A077446 -id:A077446 - cp's OEIS frontend

A364319 a(n) = (A077446(n) + 1)/2 for n >= 0.

0, 1, 3, 6, 16, 33, 91, 190, 528, 1105, 3075, 6438, 17920, 37521, 104443, 218686, 608736, 1274593, 3547971, 7428870, 20679088, 43298625, 120526555, 252362878, 702480240, 1470878641, 4094354883, 8572908966, 23863649056, 49966575153, 139087539451

Offset: 0

Wolfdieter Lang, Aug 15 2023

a(n) and b(n) = A006452(n+1), for n >= 0, give the nonnegative solution of the equation binomial(a(n), 2) = b(n)^2 - 1.

This shows that the number of independent elements of an antisymmetric a(n) X a(n) matrix coincides with the number of independent elements of a traceless b(n) X b(n) matrix. The n = 0 case is trivial: 0 = 0. (The question about this coincidence was posed to W. L. by Martin Bordemann, Mar 03 1991.)

			The solutions (a(n), b(n)) begin:
  n: 0 1 2 3  4  5  6   7   8    9   10   11    12    13     14     15 ...
  ------------------------------------------------------------------------
  a: 0 1 3 6 16 33 91 190 528 1105 3075 6438 17920 37521 104443 218686 ...
  b: 1 1 2 4 11 23 64 134 373  781 2174 4552 12671 26531  73852 154634 ...

Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (1,6,-6,-1,1).

Cf. A001109, A001541, A006452, A049310, A077446.

LinearRecurrence[{1, 6, -6, -1, 1}, {0, 1, 3, 6, 16}, 31] (* Robert P. P. McKone, Aug 29 2023 *)

a(n) = 6*a(n-2) - a(n-4) - 2, for n >= 0, with a(-4) = -32, a(-3) = -15, a(-2) = -5, a(-1) = -2.
O.g.f.: G(x) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 6*x^2 + x^4)) = x*(1 + 2*x - 3*x^2 - 2*x^3)/((1 - x)*(1 - 2*x - x^2)*(1 + 2*x - x^2)).
Bisection: a(2*k) = (5*S(k-1, 6) + S(k-2, 6) + 1)/2 and a(2*k+1) = (S(k, 6) + 5*S(k-1, 6) + 1)/2, for k >= 0, with the Chebyshev polynomials S(n, x) (A049310) with S(-2, x) = -1, S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n+1).
Bisection: a(2*k) = (1 + 8*q(k) - p(k))/2 and a(2*k+1) = (1 + 8*q(k) + p(k))/2, for k >= 0, with p(k) = A001541(k) = S(k, 6) - 3*S(k-1, 6) and q(k) = A001109(k) = S(k-1, 6).
E.g.f.: (cosh(x) - cosh(sqrt(2)*x)*(cosh(x) - 3*sinh(x)) + sinh(x) - sqrt(2)*(cosh(x) - 2*sinh(x))*sinh(sqrt(2)*x))/2. - Stefano Spezia, Aug 29 2023

A006452 a(n) = 6*a(n-2) - a(n-4).

Original entry on oeis.org

1, 1, 2, 4, 11, 23, 64, 134, 373, 781, 2174, 4552, 12671, 26531, 73852, 154634, 430441, 901273, 2508794, 5253004, 14622323, 30616751, 85225144, 178447502, 496728541, 1040068261, 2895146102, 6061962064, 16874148071, 35331704123

Offset: 0

N. J. A. Sloane, Jeffrey Shallit

Solution to a Diophantine equation.
Integers k such that k^2-1 is a triangular number. - Benoit Cloitre, Apr 05 2002
For all elements "x" of the sequence, 8*x^2 - 7 is a square. - Gregory V. Richardson, Oct 07 2002
a(n) mod 10 is a sequence of period 12: repeat (1, 1, 2, 4, 1, 3, 4, 4, 3, 1, 4, 2). - Paul Curtz, Dec 07 2012
a(n)^2 - 1 = A006454(n - 1) is a Sophie Germain triangular number of the second kind as defined in A217278. - Raphie Frank, Feb 08 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 6xy + y^2 + 7 = 0. - Colin Barker, Feb 04 2014
Except for the first term, positive values of x (or y) satisfying x^2 - 34xy + y^2 + 252 = 0. - Colin Barker, Mar 04 2014
From Wolfdieter Lang, Feb 26 2015: (Start)
a(n+1), for n >= 0, gives one half of all positive y solutions of the Pell equation x^2 - 2*y^2 = -7. The corresponding x-solutions are x(n) = A077446(n+1).
See a comment on A077446 for the first and second class solutions separately, and the connection to the Pell equation X^2 - 2*Y^2 = 14. (End)
For n > 0, a(n) is the n-th almost balancing number of second type (see Tekcan and Erdem). - Stefano Spezia, Nov 26 2022

			n = 3: 11^2 - 2*(2*4)^2 = -7 (see the Pell comment above);
(4*4)^2 - 2*11^2 = +14. - _Wolfdieter Lang_, Feb 26 2015

A. J. Gottlieb, How four dogs meet in a field, etc., Technology Review, Jul/Aug 1973 pp. 73-74.
Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. J. Gottlieb, How four dogs meet in a field, etc. (scanned copy)
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
J. Shallit, Letter to N. J. A. Sloane, Oct. 1975
Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A001333, A006451, A006452, A006454, A038723, A077446, A216134, A217278.

I:=[1,1,2,4]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 09 2013

A006452:=-(z-1)*(z**2+3*z+1)/(z**2+2*z-1)/(z**2-2*z-1); # conjectured by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

s=0;lst={1}; Do[s+=n;If[Sqrt[s+1]==Floor[Sqrt[s+1]],AppendTo[lst, Sqrt[s+1]]], {n,0,8!}]; lst (* Vladimir Joseph Stephan Orlovsky, Apr 02 2009 *)
a[0]=a[1]= 1; a[2]=2; a[3]=4; a[n_]:= 6*a[n-2] -a[n-4]; Array[a, 30, 0] (* Robert G. Wilson v, Jun 11 2010 *)
CoefficientList[Series[(1+x-4x^2-2x^3)/((1-2x-x^2)(1+2x-x^2)), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 09 2013 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;1;2;4])[1,1] \\ Charles R Greathouse IV, May 10 2016

def A001333(n): return lucas_number2(n, 2, -1)/2
def A006452(n): return (A001333(n+1) + (-1)^n *A001333(n-2))/4
[A006452(n) for n in range(41)] # G. C. Greubel, Jan 22 2023

Bisection: a(2n) = A006452(n). a(2n+1) = A038723(n).
G.f.: ( 1+x-4*x^2-2*x^3 ) / ( (1-2*x-x^2)*(1+2*x-x^2) ).
From Gregory V. Richardson, Oct 07 2002: (Start)
For n (even), a(n) = ( ((3 + sqrt(8))^((n/2)+1) - (3 - sqrt(8))^((n/2)+1)) - 2*((3 + sqrt(8))^((n/2)-1) - (3 - sqrt(8))^((n/2)-1)) ) / (6*sqrt(8)).
For n (odd), a(n) = ( ((3 + sqrt(8))^((n+1)/2) - (3 - sqrt(8))^((n+1)/2)) - 2*((3 + sqrt(8))^((n-1)/2) - (3 - sqrt(8))^((n-1)/2)) ) / (2*sqrt(8)).
Limit_{n->oo} a(n)/a(n-2) = 3 + sqrt(8).
If n is odd, lim_{n->oo} a(n)/a(n-1) = (9 + 2*sqrt(8))/7.
If n is even, lim_{n->oo} a(n)/a(n-1) = (11 + 3*sqrt(8))/7. (End)
a(n+2) = (A001333(n+3) + (-1)^n *A001333(n))/4. - Paul Curtz, Dec 06 2012
a(n+2) = sqrt(17*a(n)^2 + 6*(sqrt(8*a(n)^2 - 7))*a(n)*sgn(2*n - 1) - 7) with a(0) = 1, a(1) = 1. - Raphie Frank, Feb 08 2013
a(n+2) = (A216134(n+2) - A216134(n))/2. - Raphie Frank, Feb 11 2013
E.g.f.: (2*cosh(sqrt(2)*x)*(2*cosh(x) - sinh(x)) + sqrt(2)*(3*cosh(x) - sinh(x))*sinh(sqrt(2)*x))/4. - Stefano Spezia, Nov 26 2022

More terms from James Sellers, May 03 2000

A255236 All positive solutions x of the second class of the Pell equation x^2 - 2*y^2 = -7.

Original entry on oeis.org

5, 31, 181, 1055, 6149, 35839, 208885, 1217471, 7095941, 41358175, 241053109, 1404960479, 8188709765, 47727298111, 278175078901, 1621323175295, 9449763972869, 55077260661919, 321013799998645, 1871005539329951, 10905019435981061, 63559111076556415

Offset: 0

Wolfdieter Lang, Feb 26 2015

For the corresponding y = y2 terms see 2*A038725(n+1).
The Pell equation x^2 - 2*y^2 = 7 has two classes of solutions. See, e.g., the Nagell reference and comments under A254938 and A255233. Here the positive solutions based on the fundamental solution (5, 4) (the second largest positive solution) are considered.
The positive solutions of the first class are given in (A054490(n), 2*A038723(n)), n >= 0.
The combined solutions of both classes are given in (A077446, 4*A077447).
The solutions (x(n), y(n)) of x^2 - 2*y^2 = -7 translate to the solutions (X(n), Y(n)) = (2*y(n) , x(n)) of the Pell equation X^2 - 2*Y^2 = 14.

			n = 2: 181^2 - 2*(2*64)^2  = -7; (4*64)^2 - 2*181^2 = 14.
n = 2: 2*53 + 75 = 181. - _Wolfdieter Lang_, Mar 19 2015

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A038725, A054490, A038723, A077446, 4*A077447, A254938, A255233.

I:=[5,31]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

CoefficientList[Series[(5 + x) / (1 - 6 x + x^2), {x, 0, 30}], x] (* Vincenzo Librandi, Mar 20 2015 *)

Vec((5 + x)/(1 - 6*x + x^2) + O(x^30)) \\ Michel Marcus, Mar 20 2015

a(n) = 5*S(n, 6) + S(n-1, 6), n >= 0, with the Chebyshev polynomials S(n, x) (A049310), with S(-1, x) = 0, evaluated at x = 6. S(n, 6) = A001109(n-1).
G.f.: (5 + x)/(1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2), n >= 2, with a(-1) = -1 and a(0) = 5.
a(n) = 2*A038761(n) + A038762(n), n >= 0. See the Mar 19 comment on A054490. - Wolfdieter Lang, Mar 19 2015
a(n) = ((3-2*sqrt(2))^n*(-8+5*sqrt(2)) + (3+2*sqrt(2))^n*(8+5*sqrt(2))) / (2*sqrt(2)). - Colin Barker, Oct 13 2015

A217975 Integers k such that 2*k^2 - 7 is a square.

Original entry on oeis.org

2, 4, 8, 22, 46, 128, 268, 746, 1562, 4348, 9104, 25342, 53062, 147704, 309268, 860882, 1802546, 5017588, 10506008, 29244646, 61233502, 170450288, 356895004, 993457082, 2080136522, 5790292204, 12123924128, 33748296142, 70663408246, 196699484648

Offset: 1

Sture Sjöstedt, Oct 16 2012

a(n) gives y-values solving the Diophantine equation x^2 + 7 = 2*y^2. A077446(n) gives the x-values. - Sture Sjöstedt, Oct 16 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 28 = 0. - Colin Barker, Feb 08 2014

			Since 2(4^2) - 7 = 25 = 5^2, and 4 is the second number with this property, a(2) = 4.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077442 (2*n^2 + 7 is a square).

I:=[2, 4, 8, 22]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..31]]; // Vincenzo Librandi, Oct 16 2012

LinearRecurrence[{0, 6, 0, -1}, {2, 4, 8, 22}, 50] (* Sture Sjöstedt, Oct 16 2012 *)

Vec(2*x*(1-x)*(x^2+3*x+1)/(x^2-2*x-1)/(x^2+2*x-1)+O(x^99)) \\ Charles R Greathouse IV, Oct 24 2012

a(n) = 6*a(n - 2) - a(n - 4) with a(1)=2, a(2)=4, a(3)=8, a(4)=22. - Sture Sjöstedt, Oct 16 2012
a(n)*a(n+3)-a(n+1)*a(n+2) = 10-2*(-1)^n. - Bruno Berselli, Oct 25 2012
a(n) = 2*A006452(n). - R. J. Mathar, Oct 17 2012
G.f.: -2*x*(x - 1)*(x^2 + 3*x + 1)/((x^2 - 2*x - 1)*(x^2 + 2*x - 1)). - Colin Barker, Oct 24 2012
a(n) = a(-n+1) = ((4+sqrt(2))*(1-(-1)^n*sqrt(2))^(2*floor(n/2))+(4-sqrt(2))*(1+(-1)^n*sqrt(2))^(2*floor(n/2)))/4. - Bruno Berselli, Oct 25 2012
a(2n-1) = A078343(2n-1), a(2n) = A100525(n-1). - Bruno Berselli, Oct 25 2012

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.

Original entry on oeis.org

1, 7, 43, 253, 1477, 8611, 50191, 292537, 1705033, 9937663, 57920947, 337588021, 1967607181, 11468055067, 66840723223, 389576284273, 2270616982417, 13234125610231, 77134136678971, 449570694463597, 2620290030102613, 15272169486152083, 89012726886809887, 518804191834707241

Offset: 1

Stefano Spezia, Nov 26 2022

a(n) is the n-th almost cobalancing number of second type (see Tekcan and Erdem).

			a(2) = 7 is a term since 8*7^2 + 8*7 - 7 = 441 = 21^2.

Ahmet Tekcan and Alper Erdem, General Terms of All Almost Balancing Numbers of First and Second Type, arXiv:2211.08907 [math.NT], 2022.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A002315, A006452, A077443, A077446, A106328, A106329, A216134.

Mathematica
```
LinearRecurrence[{7,-7,1},{1,7,43},24]
```

a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) for n > 3.
a(n) = (3*(3 - 2*sqrt(2))^n*(2 + sqrt(2)) + 3*(2 - sqrt(2))*(3 + 2*sqrt(2))^n - 4)/8.
O.g.f.: x*(1 + x^2)/((1 - x)*(1 - 6*x + x^2)).
E.g.f.: (3*(2 + sqrt(2))*(cosh(3*x - 2*sqrt(2)*x) + sinh(3*x - 2*sqrt(2)*x)) + 3*(2 - sqrt(2))*(cosh(3*x + 2*sqrt(2)*x) + sinh(3*x + 2*sqrt(2)*x)) - 4*(cosh(x) + sinh(x)) - 8)/8.
a(n) = 3*A011900(n) - 2 = 6*A053142(n) + 1. - Hugo Pfoertner, Nov 26 2022

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.

Original entry on oeis.org

9, 15, 55, 89, 321, 519, 1871, 3025, 10905, 17631, 63559, 102761, 370449, 598935, 2159135, 3490849, 12584361, 20346159, 73347031, 118586105, 427497825, 691170471

Offset: 1

Ctibor O. Zizka, Nov 09 2009

Let (ceiling(sqrt(n*n/2)))^2 - n*n/2 = i. Then for i=1/2 we have A002315, for i=1 we have A005319, for i=2 we have A077444, for i=7/2 we have A077446, for i=4 we have A081554.

Conjecture: a(n) = 6*a(n-2) - a(n-4). - Charles R Greathouse IV, Apr 30 2016

Cf. A005319, A077444, A081554, A002315, A077446,

lista(nn)=for (n=1, nn, if ((ceil(sqrt(n*n/2)))^2 - n*n/2 == 17/2, print1(n, ", "));); \\ Michel Marcus, Jun 02 2013

forstep(n=9,1e9,2, if((sqrtint(n^2\2)+1)^2==(n^2+17)/2, print1(n", "))) \\ Charles R Greathouse IV, Apr 30 2016

More terms from Michel Marcus, Jun 02 2013

a(17)-a(22) from Charles R Greathouse IV, Apr 30 2016

cp's OEIS Frontend

A364319 a(n) = (A077446(n) + 1)/2 for n >= 0.

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A006452 a(n) = 6*a(n-2) - a(n-4).

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A255236 All positive solutions x of the second class of the Pell equation x^2 - 2*y^2 = -7.

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A217975 Integers k such that 2*k^2 - 7 is a square.

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Magma

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A358682 Numbers k such that 8*k^2 + 8*k - 7 is a square.

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A175033 Numbers n such that (ceiling(sqrt(n*n/2)))^2 - n*n/2 = 17/2.

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PARI

PARI

Extensions

A358682 Numbers k such that 8k^2 + 8k - 7 is a square.

A175033 Numbers n such that (ceiling(sqrt(nn/2)))^2 - nn/2 = 17/2.