 07 2003
For n > 0, one more than square of adjacent even-indexed Pell numbers (A000129). - Charlie Marion, Mar 09 2005
a(n) = A001652(n-1) + 2*A001652(n-1)*A001652(n-2) + A001652(n-2) + 2. - Charlie Marion, Nov 24 2018

A278438 Numbers m such that T(m) + 2*T(m+1) is a square, where T = A000217.

Original entry on oeis.org

7, 799, 78407, 7683199, 752875207, 73774087199, 7229107670407, 708378777612799, 69413891098384007, 6801852948864019999, 666512175097575576007, 65311391306613542428799, 6399849835873029582446407, 627119972524250285537319199, 61451357457540654953074835207
Offset: 1

Views

Author

Bruno Berselli, Nov 23 2016

Keywords

Comments

It is well known that T(m) + k*T(m+1) is always a square for k=1. For k=3, the nonnegative values of m are the terms of A278310.
Square roots of T(m) + 2*T(m+1) are listed by A168520 (after 0).
Negative values of m for which T(m) + 2*T(m+1) is a square: -1, -2, -82, -7922, -776162, ...

Links

Crossrefs

Subsequence of A056220.
Cf. A000217, A001078, A001079, A122652, A168520, A278620.
Cf. A278310: numbers m such that T(m) + 3*T(m+1) is a square.

Programs

  • Magma
    Iv:=[7, 799]; [n le 2 select Iv[n] else 98*Self(n-1)-Self(n-2)+112: n in [1..20]];
  • Maple
    P:=proc(q) local n; for n from 1 to q do if type(sqrt((3*n^2+7*n+4)/2),integer) then print(n); fi; od; end: P(10^9); #  Paolo P. Lava, Nov 25 2016
  • Mathematica
    Table[((5 + 2 Sqrt[6])^(2 n) + (5 - 2 Sqrt[6])^(2 n))/12 - 7/6, {n, 1, 20}]
RecurrenceTable[{a[1] == 7, a[2] == 799, a[n] == 98 a[n - 1] - a[n - 2] + 112}, a, {n, 1, 20}]
LinearRecurrence[{99,-99,1},{7,799,78407},20] (* Harvey P. Dale, Oct 18 2024 *)
  • PARI
    Vec(x*(7 + 106*x - x^2)/((1 - x)*(1 - 98*x + x^2)) + O(x^20)) \\ Colin Barker, Nov 27 2016
  • Sage
    def A278438():
    a, b = 7, 799
    yield a
    while True:
        yield b
        a, b = b, 98*b - a + 112
a = A278438(); print([next(a) for  in range(15)]) # _Peter Luschny, Nov 24 2016

Formula

O.g.f.: x*(7 + 106*x - x^2)/((1 - x)*(1 - 98*x + x^2)).
E.g.f.: (exp((5-2*sqrt(6))^2*x) + exp((5+2*sqrt(6))^2*x) - 14*exp(x))/12 + 1.
a(n) = 99*a(n-1) - 99*a(n-2) + a(n-3) for n>3.
a(n) = 98*a(n-1) - a(n-2) + 112 for n>2.
a(n) = a(-n) = ((5 + 2*sqrt(6))^(2*n) + (5 - 2*sqrt(6))^(2*n))/12 - 7/6.
a(n) = A001079(2*n)/6 - 7/6.
a(n) = 2*A001078(n)^2 - 1 = A122652(n)^2/2 - 1.
a(n) = -A278620(n+1) + 106*A278620(n) + 7*A278620(n-1).
Lim_{n -> infinity} a(n)/a(n-1) = (5 + 2*sqrt(6))^2.
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