A001541 -id:A001541 - cp's OEIS frontend

A011900 a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.

1, 3, 15, 85, 493, 2871, 16731, 97513, 568345, 3312555, 19306983, 112529341, 655869061, 3822685023, 22280241075, 129858761425, 756872327473, 4411375203411, 25711378892991, 149856898154533, 873430010034205, 5090723162050695, 29670908962269963, 172934730611569081

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

Members of Diophantine pairs.
Solution to b*(b-1) = 2*a*(a-1) in natural numbers; a = a(n), b = b(n) = A046090(n).
Also the indices of centered octagonal numbers which are also centered square numbers. - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 4*x^2 - 8*y^2 - 4*x + 8*y = 0. - Colin Barker, Jan 01 2015
Also the number of perfect matchings on a triangular lattice of width 3 and length n. - Sergey Perepechko, Jul 11 2019

			G.f. = 1 + 3*x + 15x^2 + 85*x^3 + 493*x^4 + 2871*x^5 + 16731*x^6 + ... - _Michael Somos_, Feb 23 2019

Mario Velucchi "The Pell's equation ... an amusing application" in Mathematics and Informatics Quarterly, to appear 1997.

Colin Barker, Table of n, a(n) for n = 0..1000
H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
S. N. Perepechko, Number of perfect matchings on triangular lattices of fixed width, DIMA'2015 slides. [see: page 12]
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000194, A001109, A001541, A001653, A002024, A006062, A011906, A029549, A046090, A053141, A075528, A156035.

I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1) - Self(n-2) - 2: n in [1..40]]; // Vincenzo Librandi, Dec 05 2015

f:= gfun:-rectoproc({a(n)=6*a(n-1)-a(n-2)-2,a(0)=1,a(1)=3},a(n),remember):
seq(f(n),n=0..40); # Robert Israel, Dec 16 2015

a[0] = 1; a[1] = 3; a[n_]:= a[n]= 6 a[n-1] -a[n-2] -2; Table[a@ n, {n,0,40}] (* Michael De Vlieger, Dec 05 2015 *)
Table[(Fibonacci[2n + 1, 2] + 1)/2, {n, 0, 40}] (* Vladimir Reshetnikov, Sep 16 2016 *)
LinearRecurrence[{7,-7,1},{1,3,15},40] (* Harvey P. Dale, Feb 16 2017 *)
a[ n_] := (4 + ChebyshevT[n, 3] + ChebyshevT[n + 1, 3])/8; (* Michael Somos, Feb 23 2019 *)

Vec((1-4*x+x^2)/((1-x)*(1-6*x+x^2)) + O(x^50)) \\ Altug Alkan, Dec 06 2015

[(1+lucas_number1(2*n+1,2,-1))//2 for n in range(41)] # G. C. Greubel, Oct 17 2024

a(n) = (A001653(n+1) + 1)/2.
a(n) = (((1+sqrt(2))^(2*n-1) - (1-sqrt(2))^(2*n-1))/sqrt(8) + 1)/2.
a(n) = 7*(a(n-1) - a(n-2)) + a(n-3); a(1) = 1, a(2) = 3, a(3) = 15. Also a(n) = 1/2 + ( (1-sqrt(2))/(-4*sqrt(2)) )*(3-2*sqrt(2))^n + ( (1+sqrt(2))/(4*sqrt(2)) )*(3+2*sqrt(2))^n. - Antonio Alberto Olivares, Dec 23 2003
Sqrt(2) = Sum_{n>=0} 1/a(n); a(n) = a(n-1) + floor(1/(sqrt(2) - Sum_{k=0..n-1} 1/a(k))) (n>0) with a(0)=1. - Paul D. Hanna, Jan 25 2004
For n>k, a(n+k) = A001541(n)*A001653(k) - A053141(n-k-1); e.g., 493 = 99*5 - 2. For n<=k, a(n+k)=A001541(n)*A001653(k) - A053141(k-n); e.g., 85 = 3*29 - 2. - Charlie Marion, Oct 18 2004
a(n+1) = 3*a(n) - 1 + sqrt(8*a(n)^2 - 8*a(n) + 1), a(1)=1. - Richard Choulet, Sep 18 2007
a(n+1) = a(n) * (a(n) + 2) / a(n-1) for n>=1 with a(0)=1 and a(1)=3. - Paul D. Hanna, Apr 08 2012
G.f.: (1 - 4*x + x^2)/((1-x)*(1 - 6*x + x^2)). - R. J. Mathar, Oct 26 2009
Sum_{k=a(n)..A001109(n+1)} k = a(n)*A001109(n+1) = A011906(n+1). Example n=2, 3+4+5+6=18, 3*6=18. - Paul Cleary, Dec 05 2015
a(n) = (sqrt(1+8*A001109(n+1)^2)+1)/2 - A001109(n+1). - Robert Israel, Dec 16 2015
a(n) = a(-1-n) for all n in Z. - Michael Somos, Feb 23 2019
E.g.f.: (2*exp(x) + exp(3*x)*(2*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/4. - Stefano Spezia, Mar 16 2024
a(n) = A053141(n) + 1 = A000194(A029549(n))+1 = A002024(A075528(n))+1. - Pontus von Brömssen, Sep 11 2024

More terms and comments from Wolfdieter Lang

A204502 Numbers such that floor[a(n)^2 / 9] is a square.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99, 102, 105, 108, 111, 114, 117, 120, 123, 126, 129, 132, 135, 138, 141, 144, 147, 150, 153, 156, 159, 162, 165, 168, 171, 174, 177

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

Or, numbers n such that n^2, with its last base-9 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 9.)

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.

The squares are in A204503, the squares with last base-9 digit dropped in A204504, and the square roots of the latter in A028310.

Cf. A031149=sqrt(A023110) (base 10), A204514=sqrt(A055872) (base 8), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Select[Range[0,200],IntegerQ[Sqrt[Floor[#^2/9]]]&] (* Harvey P. Dale, May 05 2018 *)

b=9;for(n=0,200,issquare(n^2\b) & print1(n","))

Conjecture: a(n) = 3*n-12 for n>5. G.f.: x^2*(x^2+x+1)*(x^3-x+1)/(x-1)^2. [Colin Barker, Nov 23 2012]

A204514 Numbers such that floor(a(n)^2 / 8) is again a square.

Original entry on oeis.org

0, 1, 2, 3, 6, 17, 34, 99, 198, 577, 1154, 3363, 6726, 19601, 39202, 114243, 228486, 665857, 1331714, 3880899, 7761798, 22619537, 45239074, 131836323, 263672646, 768398401, 1536796802, 4478554083, 8957108166, 26102926097, 52205852194, 152139002499, 304278004998, 886731088897

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-8 digit dropped, is again a square. (Except maybe for the 3 initial terms whose square has only 1 digit in base 8.)

See A204504 for the squares resulting from truncation of a(n)^2, and A204512 for their square roots. - M. F. Hasler, Sep 28 2014

M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Cf. A031149=sqrt(A023110) (base 10), A204502=sqrt(A204503) (base 9), A204516=sqrt(A055859) (base 7), A204518=sqrt(A055851) (base 6), A204520=sqrt(A055812) (base 5), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

Cf. A003499, A055872, A204504, A204512.

A204514 := proc(n) coeftayl((x^2+2*x^3-3*x^4-6*x^5)/(1-6*x^2+x^4), x=0, n); end proc: seq(A204514(n), n=1..30); # Wesley Ivan Hurt, Sep 28 2014

CoefficientList[Series[(x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(x (1 - 6*x^2 + x^4)), {x, 0, 30}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
LinearRecurrence[{0,6,0,-1},{0,1,2,3,6},40] (* Harvey P. Dale, Nov 23 2022 *)

b=8;for(n=0,1e7,issquare(n^2\b) & print1(n","))

A204514(n)=polcoeff((x + 2*x^2 - 3*x^3 - 6*x^4)/(1 - 6*x^2 + x^4+O(x^(n+!n))),n-1,x)

G.f. = (x^2 + 2*x^3 - 3*x^4 - 6*x^5)/(1 - 6*x^2 + x^4).
a(n) = sqrt(A055872(n)). - M. F. Hasler, Sep 28 2014
a(2n) = A001541(n-1). a(2n+1) = A003499(n-1). - R. J. Mathar, Feb 05 2020

A204516 Numbers such that floor(a(n)^2 / 7) is a square.

Original entry on oeis.org

0, 1, 2, 3, 8, 16, 45, 127, 254, 717, 2024, 4048, 11427, 32257, 64514, 182115, 514088, 1028176, 2902413, 8193151, 16386302, 46256493, 130576328, 261152656, 737201475, 2081028097, 4162056194, 11748967107, 33165873224, 66331746448

Offset: 1

M. F. Hasler, Jan 15 2012

Or: Numbers whose square, with its last base-7 digit dropped, is again a square (where for the first 3 terms, dropping the digit is meant to yield zero).

Harvey P. Dale, Table of n, a(n) for n = 1..1000
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (0,0,16,0,0,-1).

Cf. A031149 (base 10), A204502 (base 9), A204514 (base 8), A204518 (base 6), A004275=sqrt(A055808) (base 4), A001075=sqrt(A055793) (base 3), A001541=sqrt(A055792) (base 2).

LinearRecurrence[{0,0,16,0,0,-1},{0,1,2,3,8,16,45},30] (* or *) CoefficientList[Series[ (x+2x^2+3x^3-8x^4-16x^5-3x^6)/(1-16x^3+x^6),{x,0,30}],x] (* Harvey P. Dale, Apr 22 2023 *)

b=7;for(n=0,2e9,issquare(n^2\b) & print1(n","))

G.f. = (x + 2*x^2 + 3*x^3 - 8*x^4 - 16*x^5 - 3*x^6 )/(1 - 16*x^3 + x^6).

floor(a(n)^2 / 7) = A204517(n)^2.

A204520 Numbers such that floor(a(n)^2 / 5) is a square.

Original entry on oeis.org

0, 1, 2, 3, 7, 9, 18, 47, 123, 161, 322, 843, 2207, 2889, 5778, 15127, 39603, 51841, 103682, 271443, 710647, 930249, 1860498, 4870847, 12752043, 16692641, 33385282

Offset: 1

M. F. Hasler, Jan 15 2012

nonn

Also: Numbers whose square, with its last base-5 digit dropped, is again a square. (For the three initial terms whose squares have only one digit in base 5, it is then understood that this yields zero.)

Cf. A031149, A055812, A204502, A204514, A204516, A204518 and A004275, A001075, A001541 for the analog in bases 10,...,6 and 4, 3, 2.

Select[Range[0,5*10^6],IntegerQ[Sqrt[Floor[#^2/5]]]&] (* The program generates the first 24 terms of the sequence. *) (* Harvey P. Dale, Jul 15 2025 *)

b=5;for(n=0,2e9,issquare(n^2\b) && print1(n","))

a(n) = sqrt(A055812(n)).

Empirical g.f.: -x^2*(x+1)*(3*x^6 + 4*x^5 + 14*x^4 - 5*x^3 - 2*x^2 - x-1) / ((x^4 - 4*x^2 - 1)*(x^4 + 4*x^2 - 1)). - Colin Barker, Sep 15 2014

A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).

Original entry on oeis.org

1, 2, 7, 12, 41, 70, 239, 408, 1393, 2378, 8119, 13860, 47321, 80782, 275807, 470832, 1607521, 2744210, 9369319, 15994428, 54608393, 93222358, 318281039, 543339720, 1855077841, 3166815962, 10812186007, 18457556052, 63018038201, 107578520350

Offset: 1

Benoit Cloitre, May 10 2003

The upper principal and intermediate convergents to 2^(1/2), beginning with 2/1, 3/2, 10/7, 17/12, 58/41, form a strictly decreasing sequence; essentially, numerators=A143609 and denominators=A084068. - Clark Kimberling, Aug 27 2008
From Peter Bala, Mar 23 2018: (Start)
Define a binary operation o on the real numbers by x o y = x*sqrt(1 + y^2) + y*sqrt(1 + x^2). The operation o is commutative and associative with identity 0. We have
  a(2*n + 1) = 1 o 1 o ... o 1 (2*n + 1 terms) and
  a(2*n) = (1/sqrt(2))*(1 o 1 o ... o 1) (2*n terms). Cf. A049629, A108412 and A143608.
This is a fourth-order divisibility sequence. Indeed, a(2*n) = U(2*n)/sqrt(2) and a(2*n+1) = U(2*n+1), where U(n) is the Lehmer sequence [Lehmer, 1930] defined by the recurrence U(n) = 2*sqrt(2)*U(n-1) - U(n-2) with U(0) = 0 and U(1) = 1. The solution to the recurrence is U(n) = (1/2)*( (sqrt(2) + 1)^n - (sqrt(2) - 1)^n ).
It appears that this sequence consists of those numbers m such that 2*m^2 = floor( m*sqrt(2) * ceiling(m*sqrt(2)) ). Cf. A084069. (End)
Conjecture: a(n) is the earliest occurrence of n in A348295, which is to say, a(n) is the least m such that Sum_{k=1..m} (-1)^(floor(k*(sqrt(2)-1))) = Sum_{k=1..m} (-1)^A097508(k) = n. This has been confirmed for the first 32 terms by Chai Wah Wu, Oct 21 2021. - Jianing Song, Jul 16 2022

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.

Indranil Ghosh, Table of n, a(n) for n = 1..2608
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
D. H. Lehmer, An extended theory of Lucas' functions, Annals of Mathematics, Second Series, Vol. 31, No. 3 (Jul., 1930), pp. 419-448.
Eric Weisstein's World of Mathematics, Lehmer Number
Index entries for linear recurrences with constant coefficients, signature (0,6,0,-1).

Bisections are A001542 and A002315.

Cf. A084069, A084070, A133566, A079496, A001541, A001653, A008844, A055792, A049629, A108412, A143608.

a := proc (n) if `mod`(n, 2) = 1 then (1/2)*(sqrt(2) + 1)^n - (1/2)*(sqrt(2) - 1)^n else (1/2)*((sqrt(2) + 1)^n - (sqrt(2) - 1)^n)/sqrt(2) end if;
end proc:
seq(simplify(a(n)), n = 1..30); # Peter Bala, Mar 25 2018

a[n_] := ((Sqrt[2]+1)^n - (Sqrt[2]-1)^n) ((-1)^n(Sqrt[2]-2) + (Sqrt[2]+2))/8;
Table[Simplify[a[n]], {n, 30}] (* after Paul Barry, Peter Luschny, Mar 29 2018 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; -1,0,6,0]^(n-1)*[1;2;7;12])[1,1] \\ Charles R Greathouse IV, Jun 20 2015

"A Diofloortin equation": n such that 2*n^2=floor(n*sqrt(2)*ceiling(n*sqrt(2))).
a(n)*a(n+3) = -2 + a(n+1)*a(n+2).
From Paul Barry, Jun 06 2006: (Start)
G.f.: x*(1+x)^2/(1-6*x^2+x^4);
a(n) = ((sqrt(2)+1)^n-(sqrt(2)-1)^n)*((sqrt(2)/8-1/4)*(-1)^n+sqrt(2)/8+1/4);
a(n) = Sum_{k=0..floor(n/2)} 2^k*(C(n,2*k)-C(n-1,2*k+1)*(1+(-1)^n)/2). (End)
A000129(n+1) = A079496(n) + a(n). - Gary W. Adamson, Sep 18 2007
Equals A133566 * A000129, where A000129 = the Pell sequence. - Gary W. Adamson, Sep 18 2007
From Peter Bala, Mar 23 2018: (Start)
a(2*n + 2) = a(2*n + 1) + sqrt( (1 + a(2*n + 1)^2)/2 ).
a(2*n + 1) = 2*a(2*n) + sqrt( (1 + 2*a(2*n)^2) ).
More generally,
a(2*n+2*m+1) = sqrt(2)*a(2*n) o a(2*m+1), where o is the binary operation defined above, that is,
a(2*n+2*m+1) = sqrt(2)*a(2*n)*sqrt(1 + a(2*m+1)^2) + a(2*m+1)*sqrt(1 + 2*a(2*n)^2).
sqrt(2)*a(2*(n + m)) = (sqrt(2)*a(2*n)) o (sqrt(2)*a(2*m)), that is,
a(2*n+2*m) = a(2*n)*sqrt(1 + 2*a(2*m)^2) + a(2*m)*sqrt(1 + 2*a(2*n)^2).
sqrt(1 + 2*a(2*n)^2) = A001541(n).
1 + 2*a(2*n)^2 = A055792(n+1).
a(2*n) - a(2*n-1) = A001653(n).
(1 + a(2*n+1)^2)/2 = A008844(n). (End)
a(n) = A000129(n) for even n and A001333(n) for odd n. - R. J. Mathar, Oct 15 2021

A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 23, 134, 781, 4552, 26531, 154634, 901273, 5253004, 30616751, 178447502, 1040068261, 6061962064, 35331704123, 205928262674, 1200237871921, 6995498968852, 40772755941191, 237641036678294, 1385073464128573, 8072799748093144, 47051725024430291, 274237550398488602

Offset: 0

Barry E. Williams, May 02 2000

This sequence gives one half of all positive solutions y = y1 = a(n) of the first class of the Pell equation x^2 - 2*y^2 = -7. For the corresponding x=x1 terms see A054490(n). Therefore it also gives one fourth of all positive solutions x = x1 of the first class of the Pell equation x^2 - 2*y^2 = 14, with the y=y1 terms given by A054490. - Wolfdieter Lang, Feb 26 2015

			n = 2: A054490(2)^2 - 2*(2*a(2))^2 =
       65^2 - 2*(2*23)^2 = -7,
      (4*a(2))^2 - 2*A054490(2)^2 =
      (4*23)^2 - 2*65^2 = 14. - _Wolfdieter Lang_, Feb 26 2015
a(2) = (A253811(1) + A101386(1))/2 = (19 + 27)/2 = 23. - _Wolfdieter Lang_, Mar 19 2015

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

Stefano Spezia, Table of n, a(n) for n = 0..1300
I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), pp. 181-193.
Seyed Hassan Alavi, Ashraf Daneshkhah, and Cheryl E Praeger, Symmetries of biplanes, arXiv:2004.04535 [math.GR], 2020. See Lemma 7.9 p. 21.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 55, 56.
E. I. Emerson, Recurrent Sequences in the Equation DQ^2=R^2+N, Fib. Quart., 7 (1969), pp. 231-242.
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, A001653, A054490.

A038725(n) = a(-n).

a[0]:=1: a[1]:=4: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006

LinearRecurrence[{6,-1},{1,4},30] (* Harvey P. Dale, Aug 06 2020 *)

{a(n) = real((3 + 2*quadgen(8))^n * (1 + quadgen(8) / 4))} /* Michael Somos, Sep 28 2008 */

{a(n) = polchebyshev(n, 1, 3) + polchebyshev(n-1, 2, 3)} /* Michael Somos, Sep 28 2008 */

a(n) = ((4+sqrt(2))/8)*(3+2*sqrt(2))^(n-1) + ((4-sqrt(2))/8)*(3-2*sqrt(2))^(n-1). - Antonio Alberto Olivares, Mar 29 2008
a(n) = A001653(n+1) - A001109(n). - Antonio Alberto Olivares, Mar 29 2008
Sequence satisfies -7 = f(a(n), a(n+1)) where f(u, v) = u^2 + v^2 - 6*u*v. - Michael Somos, Sep 28 2008
G.f.: (1 - 2*x) / (1 - 6*x + x^2). a(n) = (7 + a(n-1)^2) / a(n-2). - Michael Somos, Sep 28 2008
a(n) = Sum_{k = 0..n} A238731(n,k)*3^k. - Philippe Deléham, Mar 05 2014
a(n) = S(n,6) - 2*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). See the g.f. and the Pell comment above. - Wolfdieter Lang, Feb 26 2015
a(0) = -(A038761(0) - A038762(0))/2, a(n) = (A253811(n-1) + A101386(n-1))/2, n >= 1. See the Mar 19 2015 comment on A054490. - Wolfdieter Lang, Mar 19 2015
E.g.f.: exp(3*x)*(4*cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x))/4. - Stefano Spezia, Apr 30 2020

More terms from James Sellers, May 03 2000

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

A072221 a(n) = 6*a(n-1) - a(n-2) + 2, with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 25, 148, 865, 5044, 29401, 171364, 998785, 5821348, 33929305, 197754484, 1152597601, 6717831124, 39154389145, 228208503748, 1330096633345, 7752371296324, 45184131144601, 263352415571284, 1534930362283105, 8946229758127348, 52142448186480985

Offset: 0

Lekraj Beedassy, Jul 04 2002

The product of three consecutive triangular numbers with middle term A000217(m) where m is in this sequence is a square.
k is in this sequence iff the triangle with sides 3,k,k+1 has integer area. Equivalently, numbers k such that 2*(k+2)*(k-1) is a square. - James R. Buddenhagen, Oct 19 2008
Triangular numbers that are equal to a square plus one have this sequence as indices. For example, the 25th triangular number is 25*26/2 = 325 = 18^2 + 1. - Tanya Khovanova and Alexey Radul, Aug 08 2009
The triangle with sides 3, a(n), a(n)+1 has area A075848(n) if n>=0. - Michael Somos, Dec 25 2018
Compare with A016064 for integers m with triangles with sides 4, m, m+2 and integer area. - Michael Somos, May 11 2019

			From _Michael Somos_, Dec 25 2018: (Start)
For n=1, the triangle (3, 4, 5) has area 6 = A075848(1).
For n=2, the triangle (3, 25, 26) has area 36 = A075848(2). (End)

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Sylvester Robins, Certain Series of Integral, Rational, Scalene Triangles, The American Mathematical Monthly, Vol. 1, No. 1 (Jan., 1894), pp. 13-14. See Example I.
Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

Cf. A000217, A001108, A001541, A016064, A075848.

a072221 n = a072221_list !! n
a072221_list = 1 : 4 : (map (+ 2) $
   zipWith (-) (map (* 6) $ tail a072221_list) a072221_list)
-- Reinhard Zumkeller, Apr 27 2012

a[n_] := a[n] = 6a[n - 1] - a[n - 2] + 2; a[0] = 1; a[1] = 4; Table[ a[n], {n, 0, 20}]
LinearRecurrence[{7, -7, 1}, {1, 4, 25}, 25] (* T. D. Noe, Dec 09 2013 *)
a[ n_] := (3 ChebyshevT[ n, 3] - 1) / 2; (* Michael Somos, Dec 25 2018 *)

{a(n) = (3 * polchebyshev( n, 1, 3) - 1) / 2}; /* Michael Somos, Dec 25 2018 */

a(n) = (3*A001541(n) - 1)/2.
a(n) = 3*A001108(n) + 1. - David Scheers, Dec 25 2006
From Franz Vrabec, Aug 21 2006: (Start)
a(n) = -1/2 + (3/4)*((3+sqrt(8))^n + (3-sqrt(8))^n) for n >= 0.
a(n) = floor((3/4)*(3+sqrt(8))^n) for n > 0. (End)
G.f.: (1-3x+4x^2)/((1-x)(1-6x+x^2)). - R. J. Mathar, Sep 09 2008
a(n) = a(-n) for all n in Z. - Michael Somos, Dec 25 2018

Edited by Robert G. Wilson v, Jul 08 2002

A077444 Numbers k such that (k^2 + 4)/2 is a square.

Original entry on oeis.org

2, 14, 82, 478, 2786, 16238, 94642, 551614, 3215042, 18738638, 109216786, 636562078, 3710155682, 21624372014, 126036076402, 734592086398, 4281516441986, 24954506565518, 145445522951122, 847718631141214, 4940866263896162, 28797478952235758, 167844007449518386

Offset: 1

Gregory V. Richardson, Nov 09 2002

The equation "(k^2 + 4)/2 is a square" is a version of the generalized Pell Equation x^2 - D*y^2 = C where x^2 - 2*y^2 = -4.
Sequence of all positive integers k such that continued fraction [k,k,k,k,k,k,...] belongs to Q(sqrt(2)). - Thomas Baruchel, Sep 15 2003
Equivalently, 2*n^2 + 8 is a square.
Numbers n such that (ceiling(sqrt(n*n/2)))^2 = 2 + n^2/2. - Ctibor O. Zizka, Nov 09 2009
The continued fraction [a(n);a(n),a(n),...] = (1 + sqrt(2))^(2*n-1). - Thomas Ordowski, Jun 07 2013
a((p+1)/2) == 2 (mod p) where p is an odd prime. - Altug Alkan, Mar 17 2016

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.
L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
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.

Amiram Eldar, Table of n, a(n) for n = 1..1306
Sergio Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics, 2014, 5, 2226-2234.
Tanya Khovanova, Recursive Sequences
Prabha Sivaraman Nair and Rejikumar Karunakaran, On k-Fibonacci Brousseau Sums, J. Int. Seq. (2024) Art. No. 24.6.4. See p. 3.
J. J. O'Connor and E. F. Robertson, Pell's Equation. [Broken link]
Eric Weisstein's World of Mathematics, Pell Equation.
Eric Weisstein's World of Mathematics, NSW Number.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

(A077445(n))^2 - 2*a(n) = 8.
First differences of A001541.
Pairwise sums of A001542.
Bisection of A002203 and A080039.
Cf. A001653.

[n: n in [0..10^8] | IsSquare((n^2 + 4) div 2)]; // Vincenzo Librandi, Jun 20 2015

LinearRecurrence[{6,-1},{2,14},30] (* Harvey P. Dale, Jul 25 2018 *)

for(n=1,20,q=(1+sqrt(2))^(2*n-1);print1(contfrac(q)[1],", ")) \\ Derek Orr, Jun 18 2015

Vec(2*x*(1+x)/(1-6*x+x^2) + O(x^100)) \\ Altug Alkan, Mar 17 2016

a(n) = (((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n) + ((3 + 2*sqrt(2))^(n-1) - (3 - 2*sqrt(2))^(n-1))) / (2*sqrt(2)).
a(n) = 2*A002315(n-1).
Recurrence: a(n) = 6*a(n-1) - a(n-2), starting 2, 14.
Offset 0, with a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = a^((2n+1)/2) - b^((2n+1)/2). a(n) = 2*(A001109(n+1) + A001109(n)) = (A003499(n+1) - A003499(n))/2 = 2*sqrt(A001108(2n+1)) = sqrt(A003499(2n+1)-2). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
Limit_{n->oo} a(n)/a(n-1) = 5.82842712474619009760... = 3 + 2*sqrt(2). See A156035.
From R. J. Mathar, Nov 16 2007: (Start)
G.f.: 2*x*(1+x)/(1-6*x+x^2).
a(n) = 2*(7*A001109(n) - A001109(n+1)). (End)
a(n) = (1+sqrt(2))^(2*n-1) - (1+sqrt(2))^(1-2*n). - Gerson Washiski Barbosa, Sep 19 2010
a(n) = floor((1 + sqrt(2))^(2*n-1)). - Thomas Ordowski, Jun 07 2013
a(n) = sqrt(2*A075870(n)^2-4). - Derek Orr, Jun 18 2015
a(n) = 2*sqrt((2*A001653(n)^2)-1). - César Aguilera, Jul 13 2023
E.g.f.: 2*(1 + exp(3*x)*(sqrt(2)*sinh(2*sqrt(2)*x) - cosh(2*sqrt(2)*x))). - Stefano Spezia, Aug 27 2024

cp's OEIS Frontend

A011900 a(n) = 6*a(n-1) - a(n-2) - 2 with a(0) = 1, a(1) = 3.

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A204502 Numbers such that floor[a(n)^2 / 9] is a square.

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A204514 Numbers such that floor(a(n)^2 / 8) is again a square.

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A204516 Numbers such that floor(a(n)^2 / 7) is a square.

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A204520 Numbers such that floor(a(n)^2 / 5) is a square.

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A084068 a(1) = 1, a(2) = 2; a(2*k) = 2*a(2*k-1) - a(2*k-2), a(2*k+1) = 4*a(2*k) - a(2*k-1).

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A038723 a(n) = 6*a(n-1) - a(n-2), n >= 2, a(0)=1, a(1)=4.

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A084068 a(1) = 1, a(2) = 2; a(2k) = 2a(2k-1) - a(2k-2), a(2k+1) = 4a(2k) - a(2k-1).