A001652 -id:A001652 - cp's OEIS frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

1, 9, 53, 309, 1801, 10497, 61181, 356589, 2078353, 12113529, 70602821, 411503397, 2398417561, 13979001969, 81475594253, 474874563549, 2767771787041, 16131756158697, 94022765165141, 548004834832149, 3194006243827753, 18616032628134369, 108502189524978461

Offset: 0

Barry E. Williams, May 02 2000

Bisection of A048654. - Lambert Klasen (lambert.klasen(AT)gmx.de), Nov 24 2004

This gives part of the (increasingly sorted) positive solutions y to the Pell equation x^2 - 2*y^2 = +7. For the x solutions see A038762. For the other part of solutions see A101386 and A253811. - Wolfdieter Lang, Feb 05 2015

			A038762(3)^2 - 2*a(4)^2 = 2547^2 - 2*1801^2 = +7. - _Wolfdieter Lang_, Feb 05 2015

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 122-125, 194-196.

Vincenzo Librandi, Table of n, a(n) for n = 0..400
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, A001541, A001542, A001652, A001653, A038762, A046090, A048654, A055997, A101386, A124124, A253811.

I:=[1, 9]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..40]]; // Vincenzo Librandi, Nov 16 2011

a[0]:=1: a[1]:=9: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..19); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,9},40] (* Vincenzo Librandi, Nov 16 2011 *)

a(n)=([0,1; -1,6]^n*[1;9])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

a(n) = (9*((3+2*sqrt(2))^n -(3-2*sqrt(2))^n)-((3+2*sqrt(2))^(n-1) - (3-2*sqrt(2))^(n-1)))/(4*sqrt(2)).
a(n) = sqrt(2*(A038762(n))^2-14)/2.
For n>1, a(n)-4a(n-1)=A001541(n)-A001542(n-2); e.g. 309-4*53=97=99-2. - Charlie Marion, Nov 12 2003
For n>0, a(n)=A046090(n)+A001653(n)+A001652(n-1)=A055997(n+1)+A001652(n-1); e.g., 309=120+169+20. - Charlie Marion, Oct 11 2006
G.f.: (1+3*x)/(1-6*x+x^2). - Philippe Deléham, Nov 03 2008
a(n) = third binomial transform of 1,6,8,48,64,384. - Al Hakanson (hawkuu(AT)gmail.com), Aug 15 2009
a(n)^2 + 2^2 = A124124(2*n+1)^2 + (A124124(2*n+1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
a(n) = irrational part of z(n) = (3 + sqrt(2))*(3 + 2*sqrt(2))^n, n >= 0. z(n) gives only part of the general positive solutions to the Pell equation x^2 - 2*y^2 = 7. See the Nagell reference in A038762 on how to find z(n), and a comment above. - Wolfdieter Lang, Feb 05 2015
a(n) = S(n, 6) + 3*S(n-1, 6), n >= 0, with the Chebyshev S-polynomials evaluated at x=6. See S(n-1, 6) = A001109(n). - Wolfdieter Lang, Mar 30 2015
E.g.f.: exp(3*x)*(2*cosh(2*sqrt(2)*x) + 3*sqrt(2)*sinh(2*sqrt(2)*x))/2. - Stefano Spezia, Mar 16 2024

Edited: Replaced the unspecific Pell comment. Moved a formula from the comment section to the formula section. - Wolfdieter Lang, Feb 05 2015

A075528 Triangular numbers that are half other triangular numbers.

Original entry on oeis.org

0, 3, 105, 3570, 121278, 4119885, 139954815, 4754343828, 161507735340, 5486508657735, 186379786627653, 6331426236682470, 215082112260576330, 7306460390622912753, 248204571168918457275, 8431648959352604634600, 286427860046819639119128

Offset: 0

Christian G. Bower, Sep 19 2002

This is the sequence of 1/2 the areas, x(n)*y(n)/2, of the ordered Pythagorean triples (x(n), y(n)=x(n)+1, z(n)) with x(0)=0, y(0)=1, z(0)=1, a(0)=0 and x(1)=3, y(1)=4, z(1)=5, a(1)=3. - George F. Johnson, Aug 24 2012

Colin Barker, Table of n, a(n) for n = 0..653
Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
Harvey J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021.
Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000129, A000217, A001652, A002315, A029546, A029549, A046090, A046729, A053141, A084159, A096979.

CoefficientList[ Series[ 3x/(1 - 35 x + 35 x^2 - x^3), {x, 0, 15}], x] (* Robert G. Wilson v, Jun 24 2011 *)

concat(0, Vec(3*x/((1-x)*(1-34*x+x^2)) + O(x^20))) \\ Colin Barker, Jun 18 2015

a(n) = 3*A029546(n-1) = A029549(n)/2.
G.f.: 3*x/((1-x)*(1-34*x+x^2)).
From George F. Johnson, Aug 24 2012: (Start)
a(n) = ((3+2*sqrt(2))^(2*n+1) + (3-2*sqrt(2))^(2*n+1) - 6)/64.
8*a(n)+1 = A000129(2*n+1)^2.
16*a(n)+1 = A002315(n)^2.
128*a(n)^2 + 24*a(n) + 1 is a perfect square.
a(n+1) = 17*a(n) + 3/2 + 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1) = 17*a(n) + 3/2 - 3*sqrt((8*a(n)+1)*(16*a(n)+1))/2.
a(n-1)*a(n+1) = a(n)*(a(n)-3); a(n+1) = 34*a(n) - a(n-1) + 3.
a(n+1) = 35*a(n) - 35*a(n-1) + a(n-2); a(n) = A096979(2*n)/2.
a(n) = A084159(n)*A046729(n)/4 = A001652(n)*A046090(n)/4.
Lim_{n->infinity} a(n)/a(n-1) =  17 + 12*sqrt(2).
Lim_{n->infinity} a(n)/a(n-2) = (17 + 12*sqrt(2))^2 = 577 + 408*sqrt(2).
Lim_{n->infinity} a(n)/a(n-r) = (17 + 12*sqrt(2))^r.
Lim_{n->infinity} a(n-r)/a(n) = (17 + 12*sqrt(2))^(-r) = (17 - 12*sqrt(2))^r.
(End)
a(n) = 34*a(n-1) - a(n-2) + 3, n >= 2. - R. J. Mathar, Nov 07 2015
a(n) = A000217(A053141(n)). - R. J. Mathar, Aug 16 2019
a(n) = (a(n-1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020
Sum_{n>=1} 1/a(n) = 2*(3 - 2*sqrt(2)). - Amiram Eldar, Dec 04 2024

A005319 a(n) = 6*a(n-1) - a(n-2).

Original entry on oeis.org

0, 4, 24, 140, 816, 4756, 27720, 161564, 941664, 5488420, 31988856, 186444716, 1086679440, 6333631924, 36915112104, 215157040700, 1254027132096, 7309005751876, 42600007379160, 248291038523084, 1447146223759344, 8434586304032980

Offset: 0

N. J. A. Sloane

Solutions y of the equation 2x^2-y^2=2; the corresponding x values are given by A001541. - N-E. Fahssi, Feb 25 2008
The lower intermediate convergents to 2^(1/2) beginning with 4/3, 24/17, 140/99, 816/577, form a strictly increasing sequence; essentially, numerators=A005319 and denominators=A001541. - Clark Kimberling, Aug 26 2008
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 1 + n*n/2. - Ctibor O. Zizka, Nov 09 2009
All nonnegative solutions of the indefinite binary quadratic form X^2 + 4*X*Y -4*Y^2 of discriminant 32, representing -4 are (X(n), Y(n)) = (a(n), A001653(n+1)), for n >= 0. - Wolfdieter Lang, Jun 13 2018
Also the number of edge covers in the n-triangular snake graph. - Eric W. Weisstein, Jun 08 2019
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=A001653(n+1), z=A002315(n) with 0Michael Somos, Jun 26 2022
a(n) is the sum of 4*n consecutive powers of the silver ratio 1+sqrt(2), starting at (1+sqrt(2))^(-2*n) and ending at (1+sqrt(2))^(2*n-1). - Greg Dresden and Ruxin Sheng, Jul 25 2024

			G.f. = 4*x + 24*x^2 + 140*x^3 + 816*x^4 + 4756*x^5 + ... - _Michael Somos_, Jun 26 2022

P. de la Harpe, Topics in Geometric Group Theory, Univ. Chicago Press, 2000, p. 160, middle display.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Marius A. Burtea, Table of n, a(n) for n = 0..100
K. Andersen, L. Carbone, and D. Penta, Kac-Moody Fibonacci sequences, hyperbolic golden ratios, and real quadratic fields, Journal of Number Theory and Combinatorics, Vol 2, No. 3 pp 245-278, 2011. See Section 9.
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Tanya Khovanova, Recursive Sequences
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
Soumeya M. Tebtoub, Hacène Belbachir, and László Németh, Integer sequences and ellipse chains inside a hyperbola, Proceedings of the 1st International Conference on Algebras, Graphs and Ordered Sets (ALGOS 2020), hal-02918958 [math.cs], 17-18.
Eric Weisstein's World of Mathematics, Edge Cover
Eric Weisstein's World of Mathematics, Triangular Snake Graph
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A000129, A001109, A001541, A001542, A001652, A001653, A046090, A046729, A090390.

a:=[0,4]; [n le 2 select a[n] else 6*Self(n-1) - Self(n-2):n in [1..22]]; // Marius A. Burtea, Sep 19 2019

LinearRecurrence[{6, -1}, {0, 4}, 22] (* Jean-François Alcover, Sep 26 2017 *)
Table[((3 + 2 Sqrt[2])^n - (3 - 2 Sqrt[2])^n)/Sqrt[2], {n, 20}] // Expand (* Eric W. Weisstein, Jun 08 2019 *)
CoefficientList[Series[(4 x)/(1 - 6 x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Jun 08 2019 *)
a[ n_] := 4*ChebyshevU[n-1, 3]; (* Michael Somos, Jun 26 2022 *)

{a(n) = 4*polchebyshev(n-1, 2, 3)}; /* Michael Somos, Jun 26 2022 */

G.f.: 4*x / ( 1-6*x+x^2 ). - Simon Plouffe in his 1992 dissertation.
G.f. for signed version beginning with 1: (1+2*x+x^2)/(1+6*x+x^2).
For any term n of the sequence, 2*n^2 + 4 is a perfect square. Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 06 2002
a(n) = ((3+2*sqrt(2))^n - (3-2*sqrt(2))^n) / sqrt(2). - Gregory V. Richardson, Oct 06 2002
(-1)^(n+1) = A090390(n+1) + A001542(n+1) + A046729(n) - a(n) (conjectured). - Creighton Dement, Nov 17 2004
For n > 0, a(n) = A000129(n+1)^2 - A000129(n-1)^2; a(n) = A046090(n-1) + A001652(n); e.g., 816 = 120 + 696; a(n) = A001653(n) - A001653(n-1); e.g., 816 = 985 - 169. - Charlie Marion Jul 22 2005
a(n) = 4*A001109(n). - M. F. Hasler, Mar 2009
For n > 1, a(n) is the denominator of continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the numerators, see A001653. - Greg Dresden, Sep 10 2019
1/a(n) - 1/a(n+1) = 1/(Pell(2*n+1) - 1/Pell(2*n+1)) for n >= 1, where Pell(n) = A000129(n). - Peter Bala, Aug 21 2022
E.g.f.: sqrt(2)*exp(3*x)*sinh(2*sqrt(2)*x). - Stefano Spezia, Nov 25 2022
a(n) = 2*A000129(2*n). - Tanya Khovanova and MIT PRIMES STEP senior group, Apr 17 2024

A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

Original entry on oeis.org

1, 25, 841, 28561, 970225, 32959081, 1119638521, 38034750625, 1292061882721, 43892069261881, 1491038293021225, 50651409893459761, 1720656898084610641, 58451683124983302025, 1985636569351347658201, 67453191674820837076801, 2291422880374557112953025

Offset: 0

N. J. A. Sloane

Numbers simultaneously square and centered square. E.g., a(1)=25 because 25 is the fourth centered square number and the fifth square number. - Steven Schlicker, Apr 24 2007
Solutions to A007913(x)=A007913(2x-1). - Benoit Cloitre, Apr 07 2002
From Ant King, Nov 09 2011: (Start)
Indices of positive hexagonal numbers that are also perfect squares.
As n increases, this sequence is approximately geometric with common ratio r = lim_{n -> infinity} a(n)/a(n-1) = (1 + sqrt(2))^4 = 17 + 12 * sqrt(2).
(End)
Also indices of hexagonal numbers (A000384) which are also centered octagonal numbers (A016754). - Colin Barker, Jan 25 2015
Also positive integers x in the solutions to 4*x^2 - 8*y^2 - 2*x + 8*y - 2 = 0, the corresponding values of y being A253826. - Colin Barker, Jan 25 2015
Squares that are sum of two consecutive squares: y^2 = (k + 1)^2 + k^2 is equivalent to x^2 - 2*y^2 = -1 with x = 2*k + 1. - Jean-Christophe Hervé, Nov 11 2015
Squares in the main diagonal of the natural number array, A000027. - Clark Kimberling, Mar 12 2023

			From _Ravi Kumar Davala_, May 26 2013: (Start)
A001333(0)=1, A001333(4)=17, A001333(8)=577, A000129(0)=0, A000129(2)=2, A000129(4)=12, A000129(8)=408 so clearly
a(n+m)=A001333(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)), with m=1,2 is true.
A002203(0)=2, A002203(4)=34, A002203(8)=1154 so clearly
a(n+m)=(1/2)*A002203(4*m)*a(n)-(A000129(2*m))^2+A000129(4*m)*sqrt(2*a(n)^2-a(n)) is true for m=1,2
a(n+1)*a(n-1) = (a(n)+4)^2 , with n=1 is 841*1=(25+4)^2, for n=2 , 28561*25=(841+4)^2.
(End)
1 = 1 + 0, 25 = 16 + 9, 841 = 29^2 = 21^2 + 20^2 = 441 + 400.

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 256.

Muniru A Asiru, Table of n, a(n) for n = 0..100
Daniel C. Fielder, Special integer sequences controlled by three parameters, Fibonacci Quarterly 6, 1968, 64-70.
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.
Thomas Koshy, Products Involving Reciprocals of Gibonacci Polynomials, The Fibonacci Quarterly, Vol. 60, No. 1 (2022), pp. 15-24.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Steven C. Schlicker, Numbers Simultaneously Polygonal and Centered Polygonal, Mathematics Magazine, Vol. 84, No. 5, December 2011
Eric Weisstein's World of Mathematics, Hexagonal Square Number.
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001844, A007913, A000384, A046177, A016754, A253826.

a := [1, 25, 841];; for i in [4..10^2] do a[i] := 35*a[i-1] - 35*a[i-2] + a[i-3]; od; a;  # Muniru A Asiru, Jan 17 2018

I:=[1,25,841]; [n le 3 select I[n] else 35*Self(n-1)-35*Self(n-2)+Self(n-3): n in [1..20]]; // Vincenzo Librandi, Jan 20 2018

CP := n -> 1+1/2*4*(n^2-n): N:=10: u:=3: v:=1: x:=4: y:=1: k_pcp:=[1]: for i from 1 to N do tempx:=x; tempy:=y; x:=tempx*u+8*tempy*v: y:=tempx*v+tempy*u: s:=(y+1)/2: k_pcp:=[op(k_pcp),CP(s)]: end do: k_pcp; # Steven Schlicker, Apr 24 2007

LinearRecurrence[{35, -35, 1}, {1, 25, 841}, 15] (* Ant King, Nov 09 2011 *)
CoefficientList[Series[(1 - 10 x + x^2) / ((1 - x) (1 - 34 x + x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Jan 20 2018 *)

a(n)=if(n<0,0,sqr(subst(poltchebi(n+1)+poltchebi(n),x,3)/4))

vector(40, n, n--; (([5, 2; 2, 1]^n)[1, 1])^2) \\ Altug Alkan, Nov 11 2015

From Benoit Cloitre, Jan 19 2003: (Start)
a(n) = A078522(n) + 1.
a(n) = ceiling(A*B^n) where A = (3 + 2*sqrt(2))/8 and B = 17 + 12*sqrt(2). (End)
G.f.: (1-10x+x^2)/((1-x)(1-34x+x^2)).
a(n) = ceiling(A046176(n)/sqrt(2)). - Helge Robitzsch (hrobi(AT)math.uni-goettingen.de), Jul 28 2000
a(n+1) = 17*a(n) - 4 + 12*sqrt(2*a(n)^2 - a(n)). - Richard Choulet, Sep 14 2007
Define x(n) + y(n)*sqrt(8) = (4+sqrt(8))*(3+sqrt(8))^n, s(n) = (y(n)+1)/2; then a(n) = (1/2)*(2+4*(s(n)^2 - s(n))). - Steven Schlicker, Apr 24 2007
From Ant King, Nov 09 2011: (Start)
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3).
a(n) = 34*a(n-1) - a(n-2) - 8.
a(n) = 1/8 * ((1 + sqrt(2))^(4*n-2) + (1 - sqrt(2))^(4*n-2) + 2).
a(n) = ceiling((1/8) * (1 + sqrt(2))^(4*n-2)). (End)
From Ravi Kumar Davala, May 26 2013: (Start)
a(n+2) = 577*a(n) - 144 + 408*sqrt(2*a(n)^2 - a(n)).
a(n+m) = A001333(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+m) = (1/2)*A002203(4*m)*a(n) - (A000129(2*m))^2 + A000129(4*m)*sqrt(2*a(n)^2 - a(n)).
a(n+1)*a(n-1) = (a(n)+4)^2. (End)
a(n) = A001652(n)^2 + A046090(n)^2. - César Aguilera, Jan 15 2018
Limit_{n -> infinity} a(n)/a(n-1) = A156164. - César Aguilera, Jan 28 2018
sqrt(2*a(n))-1 = A002315(n). - Ezhilarasu Velayutham, Apr 05 2019
4*a(n) = 1 +3*A077420(n). - R. J. Mathar, Mar 05 2024
Product_{n>=0} (1 + 4/a(n)) = 2*sqrt(2) + 3 (Koshy, 2022, section 3, p. 19). - Amiram Eldar, Jan 23 2025

Entry edited by N. J. A. Sloane, Sep 14 2007

A053142 a(n) = A053141(n)/2.

Original entry on oeis.org

0, 1, 7, 42, 246, 1435, 8365, 48756, 284172, 1656277, 9653491, 56264670, 327934530, 1911342511, 11140120537, 64929380712, 378436163736, 2205687601705, 12855689446495, 74928449077266, 436715005017102, 2545361581025347, 14835454481134981, 86467365305784540

Offset: 0

Wolfdieter Lang

Partial sums of A001109. - Barry Williams, May 03 2000.
Number m such that 16*m*(2*m+1)+1 is a square. - Bruno Berselli, Oct 19 2012
From Robert K. Moniot, Sep 21 2020: (Start)
Consecutive terms (a(n-1),a(n))=(u,v) give all points on the hyperbola u^2-u+v^2-v-6*u*v=0 in quadrant 1 with both coordinates an integer.
Let T(n) denote the n-th triangular number. If i, j are any two successive elements of the above sequence then (T(i-1) + T(j-1))/T(i+j-1) = 3/4.
(End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
R. C. Alperin, A nonlinear recurrence and its relations to Chebyshev polynomials, Fib. Q., Vol. 58, No. 2 (2020), 140-142.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000129, A001109, A001653, A053141, A001652, A046090, A049310.

Cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x/((1-x)*(1-6*x+x^2)))); // G. C. Greubel, Jul 15 2018

Join[{a=0,b=1}, Table[c=6*b-a+1; a=b; b=c, {n,60}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
Table[(Fibonacci[2n + 1, 2] - 1)/4, {n, 0, 20}] (* Vladimir Reshetnikov, Sep 16 2016 *)
LinearRecurrence[{7, -7, 1}, {0, 1, 7}, 30] (* G. C. Greubel, Jul 15 2018 *)

{a=1+sqrt(2); b=1-sqrt(2); P(n) = (a^n - b^n)/(a-b)};
for(n=0, 30, print1(round((P(2*n+1) - 1)/4), ", ")) \\ G. C. Greubel, Jul 15 2018

x='x+O('x^30); Vec(x/((1-x)*(1-6*x+x^2))) \\ G. C. Greubel, Jul 15 2018

a(n) = (A001653(n)-1)/4.
a(n) = 6*a(n-1)-a(n-2)+1, a(0)=0, a(1)=1.
G.f.: x/((1-x)*(1-6*x+x^2)).
From Paul Barry, Nov 14 2003: (Start)
a(n+1) = Sum_{k=0..n} S(k, 6) = Sum_{k=0..n} U(n, 3), Chebyshev polynomials of 2nd kind, A049310.
a(n+1) = (sqrt(2)-1)^(2*n)(5/8-7*sqrt(2)/16)+(sqrt(2)+1)^(2*n)*(7*sqrt(2)/16 + 5/8)-1/4. (End)
From Antonio Alberto Olivares, Jan 13 2004: (Start)
a(n) = 7*a(n-1)-7*a(n-2)+a(n-3).
a(n) = -(1/4) + (1-sqrt(2))/(-8*sqrt(2))*(3-2*sqrt(2))^n + (1+sqrt(2))/(8*sqrt(2))*(3+2*sqrt(2))^n. (End)
a(n) = Sum_{k=0..n} Sum_{j=0..2*k} (-1)^(j+1)*A000129(j)*A000129(2*k-j). Paul Barry, Oct 23 2009
a(2*k) = A001109(k)*(A001109(k) + A001109(k-1)) and a(2*k-1) = A001109(k)*(A001109(k) + A001109(k+1)). Kenneth J Ramsey, Sep 10 2010
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Robert K. Moniot, Sep 21 2020
E.g.f.: (exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - 2*exp(x))/8. - Stefano Spezia, Mar 16 2024

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

Original entry on oeis.org

0, 7, 104, 1455, 20272, 282359, 3932760, 54776287, 762935264, 10626317415, 148005508552, 2061450802319, 28712305723920, 399910829332567, 5570039304932024, 77580639439715775, 1080558912851088832, 15050244140475527879, 209622859053806301480

Offset: 0

N. J. A. Sloane

(a(n)+1)^3 - a(n)^3 is a square (that of A001570(n)).
The ratio A001570(n)/a(n) tends to sqrt(3) = 1.73205... as n increases. - Pierre CAMI, Apr 21 2005
Define a(1)=0 a(2)=7 such that 3*(a(1)^2) + 3*a(1) + 1 = j(1)^2 = 1^2 and 3*(a(2)^2) + 3*a(2) + 1 = j(2)^2 = 13^2. Then a(n) = a(n-2) + 8*sqrt(3*(a(n-1)^2) + 3*a(n-1) + 1). Another definition : a(n) such that 3*(a(n)^2) + 3*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005
a(n) = A001353(n)*A001075(n+1). For n>0, the triple {a(n), a(n)+1=A001922(n), A001570(n)} forms a near-isosceles triangle with angle 2*Pi/3 bounded by the consecutive sides. - Lekraj Beedassy, Jul 21 2006
Numbers n such that A003215(n) is a square, cf. A006051. - Joerg Arndt, Jan 02 2017

			G.f. = 7*x + 104*x^2 + 1455*x^3 + 20272*x^4 + 282359*x^5 + 3932760*x^6 + ... - _Michael Somos_, Aug 17 2018

J. D. E. Konhauser et al., Which Way Did the Bicycle Go?, MAA 1996, p. 104.
E.-A. Majol, Note #2228, L'Intermédiaire des Mathématiciens, 9 (1902), pp. 183-185. - N. J. A. Sloane, Mar 03 2022
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..200
J. Brenner and E. P. Starke, Problem E702, Amer. Math. Monthly, 53 (1946), 465.
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
Eric Weisstein's World of Mathematics, Hex Number
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001570, A001922, A006051.

Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; A233450 for k=3; A001652 for k=4; A129556 for k=5; this sequence for k=6. - Bruno Berselli, Dec 16 2013

[Round(-1/2 - (1/6)*Sqrt(3)*(7-4*Sqrt(3))^n + (1/6)*Sqrt(3)*(7+4*Sqrt(3))^n + (1/4)*(7+4*Sqrt(3))^n + (1/4)*(7-4*Sqrt(3))^n): n in [0..50]]; // G. C. Greubel, Nov 04 2017

A001921:=z*(-7+z)/(z-1)/(z**2-14*z+1); # Conjectured by Simon Plouffe in his 1992 dissertation.

t = {0, 7}; Do[AppendTo[t, 14*t[[-1]] - t[[-2]] + 6], {20}]; t (* T. D. Noe, Aug 17 2012 *)
LinearRecurrence[{15, -15, 1}, {0, 7, 104}, 19] (* Michael De Vlieger, Jan 02 2017 *)
a[ n_] := -1/2 + (ChebyshevT[n + 1, 7] - ChebyshevT[n, 7]) / 12; (* Michael Somos, Aug 17 2018 *)

concat(0, Vec(x*(x-7)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Jan 06 2015

{a(n) = -1/2 + (polchebyshev(n + 1, 1, 7) - polchebyshev(n, 1, 7)) / 12}; /* Michael Somos, Aug 17 2018 */

G.f.: x*(-7 + x)/(x - 1)/(x^2 - 14*x + 1) (see Simon Plouffe in Maple section).
a(n) = (A028230(n+1)-1)/2. - R. J. Mathar, Mar 19 2009
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3). - Colin Barker, Jan 06 2015
a(n) = -1 - a(-1-n) for all n in Z. - Michael Somos, Aug 17 2018

More terms from James Sellers, Jul 04 2000

A084158 a(n) = A000129(n) * A000129(n+1)/2.

Original entry on oeis.org

0, 1, 5, 30, 174, 1015, 5915, 34476, 200940, 1171165, 6826049, 39785130, 231884730, 1351523251, 7877254775, 45912005400, 267594777624, 1559656660345, 9090345184445, 52982414446326, 308804141493510, 1799842434514735, 10490250465594899, 61141660359054660, 356359711688733060

Offset: 0

Paul Barry, May 18 2003

May be called Pell triangles.

Vincenzo Librandi, Table of n, a(n) for n = 0..500
Paul Barry, Symmetric Third-Order Recurring Sequences, Chebyshev Polynomials, and Riordan Arrays, JIS 12 (2009) 09.8.6.
Sergio Falcon, On the Sequences of Products of Two k-Fibonacci Numbers, American Review of Mathematics and Statistics, March 2014, Vol. 2, No. 1, pp. 111-120.
Ronald Orozco López, Generating Functions of Generalized Simplicial Polytopic Numbers and (s,t)-Derivatives of Partial Theta Function, arXiv:2408.08943 [math.CO], 2024. See p. 11.
Ronald Orozco López, Simplicial d-Polytopic Numbers Defined on Generalized Fibonacci Polynomials, arXiv:2501.11490 [math.CO], 2025. See p. 6.
Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

Cf. A000129, A001652, A001654, A002203, A002315, A003499.

Cf. A041011, A046729, A079291, A084159, A084175, A163960, A383720.

[Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n)/16): n in [0..35]]; // Vincenzo Librandi, Jul 05 2011

with(combinat): a:=n->fibonacci(n,2)*fibonacci(n-1,2)/2: seq(a(n), n=1..22); # Zerinvary Lajos, Apr 04 2008

LinearRecurrence[{5,5,-1},{0,1,5},30] (* Harvey P. Dale, Sep 07 2011 *)

Pell(n)=([2, 1; 1, 0]^n)[2, 1];
a(n)=Pell(n)*Pell(n+1)/2 \\ Charles R Greathouse IV, Mar 21 2016

a(n)=([0,1,0; 0,0,1; -1,5,5]^n*[0;1;5])[1,1] \\ Charles R Greathouse IV, Mar 21 2016

[(lucas_number2(2*n+1,2,-1) -2*(-1)^n)/16 for n in (0..30)] # G. C. Greubel, Aug 18 2022

a(n) = ((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/16.
a(n) = 5*a(n-1) + 5*a(n-2) - a(n-3). - Mohamed Bouhamida, Sep 02 2006; corrected by Antonio Alberto Olivares, Mar 29 2008
a(n) = (-1/8)*(-1)^n + (( sqrt(2)+1)/16)*(3+2*sqrt(2))^n + ((-sqrt(2)+1)/16)*(3-2*sqrt(2))^n. - Antonio Alberto Olivares, Mar 30 2008
sqrt(a(n) - a(n-1)) = A000129(n). - Antonio Alberto Olivares, Mar 30 2008
O.g.f.: x/((1+x)(1-6*x+x^2)). - R. J. Mathar, May 18 2008
a(n) = A041011(n)*A041011(n+1). - R. K. Guy, May 18 2008
From Mohamed Bouhamida, Aug 30 2008: (Start)
a(n) = 6*a(n-1) - a(n-2) - (-1)^n.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3) - 2*(-1)^n. (End)
In general, for n>k+1, a(n+k) = A003499(k+1)*a(n-1) - a(n-k-2) - (-1)^n A000129(k+1)^2. - Charlie Marion, Jan 04 2012
For n>0, a(2n-1)*a(2n+1) = oblong(a(2n)); a(2n)*a(2n+2) = oblong(a(2n+1)-1). - Charlie Marion, Jan 09 2012
a(n) = A046729(n)/4. - Wolfdieter Lang, Mar 07 2012
a(n) = sum of squares of first n Pell numbers A000129 (A079291). - N. J. A. Sloane, Jun 18 2012
a(n) = (A002315(n) - (-1)^n)/8. - Adam Mohamed, Sep 05 2024
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*(sqrt(2)-1) (A163960). - Amiram Eldar, Dec 02 2024
G.f.: x * exp( Sum_{k>=1} Pell(3*k)/Pell(k) * x^k/k ). - Seiichi Manyama, May 07 2025

A055997 Numbers k such that k*(k - 1)/2 is a square.

Original entry on oeis.org

1, 2, 9, 50, 289, 1682, 9801, 57122, 332929, 1940450, 11309769, 65918162, 384199201, 2239277042, 13051463049, 76069501250, 443365544449, 2584123765442, 15061377048201, 87784138523762, 511643454094369, 2982076586042450, 17380816062160329, 101302819786919522

Offset: 1

Barry E. Williams, Jun 14 2000

Numbers k such that (k-th triangular number - k) is a square.
Gives solutions to A007913(2x)=A007913(x-1). - Benoit Cloitre, Apr 07 2002
Number of closed walks of length 2k on the grid graph P_2 X P_3. - Mitch Harris, Mar 06 2004
If x = A001109(n - 1), y = a(n) and z = x^2 + y, then x^4 + y^3 = z^2. - Bruno Berselli, Aug 24 2010
The product of any term a(n) with an even successor a(n + 2k) is always a square number. The product of any term a(n) with an odd successor a(n + 2k + 1) is always twice a square number. - Bradley Klee & Bill Gosper, Jul 22 2015
It appears that dividing even terms by two and taking the square root gives sequence A079496. - Bradley Klee, Jul 25 2015
The bisections of this sequence are a(2n - 1) = A055792(n) and a(2n) = A088920(n). - Bernard Schott, Apr 19 2020

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 193.
P. Tauvel, Exercices d'Algèbre Générale et d'Arithmétique, Dunod, 2004, Exercice 35 pages 346-347.

Colin Barker, Table of n, a(n) for n = 1..100
Dario Alpern, a^4+b^3=c^2.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 46, 56.
Phil Lafer, Discovering the square-triangular numbers, Fib. Quart., 9 (1971), 93-105.
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Kenneth Ramsey, Generalized Proof re Square Triangular Numbers, message 62 in Triangular_and_Fibonacci_Numbers Yahoo group, Oct 10, 2011.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A007913, A001541.
A001109(n-1) = sqrt{[(a(n))^2 - (a(n))]/2}.
a(n) = A001108(n-1)+1.
A001110(n-1)=a(n)*(a(n)-1)/2.
Cf. A001652, A001653, A046090.
Identical to A115599, but with additional leading term.

I:=[1,2,9]; [n le 3 select I[n] else 7*Self(n-1)-7*Self(n-2)+Self(n-3): n in [1..30]]; // Vincenzo Librandi, Mar 20 2015

A:= gfun:-rectoproc({a(n) = 6*a(n-1)-a(n-2)-2, a(1) = 1, a(2) = 2}, a(n), remember):
map(A,[$1..100]); # Robert Israel, Jul 22 2015

Table[ 1/4*(2 + (3 - 2*Sqrt[2])^k + (3 + 2*Sqrt[2])^k ) // Simplify, {k, 0, 20}] (* Jean-François Alcover, Mar 06 2013 *)
CoefficientList[Series[(1 - 5 x + 2 x^2) / ((1 - x) (1 - 6 x + x^2)), {x, 0, 40}], x] (* Vincenzo Librandi, Mar 20 2015 *)
(1 + ChebyshevT[#, 3])/2 & /@ Range[0, 20] (* Bill Gosper, Jul 20 2015 *)
a[1]=1;a[2]=2;a[n_]:=(a[n-1]+1)^2/a[n-2];a/@Range[25] (* Bradley Klee, Jul 25 2015 *)
LinearRecurrence[{7,-7,1},{1,2,9},30] (* Harvey P. Dale, Dec 06 2015 *)

Vec((1-5*x+2*x^2)/((1-x)*(1-6*x+x^2))+O(x^66)) /* Joerg Arndt, Mar 06 2013 */

t(n)=(1+sqrt(2))^(n-1);
for(k=1,24,print1(round((1/4)*(t(k)^2 + t(k)^(-2) + 2)),", ")) \\ Hugo Pfoertner, Nov 29 2019

a(n) = (1 + polchebyshev(n-1, 1, 3))/2; \\ Michel Marcus, Apr 21 2020

a(n) = 6*a(n - 1) - a(n - 2) - 2; n >= 3, a(1) = 1, a(2) = 2.
G.f.: x*(1 - 5*x + 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) - 1 + sqrt(2*a(n)*(a(n) - 1)) = A001652(n - 1). - Charlie Marion, Jul 21 2003; corrected by Michel Marcus, Apr 20 2020
a(n) = IF(mod(n; 2)=0; (((1 - sqrt(2))^n + (1 + sqrt(2))^n)/2)^2; 2*((((1 - sqrt(2))^(n + 1) + (1 + sqrt(2))^(n + 1)) - (((1 - sqrt(2))^n + (1 + sqrt(2))^n)))/4)^2). The odd-indexed terms are a(2n + 1) = [A001333(2n)]^2; the even-indexed terms are a(2n) = [A001333(2n - 1)]^2 + 1 = 2*[A001653(n)]^2. - Antonio Alberto Olivares, Jan 31 2004; corrected by Bernard Schott, Apr 20 2020
A053141(n + 1) + a(n + 1) = A001541(n + 1) + A001109(n + 1). - Creighton Dement, Sep 16 2004
a(n) = (1/2) + (1/4)*(3+2*sqrt(2))^(n-1) + (1/4)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Feb 21 2006; corrected by Michel Marcus, Apr 20 2020
a(n) = A001653(n)-A001652(n-1). - Charlie Marion, Apr 10 2006; corrected by Michel Marcus, Apr 20 2020
a(2k) = A001541(k)^2. - Alexander Adamchuk, Nov 24 2006
a(n) = 2*A001653(m)*A011900(n-m-1) +A002315(m)*A001652(n-m-1) - A001108(m) with mA001653(m)*A011900(m-n) - A002315(m)*A046090(m-n) - A001108(m). See Link to Generalized Proof re Square Triangular Numbers. - Kenneth J Ramsey, Oct 13 2011
a(n) = +7*a(n-1) -7*a(n-2) +1*a(n-3). - Joerg Arndt, Mar 06 2013
a(n) * a(n+2) = (A001108(n)-A001652(n)+3*A046090(n))^2. - Robert Israel, Jul 23 2015
sqrt(a(n+1)*a(n-1)) = a(n)+1 - Bradley Klee & Bill Gosper, Jul 25 2015
a(n) = 1 + sum{k=0..n-2} A002315(k). - David Pasino, Jul 09 2016; corrected by Michel Marcus, Apr 20 2020
E.g.f.: (2*exp(x) + exp((3-2*sqrt(2))*x) + exp((3+2*sqrt(2))*x))/4. - Ilya Gutkovskiy, Jul 09 2016
sqrt(a(n)*(a(n)-1)/2) = A001542(n)/2. - David Pasino, Jul 09 2016
Limit_{n -> infinity} a(n)/a(n-1) = A156035. - César Aguilera, Apr 07 2018
a(n) = (1/4)*(t^2 + t^(-2) + 2), where t = (1+sqrt(2))^(n-1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) + sqrt(a(n) - 1) = (1 + sqrt(2))^(n - 1). - Ridouane Oudra, Nov 29 2019
sqrt(a(n)) - sqrt(a(n) - 1) = (-1 + sqrt(2))^(n - 1). - Bernard Schott, Apr 18 2020

A118674 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 31)^2 = y^2.

Original entry on oeis.org

0, 9, 60, 93, 140, 429, 620, 893, 2576, 3689, 5280, 15089, 21576, 30849, 88020, 125829, 179876, 513093, 733460, 1048469, 2990600, 4274993, 6111000, 17430569, 24916560, 35617593, 101592876, 145224429, 207594620, 592126749, 846430076

Offset: 1

Mohamed Bouhamida, May 19 2006

Also values x of Pythagorean triples (x, x+31, y).
Corresponding values y of solutions (x, y) are in A157646.
For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a (prime) number in A066436 see A118673 or A129836.
lim_{n -> infinity} a(n)/a(n-3) = 3 + 2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (33 + 8*sqrt(2))/31 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (1539 + 850*sqrt(2))/31^2 for n mod 3 = 0.

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

cf. A157646, A066436 (primes of the form 2*n^2-1), A118673, A129836, A001652, A002193 (decimal expansion of sqrt(2)), A156035 (decimal expansion of 3 + 2*sqrt(2)), A157647 (decimal expansion of (33 + 8*sqrt(2))/31), A157648 (decimal expansion of (1539 + 850*sqrt(2))/31^2).

I:=[0,9,60,93,140,429,620]; [n le 7 select I[n] else Self(n-1) - 6*Self(n-3) - 6*Self(n-4) - Self(n-6) + Self(n-7): n in [1..50]]; // G. C. Greubel, Mar 31 2018

ClearAll[a]; Evaluate[Array[a, 6]] = {0, 9, 60, 93, 140, 429}; a[n_] := a[n] = 6*a[n-3] - a[n-6] + 62; Table[a[n], {n, 1, 31}] (* Jean-François Alcover, Dec 27 2011, after given formula *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,9,60,93,140,429,620}, 50] (* G. C. Greubel, Mar 31 2018 *)

{forstep(n=0, 850000000, [1, 3], if(issquare(2*n^2+62*n+961), print1(n, ",")))};

a(n) = 6*a(n-3) - a(n-6) + 62 for n > 6; a(1)=0, a(2)=9, a(3)=60, a(4)=93, a(5)=140, a(6)=429.
G.f.: x*(9 + 51*x + 33*x^2 - 7*x^3 - 17*x^4 - 7*x^5)/((1-x)*(1 - 6*x^3 + x^6)).
a(3*k + 1) = 31*A001652(k) for k >= 0.

Edited by Klaus Brockhaus, Mar 11 2009

A129836 Nonnegative values x of solutions (x, y) to the Diophantine equation x^2 + (x + 97)^2 = y^2.

Original entry on oeis.org

0, 15, 228, 291, 368, 1575, 1940, 2387, 9416, 11543, 14148, 55115, 67512, 82695, 321468, 393723, 482216, 1873887, 2295020, 2810795, 10922048, 13376591, 16382748, 63658595, 77964720, 95485887, 371029716, 454411923, 556532768

Offset: 1

Mohamed Bouhamida, May 21 2007

Also values x of Pythagorean triples (x, x + 97, y).
Corresponding values y of solutions (x, y) are in A157469.
For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a (prime) number in A066436, the x values are given by the sequence defined by a(n) = 6*a(n-3) - a(n-6) + 2p with a(1)=0, a(2) = 2m + 1, a(3) = 6m^2 - 10m + 4, a(4) = 3p, a(5) = 6m^2 + 10m + 4, a(6) = 40m^2 - 58m + 21 (cf. A118673).
Pairs (p, m) are (7, 2), (17, 3), (31, 4), (71, 6), (97, 7), (127, 8), (199, 10), (241, 11), (337, 13), (449, 15), (577, 17), (647, 18), (881, 21), (967, 22), ...
lim_{n -> infinity} a(n)/a(n-3) = 3 + 2*sqrt(2).
lim_{n -> infinity} a(n)/a(n-1) = (99 + 14*sqrt(2))/97 for n mod 3 = {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = (19491 + 12070*sqrt(2))/97^2 for n mod 3 = 0.
For the generic case x^2 + (x + p)^2 = y^2 with p = 2*m^2 - 1 a prime number in A066436, m>=2, Y values are given by the sequence defined by b(n) = 6*b(n-3) - b(n-6) with b(1) = p, b(2) = 2m^2 + 2m + 1, b(3) = 10m^2 - 14m + 5, b(4) = 5p, b(5) = 10m^2 + 14m + 5, b(6) = 58m^2 - 82m + 29. - Mohamed Bouhamida, Sep 09 2009

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

Cf. A157469, A066436 (primes of the form 2*n^2 - 1), A001652, A118673, A118674, A156035 (decimal expansion of 3 + 2*sqrt(2)), A157470 (decimal expansion of (99 + 14*sqrt(2))/97), A157471 (decimal expansion of (19491 + 12070*sqrt(2))/97^2).

m:=25; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(15+213*x+63*x^2-13*x^3-71*x^4-13*x^5)/((1-x)*(1-6*x^3 + x^6)))); // G. C. Greubel, May 07 2018

ClearAll[a]; Evaluate[Array[a, 6]] = {0, 15, 228, 291, 368, 1575}; a[n_] := a[n] = 6*a[n-3] - a[n-6] + 194; Table[a[n], {n, 1, 29}] (* Jean-François Alcover, Dec 27 2011, after given formula *)
LinearRecurrence[{1,0,6,-6,0,-1,1}, {0,15,228,291,368,1575,1940}, 50] (* G. C. Greubel, May 07 2018 *)

forstep(n=0, 600000000, [3, 1], if(issquare(2*n^2+194*n+9409), print1(n, ",")))

a(n) = 6*a(n-3) - a(n-6) + 194 for n > 6; a(1)=0, a(2)=15, a(3)=228, a(4)=291, a(5)=368, a(6)=1575.
G.f.: x*(15 + 213*x + 63*x^2 - 13*x^3 - 71*x^4 - 13*x^5)/((1-x)*(1 - 6*x^3 + x^6)).
a(3*k + 1) = 97*A001652(k) for k >= 0.

Edited and two terms added by Klaus Brockhaus, Mar 12 2009

cp's OEIS Frontend

A038761 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=9.

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A075528 Triangular numbers that are half other triangular numbers.

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A005319 a(n) = 6*a(n-1) - a(n-2).

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A008844 Squares of sequence A001653: y^2 such that x^2 - 2*y^2 = -1 for some x.

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A053142 a(n) = A053141(n)/2.

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Formula

A001921 a(n) = 14*a(n-1) - a(n-2) + 6 for n>1, a(0)=0, a(1)=7.

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