A077443 -id:A077443 - cp's OEIS frontend

A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

5, 27, 157, 915, 5333, 31083, 181165, 1055907, 6154277, 35869755, 209064253, 1218515763, 7102030325, 41393666187, 241259966797, 1406166134595, 8195736840773, 47768254910043, 278413792619485, 1622714500806867, 9457873212221717, 55124524772523435, 321289275422918893

Offset: 0

Creighton Dement, Jan 23 2005

A floretion-generated sequence relating to NSW numbers and numbers n such that (n^2 - 8)/2 is a square. It is also possible to label this sequence as the "tesfor-transform of the zero-sequence" under the floretion given in the program code, below. This is because the sequence "vesseq" would normally have been A046184 (indices of octagonal numbers which are also a square) using the floretion given. This floretion, however, was purposely "altered" in such a way that the sequence "vesseq" would turn into A000004. As (a(n)) would not have occurred under "natural" circumstances, one could speak of it as the transform of A000004.
Floretion Algebra Multiplication Program FAMP code: - tesforseq[ + 3'i - 2'j + 'k + 3i' - 2j' + k' - 4'ii' - 3'jj' + 4'kk' - 'ij' - 'ji' + 3'jk' + 3'kj' + 4e], Note: vesforseq = A000004, lesforseq = A002315, jesforseq = A077445
From Wolfdieter Lang, Feb 05 2015: (Start)
All positive solutions x = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the  fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding y solutions are given by y(n) = A253811(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See, e.g., the Nagell reference on how to find all solutions.
(End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.
Tanya Khovanova, Recursive Sequences
Morris Newman, Daniel Shanks, and H. C. Williams, Simple groups of square order and an interesting sequence of primes, Acta Arith., 38 (1980/1981) 129-140. MR82b:20022.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000004, A000129, A001109, A002315, A038761, A038762, A046184, A054490, A164737.

Cf. A077442, A077443(n+1), A077445, A253811.

R:=PowerSeriesRing(Integers(), 30); Coefficients(R!((5 - 3*x)/(1-6*x+x^2))); // G. C. Greubel, Jul 26 2018

A101386:= (n) -> simplify(5*ChebyshevU(n, 3) - 3*ChebyshevU(n-1, 3)); seq( A101386(n), n = 0..30); # G. C. Greubel, Mar 17 2020

CoefficientList[ Series[(5-3x)/(1-6x+x^2), {x,0,30}], x] (* Robert G. Wilson v, Jan 29 2005 *)
LinearRecurrence[{6,-1},{5,27},30] (* Harvey P. Dale, Apr 23 2016 *)

Vec((5-3*x)/(1-6*x+x^2) + O(x^30)) \\ Colin Barker, Feb 05 2015

[5*chebyshev_U(n,3) -3*chebyshev_U(n-1,3) for n in (0..30)] # G. C. Greubel, Mar 17 2020

a(n) = A002315(n) + A077445(n+1). Note: the offset of A077445 is 1.
a(n+1) - a(n) = 2*A054490(n+1).
a(n) = 6*a(n-1) - a(n-2), a(0)=5, a(1)=27. - Philippe Deléham, Nov 17 2008
From Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009: (Start)
a(n) = ((5+sqrt(18))*(3 + sqrt(8))^n + (5-sqrt(18))*(3 - sqrt(8))^n)/2.
Third binomial transform of A164737. (End)
a(n) = rational part of z(n), with z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions. See the preceding formula. - Wolfdieter Lang, Feb 05 2015
a(n) = 5*A001109(n+1) - 3*A001109(n). - G. C. Greubel, Mar 17 2020
a(n) = Pell(2*n+2) + 3*Pell(2*n+1), where Pell(n) = A000129(n). - G. C. Greubel, Apr 17 2020
E.g.f.: exp(3*x)*(5*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from Robert G. Wilson v, Jan 29 2005

A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

Original entry on oeis.org

3, 13, 75, 437, 2547, 14845, 86523, 504293, 2939235, 17131117, 99847467, 581953685, 3391874643, 19769294173, 115223890395, 671574048197, 3914220398787, 22813748344525, 132968269668363, 774995869665653, 4517006948325555, 26327045820287677, 153445267973400507

Offset: 0

Barry E. Williams, May 03 2000

This gives part of the (increasingly sorted) positive solutions x to the Pell equation x^2 - 2*y^2 = +7. For the y solutions see A038761. The other part of solutions is found in A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			a(3)^2 - 2*A038761(3)^2 = 437^2 - 2*309^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
M. J. DeLeon, Pell's Equation and Pell Number Triples, Fib. Quart., 14(1976), pp. 456-460.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (6,-1).
Index entries for sequences related to Chebyshev polynomials.

Cf. A001109, A038725, A038761, A077443, A101386, A253811.

I:=[3, 13]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

LinearRecurrence[{6,-1},{3,13},40] (* Vincenzo Librandi, Nov 16 2011 *)

x='x+O('x^30); Vec((3-5*x)/(1-6*x+x^2)) \\ G. C. Greubel, Jul 26 2018

a(n) = sqrt(2*(A038761(n))^2+7).
a(n) = (13*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-3*((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = A077443(2n) = A038725(n)+A038725(n+1).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3); a(n) = (1/2)*(3+sqrt(2))*(3+2*sqrt(2))^(n-1)+(1/2)*(3-sqrt(2))*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Apr 20 2008
G.f.: (3-5*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008, corrected by R. J. Mathar, Nov 06 2011
a(n) = -5*A001109(n) +3*A001109(n+1). - R. J. Mathar, Nov 06 2011
a(n) = rational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only one part of the positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
E.g.f.: exp(3*x)*(3*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)). - Stefano Spezia, Mar 16 2024

More terms from James Sellers, May 04 2000

Unspecific Pell comment replaced by Wolfdieter Lang, Feb 05 2015

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

Original entry on oeis.org

1, 3, 9, 19, 53, 111, 309, 647, 1801, 3771, 10497, 21979, 61181, 128103, 356589, 746639, 2078353, 4351731, 12113529, 25363747, 70602821, 147830751, 411503397, 861620759, 2398417561, 5021893803, 13979001969, 29269742059, 81475594253

Offset: 0

Gregory V. Richardson, Nov 06 2002

Lim. n -> Inf. a(n)/a(n-2) = 3 + 2*Sqrt(2) = R1*R2. Lim. k -> Inf. a(2*k-1)/a(2*k) = (9 + 4*Sqrt(2))/7 = R1 (ratio #1). Lim. k -> Inf. a(2*k)/a(2*k-1) = (11 + 6*Sqrt(2))/7 = R2 (ratio #2).

a(n) gives for n >= 0 all positive y-values solving the (generalized) Pell equation x^2 - 2*y^2 = 7. A077443(n+1) gives the corresponding x-values. See, e.g., the Nagell reference on how to find all solutions. - Wolfdieter Lang, Feb 05 2015

			a(4)^2 - 2*a(3)^2 = 27^2 - 2*19^2  = +7. - _Wolfdieter Lang_, Feb 05 2015

L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
A. H. Beiler, "The Pellian." Ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.
T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

Colin Barker, Table of n, a(n) for n = 0..1000
J. J. O'Connor and E. F. Robertson, History of Pell's Equation
J. P. Robertson, Solving the Generalized Pell Equation
Eric Weisstein's World of Mathematics, Pell Equation.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A077443, A038762, A038761, A101386, A253811.

CoefficientList[Series[(1+3 x+3 x^2+x^3)/ (1-6 x^2+x^4),{x,0,50}],x]  (* Harvey P. Dale, Mar 12 2011 *)
LinearRecurrence[{0, 6, 0, -1},{1,3,9,19},50] (* Sture Sjöstedt, Oct 08 2012 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^n*[1;3;9;19])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

Vec((x+1)^3/(x^2+2*x-1)/(x^2-2*x-1) + O(x^50)) \\ Colin Barker, Mar 27 2016

For n>0, a(2n) = A046090(n) + A001653(n) + A001652(n-1); a(2n+1) = A001652(n+1) - A001652(n-1) - A001653(n-1); e.g. 53=21+29+3; 111=119-3-5. - Charlie Marion, Aug 14 2003
The same recurrences hold for the odd and even indices respectively : a(n+2) = 6*a(n+1) - a(n), a(n+1) = 3*a(n) + 2*(2*a(n)^2+7)^0.5. - Richard Choulet, Oct 11 2007
G.f.: (x+1)^3/(x^2+2*x-1)/(x^2-2*x-1). a(n)= [ -A077985(n)-3*A077985(n-1)+3*A000129(n+1)+A000129(n)]/2. - R. J. Mathar, Nov 16 2007
a(n) = 6*a(n-2) - a(n-4) with a(1)=1, a(2)=3, a(3)=9, a(4)=19. - Sture Sjöstedt, Oct 08 2012
a(n) = ((-(-1 - sqrt(2))^n*(-2+sqrt(2)) - (-1+sqrt(2))^n*(2+sqrt(2)) + (1-sqrt(2))^n*(-4+3*sqrt(2)) + (1+sqrt(2))^n*(4+3*sqrt(2))))/(4*sqrt(2)). - Colin Barker, Mar 27 2016

Edited: n in Name replaced by a(n). Pell comments moved to comment section. - Wolfdieter Lang, Feb 05 2015

A253811 Part of the y solutions of the Pell equation x^2 - 2*y^2 = +7.

Original entry on oeis.org

3, 19, 111, 647, 3771, 21979, 128103, 746639, 4351731, 25363747, 147830751, 861620759, 5021893803, 29269742059, 170596558551, 994309609247, 5795261096931, 33777256972339, 196868280737103, 1147432427450279, 6687726283964571, 38978925276337147, 227185825374058311

Offset: 0

Wolfdieter Lang, Feb 05 2015

All positive solutions y = a(n) of the (generalized) Pell equation x^2 - 2*y^2 = +7 based on the fundamental solution (x2,y2) = (5,3) of the second class of (proper) solutions. The corresponding x solutions are given by x(n) = A101386(n).
All other positive solutions come from the first class of (proper) solutions based on the fundamental solution (x1,y1) = (3,1). These are given in A038762 and A038761.
All solutions of this Pell equation are found in A077443(n+1) and A077442(n), for n >= 0. See the Nagell reference on how to find all solutions.

			A101386(2)^2 - 2*a(2) = 157^2 - 2*111^2 = +7.

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, 1964, Theorem 109, pp. 207-208 with Theorem 104, pp. 197-198.

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

Cf. A001109, A038761, A038762, A077442, A077443, A101386.

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((x+3)/(x^2-6*x+1))); // G. C. Greubel, Jul 26 2018

LinearRecurrence[{6,-1}, {3,19}, 30] (* or *) CoefficientList[Series[ (x+3)/(x^2-6*x+1), {z, 0, 50}], x]  (* G. C. Greubel, Jul 26 2018 *)

Vec((x+3)/(x^2-6*x+1) + O(x^100)) \\ Colin Barker, Feb 05 2015

a(n) = irrational part of z(n), where z(n) = (5+3*sqrt(2))*(3+2*sqrt(2))^n, n >= 0, the general positive solutions of the second class of proper solutions.
From Colin Barker, Feb 05 2015: (Start)
a(n) = 6*a(n-1) - a(n-2).
G.f.: (x+3) / (x^2-6*x+1). (End)
a(n) = 3*A001109(n+1) + A001109(n). - R. J. Mathar, Feb 05 2015
E.g.f.: exp(3*x)*(6*cosh(2*sqrt(2)*x) + 5*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

Original entry on oeis.org

4, 7, 8, 16, 23, 28, 31, 32, 36, 47, 56, 63, 64, 68, 71, 72, 79, 92, 100, 103, 112, 119, 124, 127, 128, 136, 144, 151, 164, 167, 175, 184, 188, 191, 196, 199, 200, 207, 223, 224, 239, 248, 252, 256, 263, 271, 272, 279, 284, 287, 288, 292, 311, 316, 324, 328

Offset: 1

Colin Barker, Feb 10 2014

nonn

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

			4 is in the sequence because x^2 - 6xy + y^2 + 4 = 0 has integer solutions, for example (x, y) = (1, 5).

N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

Cf. A001653 (k = 4), A006452 (k = 7), A001541 (k = 8), A075870 (k = 16), A156066 (k = 23), A217975 (k = 28), A003499 (k = 32), A075841 (k = 36), A077443 (k = 56).
Cf. A031363, A084917, A237351, A237606, A237609, A237610.
For primes see A007522 and A141175.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

A069017 Triangular numbers of the form k^2 + k + 1.

Original entry on oeis.org

1, 3, 21, 91, 703, 3081, 23871, 104653, 810901, 3555111, 27546753, 120769111, 935778691, 4102594653, 31788928731, 139367449081, 1079887798153, 4734390674091, 36684396208461, 160829915470003, 1246189583289511, 5463482735306001, 42333761435634903

Offset: 1

Amarnath Murthy, Apr 02 2002

G. C. Greubel, Table of n, a(n) for n = 1..1001
Index entries for linear recurrences with constant coefficients, signature (1,34,-34,-1,1).

Cf. A124174.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)) )); // G. C. Greubel, Dec 01 2018

Do[a = n(n + 1) + 1; b = Floor[Sqrt[2a]]; If[b(b + 1) == 2a, Print[a]], {n, 1, 106}] (* Robert G. Wilson v *)
Select[Table[n^2+n+1,{n,0,206*10^6}],OddQ[Sqrt[8#+1]]&] (* The program takes a long time to run. *) (* Harvey P. Dale, Sep 22 2017 *)
CoefficientList[Series[(x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)), {x,0,50}], x] (* G. C. Greubel, Dec 01 2018 *)

Vec((x^4+2*x^3-16*x^2+2*x+1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2)) +O(x^66)) /* Joerg Arndt, Mar 25 2013 */

s=((x^4 +2*x^3 -16*x^2 +2*x +1)/((1-x)*(1-6*x+x^2)*(1+6*x+x^2))).series(x, 50); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 01 2018

G.f.: (x^4 + 2*x^3 - 16*x^2 + 2*x + 1)/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2)).
From Zak Seidov, Sep 25 2010: (Start)
a(n) = 34*a(n-2) - a(n-4) - 11.
a(n) = 2*A124174(n) + 1. (End)
a(n) = (A077443(n)^2 - 1)/2. - Amiram Eldar, Dec 01 2018

Program and terms from Robert G. Wilson v

a(18)-a(22) from Alex Ratushnyak, Mar 23 2013

A077242 Combined Diophantine Chebyshev sequences A077240 and A077239.

Original entry on oeis.org

5, 7, 23, 37, 133, 215, 775, 1253, 4517, 7303, 26327, 42565, 153445, 248087, 894343, 1445957, 5212613, 8427655, 30381335, 49119973, 177075397, 286292183, 1032071047, 1668633125, 6015350885, 9725506567, 35060034263, 56684406277, 204344854693, 330380931095

Offset: 0

Wolfdieter Lang, Nov 08 2002

a(n)^2 - 8*b(n)^2 = 17, with the companion sequence b(n)= A077241(n).

Because there is only one primitive Pythagorean triangle with sum of the legs L = 17 (see also A120681), namely (5,12,13), all positive solutions (x(n), y(n)) = (a(n), 2*A077241(n)) of the (generalized) Pell equation x^2 - 2*y^2 = +17 satisfy x(n) < 2*y(n), for n >= 1, only 5 = x(0) > 2*y(0) = 4. The proof runs along the same line as the one given in a comment on the L=7 case in A077443. - Wolfdieter Lang, Feb 05 2015

			23 = a(2) = sqrt(8*A077241(2)^2 + 17) = sqrt(8*8^2 + 17)= sqrt(529) = 23.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

I:=[5,7,23,37]; [n le 4 select I[n] else 6*Self(n-2)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Feb 18 2014

A077239 = Table[2*ChebyshevT[n+1, 3] + ChebyshevT[n, 3], {n, 0, 12}]; A077240 = Table[ChebyshevT[n+1, 3] + 2*ChebyshevT[n, 3], {n, 0, 12}]; Riffle[A077240, A077239] (* Jean-François Alcover, Dec 19 2013 *)
CoefficientList[Series[(1 - x) (5 + 12 x + 5 x^2)/(1 - 6 x^2 + x^4), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 18 2014 *)

Vec((1-x)*(5+12*x+5*x^2)/(1-6*x^2+x^4) + O(x^50)) \\ Colin Barker, Mar 27 2016

a(2*k) = A077240(k) and a(2*k+1) = A077239(k), k>=0.
G.f.: (1-x)*(5+12*x+5*x^2)/(1-6*x^2+x^4).
a(n) = 6*a(n-2)-a(n-4) for n>3. - Vincenzo Librandi, Feb 18 2014
a(n) = ((6-5*sqrt(2))*(1-sqrt(2))^n - (-1-sqrt(2))^n*(-4+sqrt(2)) + 4*(-1+sqrt(2))^n + sqrt(2)*(-1+sqrt(2))^n + 6*(1+sqrt(2))^n + 5*sqrt(2)*(1+sqrt(2))^n)/4. - Colin Barker, Mar 27 2016

A133216 Integers that are simultaneously triangular (A000217) and decagonal (A001107).

Original entry on oeis.org

0, 1, 10, 1540, 11935, 1777555, 13773376, 2051297326, 15894464365, 2367195337045, 18342198104230, 2731741367653000, 21166880717817451, 3152427171076225351, 24426562006163234620, 3637898223680596402450, 28188231388231654934425, 4198131397700237172202345

Offset: 1

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

nonn

Positive terms are of the form (m^2-9)/16 where m runs over the elements of A077443 that are congruent to 5 modulo 8. Correspondingly, for n>1, sqrt(16*a(n)+9) form a subsequence of A077443, while sqrt(8*a(n)+1) form a subsequence of A077442 with indices congruent to 2,3 modulo 4. [Max Alekseyev]

			The initial terms of the sequences of triangular (A000217) and decagonal (A001107) numbers are 0, 1, 3, 6, 10, 15, ... and 0, 1, 10, 27, ... respectively. As the third number which is common to both sequences is 10, we have a(3) = 10.

Index entries for linear recurrences with constant coefficients, signature (1, 1154, -1154, -1, 1).

Cf. A000217, A001107, A133217, A133218, A077443, A077442.

LinearRecurrence[{1, 1154, -1154, -1, 1} , {0, 1, 10, 1540, 11935, 1777555}, 17] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

a(n) = A000217(A133218(n)) = A001107(A133217(n)).
For n>5, a(n) = 1154*a(n-2) - a(n-4) + 396.
For n>6, a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/64 * ( (9 + 4* sqrt(2)*(-1)^n)*(1+sqrt(2))^(4*n-6) + (9 - 4* sqrt(2)*(-1)^n)*(1-sqrt(2))^(4*n-6) - 22).
a(n) = floor ( 1/64 * (9 + 4*sqrt(2)*(-1)^n) * (1+sqrt(2))^(4*n-6) ).
G.f.: (x^5 + 9*x^4 + 376*x^3 + 9*x^2 + x)/((1 - x)*(x^2 - 34*x + 1)*(x^2 + 34*x + 1)). [corrected by Peter Luschny, Apr 04 2019]
Lim (n -> Infinity, a(2n+1)/a(2n)) = (1/49)*(3649+2580*sqrt(2)).
Lim (n -> Infinity, a(2n)/a(2n-1)) = (1/49)*(193+132*sqrt(2)).

Entry revised by N. J. A. Sloane, Nov 06 2011

Term 0 prepended and entry revised accordingly by Max Alekseyev, Nov 06 2011

A133217 Indices of decagonal numbers (A001107) that are also triangular (A000217).

Original entry on oeis.org

0, 1, 2, 20, 55, 667, 1856, 22646, 63037, 769285, 2141390, 26133032, 72744211, 887753791, 2471161772, 30157495850, 83946756025, 1024467105097, 2851718543066, 34801724077436, 96874483708207, 1182234151527715, 3290880727535960, 40161159427864862

Offset: 1

Richard Choulet, Oct 11 2007; Ant King, Nov 04 2011

nonn

For n>0, a(n) = (A055979(n) - A056161(n))/2, with those two sequences related through the Diophantine equation 2x^2 + 3x + 2 = r^2. - Richard R. Forberg, Nov 24 2013

			The third number which is both decagonal (A001107) and triangular (A000217) is A133216(3)=10. As this is the second decagonal number, we have a(3) = 2.

Index entries for linear recurrences with constant coefficients, signature (1, 34, -34, -1, 1).

Cf. A000217, A001107, A077443, A077442, A133216, A133218.

LinearRecurrence[{1, 34, -34, -1, 1} , {0, 1, 2, 20, 55, 667}, 24] (* first term 0 corrected by Georg Fischer, Apr 02 2019 *)

For n>5, a(n) = 34*a(n-2) - a(n-4) - 12.
For n>6, a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5).
For n>1, a(n) = 1/16 * ((2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3) - (2*sqrt(2) - (-1)^n)*(1 - sqrt(2))^(2*n - 3) + 6).
For n>1, a(n) = ceiling (1/16*(2*sqrt(2) + (-1)^n)*(1 + sqrt(2))^(2*n - 3)).
G.f.: ( 1 - 33*x^2 + 18*x^3 + 2*x^4 ) / ((1 - x ) * (1 - 6*x + x^2 ) * (1 + 6*x + x^2)).
lim (n -> Infinity, a(2n+1)/a(2n)) = 1/7*(43 + 30*sqrt(2)).
lim (n -> Infinity, a(2n)/a(2n-1)) = 1/7*(11 + 6*sqrt(2)).

Entry revised by Max Alekseyev, Nov 06 2011

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

cp's OEIS Frontend

A101386 Expansion of g.f.: (5 - 3*x)/(1 - 6*x + x^2).

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A038762 a(n) = 6*a(n-1) - a(n-2) for n >= 2, with a(0)=3, a(1)=13.

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

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A253811 Part of the y solutions of the Pell equation x^2 - 2*y^2 = +7.

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A237599 Positive integers k such that x^2 - 6xy + y^2 + k = 0 has integer solutions.

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A069017 Triangular numbers of the form k^2 + k + 1.

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A101386 Expansion of g.f.: (5 - 3x)/(1 - 6x + x^2).

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