A001075 -id:A001075 - cp's OEIS frontend

A001570 Numbers k such that k^2 is centered hexagonal.

1, 13, 181, 2521, 35113, 489061, 6811741, 94875313, 1321442641, 18405321661, 256353060613, 3570537526921, 49731172316281, 692665874901013, 9647591076297901, 134373609193269601, 1871582937629476513, 26067787517619401581, 363077442309042145621

Offset: 1

N. J. A. Sloane

Chebyshev T-sequence with Diophantine property. - Wolfdieter Lang, Nov 29 2002
a(n) = L(n,14), where L is defined as in A108299; see also A028230 for L(n,-14). - Reinhard Zumkeller, Jun 01 2005
Numbers x satisfying x^2 + y^3 = (y+1)^3. Corresponding y given by A001921(n)={A028230(n)-1}/2. - Lekraj Beedassy, Jul 21 2006
Mod[ a(n), 12 ] = 1. (a(n) - 1)/12 = A076139(n) = Triangular numbers that are one-third of another triangular number. (a(n) - 1)/4 = A076140(n) = Triangular numbers T(k) that are three times another triangular number. - Alexander Adamchuk, Apr 06 2007
Also numbers n such that RootMeanSquare(1,3,...,2*n-1) is an integer. - Ctibor O. Zizka, Sep 04 2008
a(n), with n>1, is the length of the cevian of equilateral triangle whose side length is the term b(n) of the sequence A028230. This cevian divides the side (2*x+1) of the triangle in two integer segments x and x+1. - Giacomo Fecondo, Oct 09 2010
For n>=2, a(n) equals the permanent of the (2n-2)X(2n-2) tridiagonal matrix with sqrt(12)'s along the main diagonal, and 1's along the superdiagonal and the subdiagonal. - John M. Campbell, Jul 08 2011
Beal's conjecture would imply that set intersection of this sequence with the perfect powers (A001597) equals {1}. In other words, existence of a nontrivial perfect power in this sequence would disprove Beal's conjecture. - Max Alekseyev, Mar 15 2015
Numbers n such that there exists positive x with x^2 + x + 1 = 3n^2. - Jeffrey Shallit, Dec 11 2017
Given by the denominators of the continued fractions [1,(1,2)^i,3,(1,2)^{i-1},1]. - Jeffrey Shallit, Dec 11 2017
A near-isosceles integer-sided triangle with an angle of 2*Pi/3 is a triangle whose sides (a, a+1, c) satisfy Diophantine equation (a+1)^3 - a^3 = c^2. For n >= 2, the largest side c is given by a(n) while smallest and middle sides (a, a+1) = (A001921(n-1), A001922(n-1)) (see Julia link). - Bernard Schott, Nov 20 2022

			G.f. = x + 13*x^2 + 181*x^3 + 2521*x^4 + 35113*x^5 + 489061*x^6 + 6811741*x^7 + ...

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).

G. C. Greubel, Table of n, a(n) for n = 1..870 (terms 1..101 from T. D. Noe)
Alex Fink, Richard K. Guy, and Mark Krusemeyer, Partitions with parts occurring at most thrice, Contributions to Discrete Mathematics, Vol 3, No 2 (2008), pp. 76-114. See Section 13.
G. Julia, Triangles dont un angle mesure 120 degrés, Problème Capes, part C (in French).
Tanya Khovanova, Recursive Sequences
J.-C. Novelli and J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962 [math.CO], 2014.
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.
Sociedad Magic Penny Patagonia, Leonardo en Patagonia
V. Thebault, Consecutive cubes with difference a square, Amer. Math. Monthly, 56 (1949), 174-175.
Eric Weisstein's World of Mathematics, Hex Number
Wikipedia, Beal's conjecture
Index entries for sequences related to Chebyshev polynomials.
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (14,-1).

Bisection of A003500/4. Cf. A006051, A001921, A001922.
One half of odd part of bisection of A001075. First differences of A007655.
Cf. A077417 with companion A077416.
Row 14 of array A094954.
Cf. A076139, A076140, A102871.
A122571 is another version of the same sequence.
Row 2 of array A188646.
Cf. similar sequences listed in A238379.
Cf. A028231, which gives the corresponding values of x in 3n^2 = x^2 + x + 1.
Similar sequences of the type cosh((2*m+1)*arccosh(k))/k are listed in A302329. This is the case k=2.

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

A001570:=-(-1+z)/(1-14*z+z**2); # Simon Plouffe in his 1992 dissertation.

NestList[3 + 7*#1 + 4*Sqrt[1 + 3*#1 + 3*#1^2] &, 0, 24] (* Zak Seidov, May 06 2007 *)
f[n_] := Simplify[(2 + Sqrt@3)^(2 n - 1) + (2 - Sqrt@3)^(2 n - 1)]/4; Array[f, 19] (* Robert G. Wilson v, Oct 28 2010 *)
a[c_, n_] := Module[{},
   p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
   d := Denominator[Convergents[Sqrt[c], n p]];
   t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
   Return[t];
  ] (* Complement of A041017 *)
a[12, 20] (* Gerry Martens, Jun 07 2015 *)
LinearRecurrence[{14, -1}, {1, 13}, 19] (* Jean-François Alcover, Sep 26 2017 *)
CoefficientList[Series[x (1-x)/(1-14x+x^2),{x,0,20}],x] (* Harvey P. Dale, Sep 18 2024 *)

{a(n) = real( (2 + quadgen( 12)) ^ (2*n - 1)) / 2}; /* Michael Somos, Feb 15 2011 */

a(n) = ((2 + sqrt(3))^(2*n - 1) + (2 - sqrt(3))^(2*n - 1)) / 4. - Michael Somos, Feb 15 2011
G.f.: x * (1 - x) / (1 -14*x + x^2). - Michael Somos, Feb 15 2011
Let q(n, x) = Sum_{i=0, n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 12). - Benoit Cloitre, Dec 10 2002
a(n) = S(n, 14) - S(n-1, 14) = T(2*n+1, 2)/2 with S(n, x) := U(n, x/2), resp. T(n, x), Chebyshev's polynomials of the second, resp. first, kind. See A049310 and A053120. S(-1, x)=0, S(n, 14)=A007655(n+1) and T(n, 2)=A001075(n). - Wolfdieter Lang, Nov 29 2002
a(n) = A001075(n)*A001075(n+1) - 1 and thus (a(n)+1)^6 has divisors A001075(n)^6 and A001075(n+1)^6 congruent to -1 modulo a(n) (cf. A350916). - Max Alekseyev, Jan 23 2022
4*a(n)^2 - 3*b(n)^2 = 1 with b(n)=A028230(n+1), n>=0.
a(n)*a(n+3) = 168 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
a(n) = 14*a(n-1) - a(n-2), a(0) = a(1) = 1. a(1 - n) = a(n) (compare A122571).
a(n) = 12*A076139(n) + 1 = 4*A076140(n) + 1. - Alexander Adamchuk, Apr 06 2007
a(n) = (1/12)*((7-4*sqrt(3))^n*(3-2*sqrt(3))+(3+2*sqrt(3))*(7+4*sqrt(3))^n -6). - Zak Seidov, May 06 2007
a(n) = A102871(n)^2+(A102871(n)-1)^2; sum of consecutive squares. E.g. a(4)=36^2+35^2. - Mason Withers (mwithers(AT)semprautilities.com), Jan 26 2008
a(n) = sqrt((3*A028230(n+1)^2 + 1)/4).
a(n) = A098301(n+1) - A001353(n)*A001835(n).
a(n) = A000217(A001571(n-1)) + A000217(A133161(n)), n>=1. - Ivan N. Ianakiev, Sep 24 2013
a(n)^2 = A001922(n-1)^3 - A001921(n-1)^3, for n >= 1. - Bernard Schott, Nov 20 2022
a(n) = 2^(2*n-3)*Product_{k=1..2*n-1} (2 - sin(2*Pi*k/(2*n-1))). Michael Somos, Dec 18 2022
a(n) = A003154(A101265(n)). - Andrea Pinos, Dec 19 2022

A092184 Sequence S_6 of the S_r family.

Original entry on oeis.org

0, 1, 6, 25, 96, 361, 1350, 5041, 18816, 70225, 262086, 978121, 3650400, 13623481, 50843526, 189750625, 708158976, 2642885281, 9863382150, 36810643321, 137379191136, 512706121225, 1913445293766, 7141075053841, 26650854921600, 99462344632561, 371198523608646

Offset: 0

Rainer Rosenthal, Apr 03 2004

The r-family of sequences is S_r(n) = 2*(T(n,(r-2)/2) - 1)/(r-4) provided r is not equal to 4 and S_4(n) = n^2 = A000290(n). Here T(n,x) are Chebyshev's polynomials of the first kind. See their coefficient triangle A053120. See also the R. Stephan link for the explicit formula for s_k(n) for k not equal to 4 (Stephan's s_k(n) is identical with S_r(n)).
An integer n is in this sequence iff mutually externally tangent circles with radii n, n+1, n+2 have Soddy circles (i.e., circles tangent to all three) of rational radius. - James R. Buddenhagen, Nov 16 2005
This sequence is a divisibility sequence, i.e., a(n) divides a(m) whenever n divides m. It is the case P1 = 6, P2 = 8, Q = 1 of the 3-parameter family of 4th-order linear divisibility sequences found by Williams and Guy. - Peter Bala, Mar 25 2014
a(n) is the block size of the (n-1)-th design in a sequence of multi-set designs with 2 blocks, see A335649. - John P. McSorley, Jun 22 2020

			a(3)=25 because a(1)=1 and a(2)=6 and a(1)*a(3) = 1*25 = (6-1)^2 = (a(2)-1)^2.

Hugo Pfoertner, Table of n, a(n) for n = 0..250
Marco Abrate, Stefano Barbero, Umberto Cerruti, Nadir Murru, Polynomial sequences on quadratic curves, Integers, Vol. 15, 2015, #A38.
Peter Bala, Linear divisibility sequences and Chebyshev polynomials
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
R. Stephan, Boring proof of a nonlinearity
H. C. Williams and R. K. Guy, Some fourth-order linear divisibility sequences, Intl. J. Number Theory 7 (5) (2011) 1255-1277.
H. C. Williams and R. K. Guy, Some Monoapparitic Fourth Order Linear Divisibility Sequences Integers, Volume 12A (2012) The John Selfridge Memorial Volume.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

See A001110=S_36 for further references to S_r sequences.

Other members of this r-family are: A007877 (r=2), |A078070| (r=3), A004146 (r=5), A054493 (r=7). A098306, A100047. A001353, A001834. A001350, A052530.

[Floor(1/2*(-2+(2+Sqrt(3))^n+(2-Sqrt(3))^n)): n in [0..30]]; // Vincenzo Librandi, Oct 14 2015

A092184 := proc(n)
    option remember;
    if n <= 1 then
        n;
    else
        4*procname(n-1)-procname(n-2)+2 ;
    end if ;
end proc:
seq(A092184(n),n=0..10) ;# Zerinvary Lajos, Mar 09 2008

Table[Simplify[ -((2 + Sqrt[3])^n - 1)*((2 - Sqrt[3])^n - 1)]/2, {n, 0, 26}] (* Stefan Steinerberger, May 15 2007 *)
LinearRecurrence[{5,-5,1},{0,1,6},27] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[x (1 + x)/(1 - 5 x + 5 x^2 - x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 14 2015 *)

Vec(x*(1+x)/(1 - 5*x + 5*x^2 - x^3) + O(x^50)) \\ Michel Marcus, Oct 14 2015

S_r type sequences are defined by a(0)=0, a(1)=1, a(2)=r and a(n-1)*a(n+1) = (a(n)-1)^2. This sequence emanates from r=6.
a(n) = 1/2*(-2 + (2+sqrt(3))^n + (2-sqrt(3))^n). - Ralf Stephan, Apr 14 2004
G.f.: x*(1+x)/(1 - 5*x + 5*x^2 - x^3) = x*(1+x)/((1-x)*(1 - 4*x + x^2)). [from the Ralf Stephan link]
a(n) = T(n, 2)-1 = A001075(n)-1, with Chebyshev's polynomials T(n, 2) of the first kind.
a(n) = b(n) + b(n-1), n >= 1, with b(n):=A061278(n) the partial sums of S(n, 4) = U(n, 2) = A001353(n+1) Chebyshev's polynomials of the second kind.
An integer k is in this sequence iff k is nonnegative and (k^2 + 2*k)/3 is a square. - James R. Buddenhagen, Nov 16 2005
a(0)=0, a(1)=1, a(n+1) = 3 + floor(a(n)*(2+sqrt(3))). - Anton Vrba (antonvrba(AT)yahoo.com), Jan 16 2007
a(n) = 4*a(n-1) - a(n-2) + 2. - Zerinvary Lajos, Mar 09 2008
From Peter Bala, Mar 25 2014: (Start)
a(2*n) = 6*A001353(n)^2; a(2*n+1) = A001834(n)^2.
a(n) = u(n)^2, where {u(n)} is the Lucas sequence in the quadratic integer ring Z[sqrt(6)] defined by the recurrence u(0) = 0, u(1) = 1, u(n) = sqrt(6)*u(n-1) - u(n-2) for n >= 2.
Equivalently, a(n) = U(n-1,sqrt(6)/2)^2, where U(n,x) denotes the Chebyshev polynomial of the second kind.
a(n) = (1/2)*( ((sqrt(6) + sqrt(2))/2)^n - ((sqrt(6) - sqrt(2))/2)^n )^2.
a(n) = bottom left entry of the 2 X 2 matrix T(n, M), where M is the 2 X 2 matrix [0, -2; 1, 3] and T(n,x) denotes the Chebyshev polynomial of the first kind. Cf. A098306.
See the remarks in A100047 for the general connection between Chebyshev polynomials of the first kind and 4th-order linear divisibility sequences. (End)
exp( Sum_{n >= 1} 2*a(n)*x^n/n ) = 1 + Sum_{n >= 1} A052530(n)*x^n. Cf. A001350. - Peter Bala, Mar 19 2015
E.g.f.: exp(2*x)*cosh(sqrt(3)*x) - cosh(x) - sinh(x). - Stefano Spezia, Oct 13 2019

Extension and Chebyshev comments from Wolfdieter Lang, Sep 10 2004

A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

Original entry on oeis.org

1, 1, 1, 1, 6, 1, 1, 15, 15, 1, 1, 28, 70, 28, 1, 1, 45, 210, 210, 45, 1, 1, 66, 495, 924, 495, 66, 1, 1, 91, 1001, 3003, 3003, 1001, 91, 1, 1, 120, 1820, 8008, 12870, 8008, 1820, 120, 1, 1, 153, 3060, 18564, 43758, 43758, 18564, 3060, 153, 1, 1, 190, 4845, 38760

Offset: 0

Philippe Deléham, Jul 26 2003

Terms have the same parity as those of Pascal's triangle.
Coefficients of polynomials (1/2)*((1 + x^(1/2))^(2n) + (1 - x^(1/2))^(2n)).
Number of compositions of 2n having k parts greater than 1; example: T(3, 2) = 15 because we have 4+2, 2+4, 3+2+1, 3+1+2, 2+3+1, 2+1+3, 1+3+2, 1+2+3, 2+2+1+1, 2+1+2+1, 2+1+1+2, 1+2+2+1, 1+2+1+2, 1+1+2+2, 3+3. - Philippe Deléham, May 18 2005
Number of binary words of length 2n - 1 having k runs of consecutive 1's; example: T(3,2) = 15 because we have 00101, 01001, 01010, 01011, 01101, 10001, 10010, 10011, 10100, 10110, 10111, 11001, 11010, 11011, 11101. - Philippe Deléham, May 18 2005
Let M_n be the n X n matrix M_n(i, j) = T(i, j-1); then for n > 0, det(M_n) = A000364(n), Euler numbers; example: det([1, 1, 0, 0; 1, 6, 1, 0; 1, 15, 15, 1; 1, 28, 70, 28 ]) = 1385 = A000364(4). - Philippe Deléham, Sep 04 2005
Equals ConvOffsStoT transform of the hexagonal numbers, A000384: (1, 6, 15, 28, 45, ...); e.g., ConvOffs transform of (1, 6, 15, 28) = (1, 28, 70, 28, 1). - Gary W. Adamson, Apr 22 2008
From Peter Bala, Oct 23 2008: (Start)
Let C_n be the root lattice generated as a monoid by {+-2*e_i: 1 <= i <= n; +-e_i +- e_j: 1 <= i not equal to j <= n}. Let P(C_n) be the polytope formed by the convex hull of this generating set. Then the rows of this array are the h-vectors of a unimodular triangulation of P(C_n) [Ardila et al.]. See A127674 for (a signed version of) the corresponding array of f-vectors for these type C_n polytopes. See A008459 for the array of h-vectors for type A_n polytopes and A108558 for the array of h-vectors associated with type D_n polytopes.
The Hilbert transform of this triangle is A142992 (see A145905 for the definition of this term).
(End)
Diagonal sums: A108479. - Philippe Deléham, Sep 08 2009
Coefficients of Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
Generalized Narayana triangle for 4^n (or cosh(2x)). - Paul Barry, Sep 28 2010
Coefficients of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A086646(n,k). - R. J. Mathar, Mar 12 2013
Let E(y) = Sum_{n>=0} y^n/(2*n)! = cosh(sqrt(y)). Then this triangle is the generalized Riordan array (E(y), y) with respect to the sequence (2*n)! as defined in Wang and Wang. Cf. A103327. - Peter Bala, Aug 06 2013
Row 6, (1,66,495,924,495,66,1), plays a role in expansions of powers of the Dedekind eta function. See the Chan link, p. 534, and A034839. - Tom Copeland, Dec 12 2016

			From _Peter Bala_, Oct 23 2008: (Start)
The triangle begins
n\k|..0.....1.....2.....3.....4.....5.....6
===========================================
0..|..1
1..|..1.....1
2..|..1.....6.....1
3..|..1....15....15.....1
4..|..1....28....70....28.....1
5..|..1....45...210...210....45.....1
6..|..1....66...495...924...495....66.....1
...
(End)
From _Peter Bala_, Aug 06 2013: (Start)
Viewed as the generalized Riordan array (cosh(sqrt(y)), y) with respect to the sequence (2*n)! the column generating functions begin
1st col: cosh(sqrt(y)) = 1 + y/2! + y^2/4! + y^3/6! + y^4/8! + ....
2nd col: 1/2!*y*cosh(sqrt(y)) = y/2! + 6*y^2/4! + 15*y^3/6! + 28*y^4/8! + ....
3rd col: 1/4!*y^2*cosh(sqrt(y)) = y^2/4! + 15*y^3/6! + 70*y^4/8! + 210*y^5/10! + .... (End)

A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 224.

Indranil Ghosh, Rows 0.. 120 of triangle, flattened
F. Ardila, M. Beck, S. Hosten, J. Pfeifle and K. Seashore, Root polytopes and growth series of root lattices arXiv:0809.5123 [math.CO], 2008.
H. Chan, S. Cooper and P. Toh, The 26th power of Dedekind's eta function Advances in Mathematics, 207 (2006) 532-543.
T. Copeland, Infinitesimal Generators, the Pascal Pyramid, and the Witt and Virasoro Algebras, 2012.
T. Copeland, Skipping over Dimensions, Juggling Zeros in the Matrix, 2020.
E. Lucas, Théorie des fonctions numériques simplement périodiques, Baltimore, 1878. See page 145 equation (153).
W. Wang and T. Wang, Generalized Riordan array, Discrete Mathematics, Vol. 308, No. 24, 6466-6500.

Cf. A098158, A000384, A081294.
Cf. A008459, A108558, A127674, A142992. - Peter Bala, Oct 23 2008
Cf. A103327 (binomial(2n+1, 2k+1)), A103328 (binomial(2n, 2k+1)), A091042 (binomial(2n+1, 2k)). -Wolfdieter Lang, Jan 06 2013
Cf. A086646 (unsigned matrix inverse), A103327.
Cf. A034839.

/* As triangle: */ [[Binomial(2*n, 2*k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Dec 14 2016

A086645:=(n,k)->binomial(2*n,2*k): seq(seq(A086645(n,k),k=0..n),n=0..12);

Table[Binomial[2 n, 2 k], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Dec 13 2016 *)

create_list(binomial(2*n,2*k),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */

PARI
```
{T(n, k) = binomial(2*n, 2*k)};
```

{T(n, k) = sum( i=0, min(k, n-k), 4^i * binomial(n, 2*i) * binomial(n - 2*i, k-i))}; /* Michael Somos, May 26 2005 */

T(n, k) = (2*n)!/((2*(n-k))!*(2*k)!) row sums = A081294. COLUMNS: A000012, A000384
Sum_{k>=0} T(n, k)*A000364(k) = A000795(n) = (2^n)*A005647(n).
Sum_{k>=0} T(n, k)*2^k = A001541(n). Sum_{k>=0} T(n, k)*3^k = 2^n*A001075(n). Sum_{k>=0} T(n, k)*4^k = A083884(n). - Philippe Deléham, Feb 29 2004
O.g.f.: (1 - z*(1+x))/(x^2*z^2 - 2*x*z*(1+z) + (1-z)^2) = 1 + (1 + x)*z +(1 + 6*x + x^2)*z^2 + ... . - Peter Bala, Oct 23 2008
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A081294(n), A001541(n), A090965(n), A083884(n), A099140(n), A099141(n), A099142(n), A165224(n), A026244(n) for x = 0,1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 08 2009
Product_{k=1..n} (cot(k*Pi/(2n+1))^2 - x) = Sum_{k=0..n} (-1)^k*binomial(2n,2k)*x^k/(2n+1-2k). - David Ingerman (daviddavifree(AT)gmail.com), Mar 30 2010
From Paul Barry, Sep 28 2010: (Start)
G.f.: 1/(1-x-x*y-4*x^2*y/(1-x-x*y)) = (1-x*(1+y))/(1-2*x*(1+y)+x^2*(1-y)^2);
E.g.f.: exp((1+y)*x)*cosh(2*sqrt(y)*x);
T(n,k) = Sum_{j=0..n} C(n,j)*C(n-j,2*(k-j))*4^(k-j). (End)
T(n,k) = 2*T(n-1,k) + 2*T(n-1,k-1) + 2*T(n-2,k-1) - T(n-2,k) - T(n-2,k-2), with T(0,0)=T(1,0)=T(1,1)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 26 2013
From Peter Bala, Sep 22 2021: (Start)
n-th row polynomial R(n,x) = (1-x)^n*T(n,(1+x)/(1-x)), where T(n,x) is the n-th Chebyshev polynomial of the first kind. Cf. A008459.
R(n,x) = Sum_{k = 0..n} binomial(n,2*k)*(4*x)^k*(1 + x)^(n-2*k).
R(n,x) = n*Sum_{k = 0..n} (n+k-1)!/((n-k)!*(2*k)!)*(4*x)^k*(1-x)^(n-k) for n >= 1. (End)

A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

Original entry on oeis.org

0, 1, 5, 20, 76, 285, 1065, 3976, 14840, 55385, 206701, 771420, 2878980, 10744501, 40099025, 149651600, 558507376, 2084377905, 7779004245, 29031639076, 108347552060, 404358569165, 1509086724601, 5631988329240, 21018866592360, 78443478040201, 292755045568445

Offset: 0

Henry Bottomley, Jun 04 2001

Indices m of triangular numbers T(m) which are one-third of another triangular number: 3*T(m) = T(k); the k's are given by A001571. - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002
On the previous comment: for m=0 this is actually one third of the same triangular number. - Zak Seidov, Apr 07 2011
Also numbers n such that the n-th centered 24-gonal number 12*n*(n+1)+1 is a perfect square A001834(n)^2, where A001834(n) is defined by the recursion: a(0) = 1, a(1) = 5, a(n) = 4*a(n-1) - a(n-2) + 1. - Alexander Adamchuk, Apr 21 2007
Also numbers n such that RootMeanSquare(5,...,6*n-1) is an integer. - Ctibor O. Zizka, Dec 17 2008 (Corrected by Robert K. Moniot, Jul 22 2020)
Also numbers n such that n*(n+1) = Sum_{i=1..x} n+i for some x. (This does not apply to the first term.). - Gil Broussard, Dec 23 2008
From John P. McSorley, May 26 2020: (Start)
Consecutive terms (a(n-1), a(n)) = (u,v) give all points on the hyperbola u^2 - u + v^2 - v - 4*u*v = 0 in quadrant I with both coordinates an integer.
Also related to the block sizes of small multi-set designs. (End)
If a(n) white balls and a(n+1) black balls are mixed in a bag, and a pair of balls is drawn without replacement, the probability that one ball of each color is drawn is exactly 1/3. These are the only integers for which the probability is 1/3. For example, if there are 20 white balls and 76 black balls, the probability of drawing one of each is (20/96)*(76/95) + (76/96)*(20/95) = 1/3. - Elliott Line, May 13 2022

			a(2)=5 and T(5)=15 which is 1/3 of 45=T(9).

Vincenzo Librandi, 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.
Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.
Brian Lawrence and Will Sawin, The Shafarevich conjecture for hypersurfaces in abelian varieties, arXiv:2004.09046 [math.NT], 2020.
Ioana-Claudia Lazăr, Lucas sequences in t-uniform simplicial complexes, arXiv:1904.06555 [math.GR], 2019.
L. A. Medina and A. Straub, On multiple and infinite log-concavity, 2013.
S. Morier-Genoud, V. Ovsienko and S. Tabachnikov, 2-frieze patterns and the cluster structure of the space of polygons, Annales de l'institut Fourier, 62 no. 3 (2012), 937-987. - From _N. J. A. Sloane_, Dec 26 2012
Robert Phillips, A triangular number result, 2009.
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.
Eric Weisstein, Centered Polygonal Number.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001571, A001834, A001835, A079935, A101265. Also cf. A212336 for more sequences with g.f. of the type 1/(1-k*x+k*x^2-x^3).

I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1) - Self(n-2) + 1: n in [1..30]]; // Vincenzo Librandi, Dec 23 2012

f:= gfun:-rectoproc({a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3),a(1)=1,a(0)=0,a(-1)=0},a(n),remember):
map(f, [$0..50]); # Robert Israel, Jun 05 2015

CoefficientList[Series[x/(1 - 5*x + 5*x^2 - x^3), {x, 0, nn}], x] (* T. D. Noe, Jun 04 2012 *)
LinearRecurrence[{5,-5,1},{0,1,5},30] (* Harvey P. Dale, Dec 23 2012 *)

M = [1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1, 30, print1(([1, 0, 0]*M^i)[3], ",")) \\ Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

a(n) = 4*a(n-1) - a(n-2) + 1.
a(n) = A001075(n) - a(n-1) - 1.
a(n) = (A001835(n+1) - 1)/2 = (A001353(n+1) - A001353(n) - 1)/2.
a(n) = a(n-1) + A001353(n), i.e., partial sum of A001353.
From Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002: (Start)
a(n+2) = 4*a(n+1) - a(n) + 1 for a(0)=0, a(1)=1.
G.f.: x/((1 - x)*(1 - 4*x + x^2)).
a(n) = (1/12)*((3 - sqrt(3))*(2 - sqrt(3))^n + (3 + sqrt(3))*(2 + sqrt(3))^n-6). (End)
a(n) = (1/12)*(A003500(n) + A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003
a(n+1) = Sum_{k=0..n} U(k, 2) = Sum_{k=0..n} S(k, 4), where U(n,x) and S(n,x) are Chebyshev polynomials. - Paul Barry, Nov 14 2003
G.f.: x/(1 - 5*x + 5*x^2 - x^3).
a(n) = A079935(n+1) + A001571(n) for n>0, a(0)=0. - Gerry Martens, Jun 05 2015
a(n)*a(n-2) = a(n-1)*(a(n-1) - 1) for n>1. - Bruno Berselli, Nov 29 2016
From John P. McSorley, May 25 2020: (Start)
a(n)^2 - a(n) + a(n-1)^2 - a(n-1) - 4*a(n)*a(n-1) = 0.
a(n) = A001834(n-1) + a(n-2). (End)
(T(a(n)-1) + T(a(n+1)-1))/T(a(n) + a(n+1) - 1) = 2/3 where T(i) is the i-th triangular number. - Robert K. Moniot, Oct 11 2020
E.g.f.: exp(x)*(exp(x)*(3*cosh(sqrt(3)*x) + sqrt(3)*sinh(sqrt(3)*x)) - 3)/6. - Stefano Spezia, Feb 05 2021
a(n) = A101265(n) - 1. - Jon E. Schoenfield, Jan 01 2022

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

A002531 a(2n) = a(2n-1) + a(2n-2), a(2n+1) = 2a(2n) + a(2*n-1); a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 5, 7, 19, 26, 71, 97, 265, 362, 989, 1351, 3691, 5042, 13775, 18817, 51409, 70226, 191861, 262087, 716035, 978122, 2672279, 3650401, 9973081, 13623482, 37220045, 50843527, 138907099, 189750626, 518408351, 708158977, 1934726305

Offset: 0

N. J. A. Sloane

Numerators of continued fraction convergents to sqrt(3), for n >= 1.
For the denominators see A002530.
Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the convergents 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to sqrt(3). Sequence contains the numerators. - Amarnath Murthy, Mar 22 2003
In the Murthy comment if we take a = 0, b = 1 then the denominator of the reduced fraction is a(n+1). A083336(n)/a(n+1) converges to sqrt(3). - Mario Catalani (mario.catalani(AT)unito.it), Apr 26 2003
If signs are disregarded, all terms of A002316 appear to be elements of this sequence. - Creighton Dement, Jun 11 2007
2^(-floor(n/2))*(1 + sqrt(3))^n = a(n) + A002530(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 10 2018
Let T(n) = A000034(n), U(n) = A002530(n), V(n) = a(n), x(n) = U(n)/V(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (x(n) + x(m))/(1 + 3*x(n)*x(m)). - Michael Somos, Nov 29 2022

			1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 19.
Convergents are 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530.
G.f. = 1 + x + 2*x^2 + 5*x^3 + 7*x^4 + 19*x^5 + 26*x^6 + 71*x^7 + ... - _Michael Somos_, Mar 22 2022

I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
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).
A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Harry J. Smith, Table of n, a(n) for n = 0..2000
MacTutor, D'Arcy Thompson on Greek irrationals
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
Albert Tarn, Approximations to certain square roots and the series of numbers connected therewith [Annotated scanned copy]
D'Arcy Thompson, Excess and Defect: Or the Little More and the Little Less, Mind, New Series, Vol. 38, No. 149 (Jan., 1929), pp. 43-55 (13 pages). See page 48.
Hein van Winkel, Q-quadrangles inscribed in a circle, 2014. See Table 1. [Reference from Antreas Hatzipolakis, Jul 14 2014]
Index entries for "core" sequences
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).
Index entries for sequences related to Chebyshev polynomials.

Bisections are A001075 and A001834.
Cf. A002530 (denominators), A048788.
Cf. A002316.
Cf. A083332, A199710, A026150, A053120.

a:=[1,1,2,5];; for n in [5..40] do a[n]:=4*a[n-2]-a[n-4]; od; a; # G. C. Greubel, Nov 16 2018

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1 +x-2*x^2+x^3)/(1-4*x^2+x^4))); // G. C. Greubel, Nov 16 2018

A002531 := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif type(n,odd) then A002531(n-1)+A002531(n-2) else 2*A002531(n-1)+A002531(n-2) fi; end; [ seq(A002531(n), n=0..50) ];
with(numtheory): tp := cfrac (tan(Pi/3),100): seq(nthnumer(tp,i), i=-1..32 ); # Zerinvary Lajos, Feb 07 2007
A002531:=(1+z-2*z**2+z**3)/(1-4*z**2+z**4); # Simon Plouffe; see his 1992 dissertation

Insert[Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[3], n]]], {n, 1, 40}], 1, 1] (* Stefan Steinerberger, Apr 01 2006 *)
Join[{1},Numerator[Convergents[Sqrt[3],40]]] (* Harvey P. Dale, Jan 23 2012 *)
CoefficientList[Series[(1 + x - 2 x^2 + x^3)/(1 - 4 x^2 + x^4), {x, 0, 30}], x] (* Vincenzo Librandi, Nov 01 2014 *)
LinearRecurrence[{0, 4, 0, -1}, {1, 1, 2, 5}, 35] (* Robert G. Wilson v, Feb 11 2018 *)
a[ n_] := ChebyshevT[n, Sqrt[-1/2]]*Sqrt[2]^Mod[n,2]/I^n //Simplify; (* Michael Somos, Mar 22 2022 *)
a[ n_] := If[n<0, (-1)^n*a[-n], SeriesCoefficient[ (1 + x - 2*x^2 + x^3) / (1 - 4*x^2 + x^4), {x, 0, n}]]; (* Michael Somos, Sep 23 2024 *)

a(n)=contfracpnqn(vector(n,i,1+(i>1)*(i%2)))[1,1]

apply( {A002531(n,w=quadgen(12))=real((2+w)^(n\/2)*if(bittest(n, 0), w-1, 1))}, [0..30]) \\ M. F. Hasler, Nov 04 2019

{a(n) = if(n<0, (-1)^n*a(-n), polcoeff( (1 + x - 2*x^2 + x^3) / (1 - 4*x^2 + x^4) + x*O(x^n), n))}; /* Michael Somos, Sep 23 2024 */

s=((1+x-2*x^2+x^3)/(1-4*x^2+x^4)).series(x,40); s.coefficients(x, sparse=False) # G. C. Greubel, Nov 16 2018

G.f.: (1 + x - 2*x^2 + x^3)/(1 - 4*x^2 + x^4).
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1), n > 0.
a(2*n) = (1/2)*((2 + sqrt(3))^n+(2 - sqrt(3))^n); a(2*n) = A003500(n)/2; a(2*n+1) = round(1/(1 + sqrt(3))*(2 + sqrt(3))^n). - Benoit Cloitre, Dec 15 2002
a(n) = ((1 + sqrt(3))^n + (1 - sqrt(3))^n)/(2*2^floor(n/2)). - Bruno Berselli, Nov 10 2011
a(n) = A080040(n)/(2*2^floor(n/2)). - Ralf Stephan, Sep 08 2013
a(2*n) = (-1)^n*T(2*n,u) and a(2*n+1) = (-1)^n*1/u*T(2*n+1,u), where u = sqrt(-1/2) and T(n,x) denotes the Chebyshev polynomial of the first kind. - Peter Bala, May 01 2012
a(n) = (-sqrt(2)*i)^n*T(n, sqrt(2)*i/2)*2^(-floor(n/2)) = A026150(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and the Chebyshev T polynomials (A053120). - Wolfdieter Lang, Feb 10 2018
From Franck Maminirina Ramaharo, Nov 14 2018: (Start)
a(n) = ((1 - sqrt(2))*(-1)^n + 1 + sqrt(2))*(((sqrt(2) - sqrt(6))/2)^n + ((sqrt(6) + sqrt(2))/2)^n)/4.
E.g.f.: cosh(sqrt(3/2)*x)*(sqrt(2)*sinh(x/sqrt(2)) + cosh(x/sqrt(2))). (End)
a(n) = (-1)^n*a(-n) for all n in Z. - Michael Somos, Mar 22 2022
a(n) = 4*a(n-2) - a(n-4). - Boštjan Gec, Sep 21 2023

Name edited (as by discussion in A002530) by M. F. Hasler, Nov 04 2019

A005246 a(n) = (1 + a(n-1)*a(n-2))/a(n-3), a(0) = a(1) = a(2) = 1.

Original entry on oeis.org

1, 1, 1, 2, 3, 7, 11, 26, 41, 97, 153, 362, 571, 1351, 2131, 5042, 7953, 18817, 29681, 70226, 110771, 262087, 413403, 978122, 1542841, 3650401, 5757961, 13623482, 21489003, 50843527, 80198051, 189750626, 299303201, 708158977, 1117014753

Offset: 0

N. J. A. Sloane

For n >= 4 we have the linear recurrence a(n) = 4*a(n-2) - a(n-4). - Ahmed Fares (ahmedfares(AT)my-deja.com), Jun 04 2001
Integer solutions to the equation floor(sqrt(3)*x^2) = x*floor(sqrt(3)*x). - Benoit Cloitre, Mar 18 2004
For n > 2, a(n) is the smallest integer > a(n-1) such that sqrt(3)*a(n) is closer to and greater than an integer than sqrt(3)*a(n-1). I.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(3)*a(n)) < frac(sqrt(3)*a(n-1)). - Benoit Cloitre, Jan 20 2003
The lower principal and intermediate convergents to 3^(1/2), beginning with 1/1, 3/2, 5/3, 12/7, 19/11, form a strictly increasing sequence; essentially, numerators=A143643 and denominators=A005246. - Clark Kimberling, Aug 27 2008
This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z. The g.f. is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n). The general form of a(n) is given by: a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (b-Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p) and a(2*m+1) = a*Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*Q^(m-2*p) + (a*b+q-a*Q)*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p). - Richard Choulet, Feb 24 2010
a(n) for n > 1 are the integer square roots of (floor(m^2/3)+1), where the values of m are given by A143643.  Also see A082630. - Richard R. Forberg, Nov 14 2013
The a(n) = (1 + a(n-1)*a(n-2))/a(n-3) recursion has the Laurent property. If a(0), a(1), a(2) are variables, then a(n) is a Laurent polynomial (a rational function with a monomial denominator). - Michael Somos, Feb 27 2019

			G.f. = 1 + x + x^2 + 2*x^3 + 3*x^4 + 7*x^5 + 11*x^6 + 26*x^7 + 41*x^8 + ...
From _Richard Choulet_, Feb 24 2010: (Start)
a(4) = 4^2 - 4^0 - 3*4^1 = 3.
a(7) = 4^3 - 4*binomial(2,1) - 2*(4^2-1) = 26. (End)

Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
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..500
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ...
Reid Barton, A combinatorial interpretation of the sequence 1, 1, 2, 6, 21, 77, ..., [Annotated scanned copy]
Peter Cameron's Blog, The ADE affair, 3, Posted 23/06/2011.
T. Crilly, Double sequences of positive integers, Math. Gaz., 69 (1985), 263-271.
Enrica Duchi, Andrea Frosini, Renzo Pinzani and Simone Rinaldi, A Note on Rational Succession Rules, J. Integer Seqs., Vol. 6, 2003.
R. K. Guy, Letter to N. J. A. Sloane, Feb 1986
Clark Kimberling, Best lower and upper approximates to irrational numbers, Elemente der Mathematik, 52 (1997) 122-126.
Valentin Ovsienko, Serge Tabachnikov, Dual numbers, weighted quivers, and extended Somos and Gale-Robinson sequences, arXiv:1705.01623 [math.CO], 2017. See p. 10.
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
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-1).

Bisections are A001835 and A001075.
Cf. A101265. Row sums of A211956.
Cf. A001353.

a005246 n = a005246_list !! n
a005246_list = 1 : 1 : 1 : map (+ 1) (zipWith div
   (zipWith (*) (drop 2 a005246_list) (tail a005246_list)) a005246_list)
-- Reinhard Zumkeller, Mar 07 2012

A005246:=-(-1-z+2*z**2+z**3)/(1-4*z**2+z**4); # Conjectured by Simon Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.
for q from 1 to 10 do :a:=1:b:=1:Q:=(a*b^2+q*b+a+q)/(a*b): for m from 0 to 15 do U(m):=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od: for m from 0 to 15 do V(m):=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)):od:for m from 0 to 15 do W(2*m):=U(m):od:for m from 0 to 14 do W(2*m+1):=V(m):od:seq(W(m),m=0..30):od; # Richard Choulet, Feb 24 2010

RecurrenceTable[{a[0]==a[1]==a[2]==1,a[n]==(1+a[n-1]a[n-2])/a[n-3]},a,{n,40}] (* Harvey P. Dale, May 28 2013 *)
a[n_] := Cosh[(n-1)*ArcSinh[1/Sqrt[2]]]*If[EvenQ[n], Sqrt[2/3], 1]; Table[a[n] // FunctionExpand, {n, 0, 34}] (* Jean-François Alcover, Dec 10 2014, after Peter Bala *)
a[ n_] := With[{m = If[ n < 0, 2 - n, n]}, SeriesCoefficient[ (1 + x - 3 x^2 - 2 x^3) / (1 - 4 x^2 + x^4), {x, 0, m}]]; (* Michael Somos, Feb 10 2017 *)

{a(n) = if( n<0, n = 2 - n); polcoeff((1 + x - 3*x^2 - 2*x^3) / (1 - 4*x^2 + x^4) + x * O(x^n), n)}; /* Michael Somos, Nov 15 2006 */

{a(n) = real( (2 + quadgen(12))^(n\2) * if( n%2, 1, 1 - 1 / quadgen(12)) )}; /* Michael Somos, May 24 2012 */

G.f.: (1 + x - 3*x^2 - 2*x^3)/(1 - 4*x^2 + x^4).
Limit_{n->oo} a(2n+1)/a(2n) = (3+sqrt(3))/3 = 1.5773502...; lim_{n->oo} a(2n)/a(2n-1) = (3+sqrt(3))/2 = 2.3660254.... - Benoit Cloitre, Aug 07 2002
A101265(n) = a(n)*a(n+1). - Franklin T. Adams-Watters, Apr 24 2006
a(n) = a(2-n) for all n in Z. - Michael Somos, Nov 15 2006
a(2*n + 1) = A001075(n). a(2*n) = A001835(n). a(2*n + 1) - a(2*n) = a(2*n + 2) - a(2*n + 1) = A001353(n). - Michael Somos, May 24 2012
For n > 2: a(n) = a(n-1) + Sum_{k=1..floor((n-1)/2)} a(2*k). - Reinhard Zumkeller, Dec 16 2007
From Richard Choulet, Feb 24 2010: (Start)
a(2*m) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 3*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p).
a(2*m+1) = Sum_{p=0..floor(m/2)} (-1)^p*binomial(m-p,p)*4^(m-2*p) - 2*Sum_{p=0..floor((m-1)/2)} (-1)^p*binomial(m-1-p,p)*4^(m-1-2*p). (End)
From Tim Monahan, Jul 01 2011: (Start)
Closed form without extra leading 1: ((sqrt(6)+3)*(sqrt(2+sqrt(3))^n+(sqrt(2-sqrt(3))^n))+(3-sqrt(6))*((-sqrt(2+sqrt(3)))^n+(-sqrt(2-sqrt(3)))^n))/12.
Closed form with extra leading 1: ((6+3*sqrt(6)-2*sqrt(3)-3*sqrt(2))*(sqrt(2+sqrt(3))^n)+(6+3*sqrt(6)+2*sqrt(3)+3*sqrt(2))*(sqrt(2-sqrt(3))^n)+(6-3*sqrt(6)-2*sqrt(3)+3*sqrt(2))*((-sqrt(2+sqrt(3)))^n)+(6-3*sqrt(6)+2*sqrt(3)-3*sqrt(2))*((-sqrt(2-sqrt(3)))^n))/24. (End)
a(2*n+2) = Sum_{k = 0..n} 2^k*binomial(n+k,2*k); a(2*n+1) = Sum_{k = 0..n} n/(n+k)*2^k*binomial(n+k,2*k) for n >= 1. Row sums of A211956. - Peter Bala, May 01 2012
a(n) = ((sqrt(2)+sqrt(3)+(-1)^n*(sqrt(2)-sqrt(3)))*sqrt(2+(2-sqrt(3))^n*(2+ sqrt(3))-(-2+sqrt(3))*(2+ sqrt(3))^n))/(4*sqrt(3)). - Gerry Martens, Jun 06 2015
0 = a(n) - 2*a(n+1) + a(n+2) if n is even, 0 = a(n) - 3*a(n+1) + a(n+2) if n is odd for all n in Z. - Michael Somos, Feb 10 2017

More terms from Michael Somos, Aug 01 2001

A023110 Squares which remain squares when the last digit is removed.

Original entry on oeis.org

0, 1, 4, 9, 16, 49, 169, 256, 361, 1444, 3249, 18496, 64009, 237169, 364816, 519841, 2079364, 4678569, 26666896, 92294449, 341991049, 526060096, 749609641, 2998438564, 6746486769, 38453641216, 133088524969, 493150849009, 758578289296, 1080936581761

Offset: 1

David W. Wilson

This A023110 = A031149^2 is the base 10 version of A001541^2 = A055792 (base 2), A001075^2 = A055793 (base 3), A004275^2 = A055808 (base 4), A204520^2 = A055812 (base 5), A204518^2 = A055851 (base 6), A204516^2 = A055859 (base 7), A204514^2 = A055872 (base 8) and A204502^2 = A204503 (base 9). - M. F. Hasler, Sep 28 2014

For the first 4 terms the square has only one digit. It is understood that deleting this digit yields 0. - Colin Barker, Dec 31 2017

R. K. Guy, Neg and Reg, preprint, Jan 2012.

Jon E. Schoenfield, Table of n, a(n) for n = 1..70 (terms 1..38 from David W. Wilson, terms 39..40 from Robert G. Wilson v, terms 41..67 from Dmitry Petukhov)
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Joshua Stucky, Pell's Equation and Truncated Squares, Number Theory Seminar, Kansas State University, Feb 19 2018.
Index to sequences related to truncating digits of squares.

Cf. A023111.
Cf. A031150, A053784, A031149, A055792, A055793, A055808, A055812, A055851, A055859, A055872.
Cf. A001541, A001075, A004275, A204520, A204518, A204516, A204514, A204502, A204503.

count:= 1: A[1]:= 0:
for n from 0 while count < 35 do
  for t in [1,4,6,9] do
    if issqr(10*n^2+t) then
       count:= count+1;
       A[count]:= 10*n^2+t;
    fi
  od
od:
seq(A[i],i=1..count); # Robert Israel, Sep 28 2014

fQ[n_] := IntegerQ@ Sqrt@ Quotient[n^2, 10]; Select[ Range@ 1000000, fQ]^2 (* Robert G. Wilson v, Jan 15 2011 *)

for(n=0,1e7, issquare(n^2\10) & print1(n^2",")) \\  M. F. Hasler, Jan 16 2012

Appears to satisfy a(n)=1444*a(n-7)+a(n-14)-76*sqrt(a(n-7)*a(n-14)) for n >= 16. For n = 15, 14, 13, ... this would require a(1) = 16, a(0) = 49, a(-1) = 169, ... - Henry Bottomley, May 08 2001; edited by Robert Israel, Sep 28 2014
a(n) = A031149(n)^2. - M. F. Hasler, Sep 28 2014
Conjectures from Colin Barker, Dec 31 2017: (Start)
G.f.: x^2*(1 + 4*x + 9*x^2 + 16*x^3 + 49*x^4 + 169*x^5 + 256*x^6 - 1082*x^7 - 4328*x^8 - 9738*x^9 - 4592*x^10 - 6698*x^11 - 6698*x^12 - 4592*x^13 + 361*x^14 + 1444*x^15 + 3249*x^16 + 256*x^17 + 169*x^18 + 49*x^19 + 16*x^20) / ((1 - x)*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)*(1 - 1442*x^7 + x^14)).
a(n) = 1443*a(n-7) - 1443*a(n-14) + a(n-21) for n>22.
(End)

More terms from M. F. Hasler, Jan 16 2012

A055793 Numbers k such that k and floor[k/3] are both squares; i.e., squares which remain squares when written in base 3 and last digit is removed.

Original entry on oeis.org

0, 1, 4, 49, 676, 9409, 131044, 1825201, 25421764, 354079489, 4931691076, 68689595569, 956722646884, 13325427460801, 185599261804324, 2585064237799729, 36005300067391876, 501489136705686529, 6984842613812219524, 97286307456665386801, 1355023461779503195684, 18873042157456379352769

Offset: 1

Henry Bottomley, Jul 14 2000

Or, squares of the form 3k^2+1.

See A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

			a(3) = 49 because 49 = 7^2 = 1211 base 3 and 121 base 3 = 16 = 4^2.

Vincenzo Librandi, Table of n, a(n) for n = 1..800
Tom C. Brown and Peter J Shiue, Squares of second-order linear recurrence sequences, Fib. Quart., 33 (1994), 352-356.
Yurii S. Bystryk, Vitalii L. Denysenko, and Volodymyr I. Ostryk, Lune and Lens Sequences, ResearchGate preprint, 2024. See pp. 48, 56.
M. F. Hasler, Truncated squares, OEIS wiki, Jan 16 2012
Giovanni Lucca, Integer Sequences and Circle Chains Inside a Circular Segment, Forum Geometricorum, Vol. 18 (2018), 47-55.
Index to sequences related to truncating digits of squares.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A023110, A098301.

Cf. also A023110, A204503, A204512, A204517, A204519, A055812, A055808 and A055792 for the analog in other bases.

I:=[0, 1, 4]; [n le 3 select I[n] else 14*Self(n-1) - Self(n-2) - 6: n in [1..30]]; // Vincenzo Librandi, Jan 27 2013

A055793 := proc(n) coeftayl(x*(1-11*x+4*x^2)/((1-x)*(1-14*x+x^2)), x=0, n); end proc: seq(A055793(n), n=0..20); # Wesley Ivan Hurt, Sep 28 2014

CoefficientList[Series[x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)), {x, 0, 20}], x] (* Wesley Ivan Hurt, Sep 28 2014 *)
LinearRecurrence[{15,-15,1},{0,1,4,49},40] (* Harvey P. Dale, Jun 19 2021 *)

sq3nsqplus1(n) = { for(x=1,n, y = 3*x*x+1; \ print1(y" ") if(issquare(y),print1(y" ")) ) }

a(n) = 3*A098301(n-2)+1. - R. J. Mathar, Jun 11 2009
a(n) = 14*a(n-1)-a(n-2)-6, with a(0)=1, a(1)=4. (See Brown and Shiue)
a(n) = (A001075(n-2))^2. - Johannes Boot Dec 16 2011, corrected by M. F. Hasler, Jan 15 2012
G.f.: x*(1 - 11*x + 4*x^2)/((1 - x)*(1 - 14*x + x^2)). - M. F. Hasler, Jan 15 2012

More terms from Cino Hilliard, Mar 01 2003

A008310 Triangle of coefficients of Chebyshev polynomials T_n(x).

Original entry on oeis.org

1, 1, -1, 2, -3, 4, 1, -8, 8, 5, -20, 16, -1, 18, -48, 32, -7, 56, -112, 64, 1, -32, 160, -256, 128, 9, -120, 432, -576, 256, -1, 50, -400, 1120, -1280, 512, -11, 220, -1232, 2816, -2816, 1024, 1, -72, 840, -3584, 6912, -6144, 2048, 13, -364, 2912, -9984, 16640, -13312, 4096

Offset: 0

N. J. A. Sloane

The row length sequence of this irregular array is A008619(n), n >= 0. Even or odd powers appear in increasing order starting with 1 or x for even or odd row numbers n, respectively. This is the standard triangle A053120 with 0 deleted. - Wolfdieter Lang, Aug 02 2014

Let T* denote the triangle obtained by replacing each number in this triangle by its absolute value. Then T* gives the coefficients for cos(nx) as a polynomial in cos x. - Clark Kimberling, Aug 04 2024

			Rows are: (1), (1), (-1,2), (-3,4), (1,-8,8), (5,-20,16) etc., since if c = cos(x): cos(0x) = 1, cos(1x) = 1c; cos(2x) = -1+2c^2; cos(3x) = -3c+4c^3, cos(4x) = 1-8c^2+8c^4, cos(5x) = 5c-20c^3+16c^5, etc.
From _Wolfdieter Lang_, Aug 02 2014: (Start)
This irregular triangle a(n,k) begins:
  n\k   0    1     2      3      4      5      6      7 ...
  0:    1
  1:    1
  2:   -1    2
  3:   -3    4
  4:    1   -8     8
  5:    5  -20    16
  6:   -1   18   -48     32
  7:   -7   56  -112     64
  8:    1  -32   160   -256    128
  9:    9 -120   432   -576    256
 10:   -1   50  -400   1120  -1280    512
 11:  -11  220 -1232   2816  -2816   1024
 12:    1  -72   840  -3584   6912  -6144   2048
 13:   13 -364  2912  -9984  16640 -13312   4096
 14:   -1   98 -1568   9408 -26880  39424 -28672   8192
 15:  -15  560 -6048  28800 -70400  92160 -61440  16384
  ...
T(4,x) = 1 - 8*x^2 + 8*x^4, T(5,x) = 5*x - 20*x^3 +16*x^5.
(End)

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
E. A. Guilleman, Synthesis of Passive Networks, Wiley, 1957, p. 593.
Yaroslav Zolotaryuk, J. Chris Eilbeck, "Analytical approach to the Davydov-Scott theory with on-site potential", Physical Review B 63, p543402, Jan. 2001. The authors write, "Since the algebra of these is 'hyperbolic', contrary to the usual Chebyshev polynomials defined on the interval 0 <= x <= 1, we call the set of functions (21) the hyperbolic Chebyshev polynomials." (This refers to the triangle T* described in Comments.)

R. J. Mathar, Table of n, a(n) for n = 0..2600
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Renato Ferreira Pinto Jr. and Nathaniel Harms, Testing Support Size More Efficiently Than Learning Histograms, arXiv:2410.18915 [cs.DS], 2024. See p. 40.
D. Foata and G.-N. Han, Nombres de Fibonacci et polynomes orthogonaux
C. Lanczos, Applied Analysis (Annotated scans of selected pages)
I. Rivin, Growth in free groups (and other stories), arXiv:math/9911076 [math.CO], 1999.
Eric Weisstein's World of Mathematics, Chebyshev Polynomial of the First Kind.
Wikipedia, Chebyshev polynomials.
Index entries for sequences related to Chebyshev polynomials.

A039991 is a row reversed version, but has zeros which enable the triangle to be seen. Columns/diagonals are A011782, A001792, A001793, A001794, A006974, A006975, A006976 etc.
Reflection of A028297. Cf. A008312, A053112.
Row sums are one. Polynomial evaluations include A001075 (x=2), A001541 (x=3), A001091, A001079, A023038, A011943, A001081, A023039, A001085, A077422, A077424, A097308, A097310, A068203.
Cf. A053120.

A008310 := proc(n,m) local x ; coeftayl(simplify(ChebyshevT(n,x),'ChebyshevT'),x=0,m) ; end: i := 0 : for n from 0 to 100 do for m from n mod 2 to n by 2 do printf("%d %d ",i,A008310(n,m)) ; i := i+1 ; od ; od ; # R. J. Mathar, Apr 20 2007
# second Maple program:
b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
T:= n-> (p-> (d-> seq(coeff(p, x, d-i), i=0..d))(degree(p)))(b(n)):
seq(T(n), n=0..15);  # Alois P. Heinz, Sep 04 2019

Flatten[{1, Table[CoefficientList[ChebyshevT[n, x], x], {n, 1, 13}]}]//DeleteCases[#, 0, Infinity]& (* or *) Flatten[{1, Table[Table[((-1)^k*2^(n-2 k-1)*n*Binomial[n-k, k])/(n-k), {k, Floor[n/2], 0, -1}], {n, 1, 13}]}] (* Eugeniy Sokol, Sep 04 2019 *)

a(n,m) = 2^(m-1) * n * (-1)^((n-m)/2) * ((n+m)/2-1)! / (((n-m)/2)! * m!) if n>0. - R. J. Mathar, Apr 20 2007
From Paul Weisenhorn, Oct 02 2019: (Start)
T_n(x) = 2*x*T_(n-1)(x) - T_(n-2)(x), T_0(x) = 1, T_1(x) = x.
T_n(x) = ((x+sqrt(x^2-1))^n + (x-sqrt(x^2-1))^n)/2. (End)
From Peter Bala, Aug 15 2022: (Start)
T(n,x) = [z^n] ( z*x + sqrt(1 + z^2*(x^2 - 1)) )^n.
Sum_{k = 0..2*n} binomial(2*n,k)*T(k,x) = (2^n)*(1 + x)^n*T(n,x).
exp( Sum_{n >= 1} T(n,x)*t^n/n ) = Sum_{n >= 0} P(n,x)*t^n, where P(n,x) denotes the n-th Legendre polynomial. (End)

Additional comments and more terms from Henry Bottomley, Dec 13 2000

Edited: Corrected Cf. A039991 statement. Cf. A053120 added. - Wolfdieter Lang, Aug 06 2014

A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

Original entry on oeis.org

1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

E. K. Lloyd

If k is in the sequence, then it has successor 7*k + 4*sqrt(3*(k^2 - 1)). - Lekraj Beedassy, Jun 28 2002
Chebyshev's polynomials T(n,x) evaluated at x=7.
a(n+1) give all (nontrivial) solutions of Pell equation a(n+1)^2 - 48*b(n+1)^2 = +1 with b(n+1)=A007655(n+2), n >= 0.
Also all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - Herbert Kociemba, Jun 05 2022
Also numbers x of the form 3j+1 such that x^2 = 3m^2+1. Also solutions of x in x^2 - 3*y^2 = 1 in A001075 if x = 3j+1, j=1,2,... - Cino Hilliard, Mar 05 2005
In addition to having integral standard deviation, these k consecutive integers also have integral mean. This question was posed by Jim Delany of Cal Poly in 1989. The solution appeared in the American Mathematical Monthly Vol. 97, No. 5, (May, 1990), pp. 432 as problem E3302. - Ronald S. Tiberio, Jun 23 2008
Lebl and Lichtblau give the formula a(d) = ((7+4*sqrt(3))^d + (7-4*sqrt(3))^d)/2 in Theorem 1.2(iii), p. 4. - Jonathan Vos Post, Aug 05 2008
In a (square pyramidal) pile of cannonballs, paint the ball at the top and the balls on 2 opposite sides of the base. The sequence, after its 1st term, gives the numbers of painted balls, k, in piles where the total number of balls is twice a square multiple of k. - Peter Munn, Feb 06 2025

P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Robert Israel, Table of n, a(n) for n = 1..788
Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, Problem E 3302: Averaging to Integers, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]
Tanya Khovanova, Recursive Sequences
Jiri Lebl and Daniel Lichtblau, Uniqueness of certain polynomials constant on a hyperplane, arXiv:0808.0284 [math.CV], 2008-2010.
E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.
Ronald S. Tiberio, Solution to Problem E 3302 [Broken link]
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).

a(n) = A001075(2n).
Row 2 of array A188644
Cf. A000330, A007654, A007655, A011944, A102344.

I:=[1,7]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

seq(orthopoly[T](n,7), n = 0..50); # Robert Israel, Jun 02 2015
a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
seq(simplify(a(n)), n=1..20); # Peter Luschny, Jul 26 2020

LinearRecurrence[{14,-1},{1,7},30] (* Harvey P. Dale, Dec 16 2013 *)
a[n_]:=1/2((7-4 Sqrt[3])^n+(7+4 Sqrt[3])^n); Table[a[n] // Simplify,{n,0,20}] (* Gerry Martens, May 30 2015 *)

PARI
```
a(n)=if(n<0,0,subst(poltchebi(n),x,7))
```

g(n) = forstep(x=1,n,3,y=(x^2-1)/3;if(issquare(y),print1(x","))) \\ Cino Hilliard, Mar 05 2005

a(n) = 14*a(n-1) - a(n-2).
a(n) = sqrt(12*A011944(n)^2 + 1).
a(n) ~ (1/2)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = T(n, 7) = (S(n, 14)-S(n-2, 14))/2 = T(2*n, 2) with S(n, x) := U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(-2, x) := -1, S(-1, x) := 0, S(n, 14)=A007655(n+2).
a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
a(n) = sqrt(48*A007655(n+1)^2 + 1).
G.f.: (1-7*x)/(1-14*x+x^2).
a(n) = cosh(2n*arcsinh(sqrt(3))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020
E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - Stefano Spezia, Dec 12 2022

Better description from Lekraj Beedassy, Jun 27 2002
Chebyshev comments from Wolfdieter Lang, Nov 08 2002
More terms from Vincenzo Librandi, Apr 19 2015

cp's OEIS Frontend

A001570 Numbers k such that k^2 is centered hexagonal.

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A092184 Sequence S_6 of the S_r family.

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A086645 Triangle read by rows: T(n, k) = binomial(2n, 2k).

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A061278 a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

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A002531 a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1); a(0) = a(1) = 1.

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A061278 a(n) = 5a(n-1) - 5a(n-2) + a(n-3) with a(1) = 1 and a(k) = 0 if k <= 0.

A002531 a(2n) = a(2n-1) + a(2n-2), a(2n+1) = 2a(2n) + a(2*n-1); a(0) = a(1) = 1.