A001075 -id:A001075 - cp's OEIS frontend

A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) for n > 3.

1, 2, 6, 21, 77, 286, 1066, 3977, 14841, 55386, 206702, 771421, 2878981, 10744502, 40099026, 149651601, 558507377, 2084377906, 7779004246, 29031639077, 108347552061, 404358569166, 1509086724602, 5631988329241, 21018866592361, 78443478040202, 292755045568446

Offset: 1

Lambert Klasen (lambert.klasen(AT)gmx.net) and Gary W. Adamson, Jan 25 2005

Let M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1 ]; then [1,0,0]*M^n = [a(n), A001353(n), A061278(n-1)] for n > 1. Further, A001353 consists of the first differences of {a(n)}, and since a(n) = A061278(n) + 1, A001353 is also the first differences of A061278. Let v(n) = [1,0,0]*M^n; then, for n >= 0, sum(v_i(n)) = A001075(n) and v_1(n) + v_3(n) = A001835(n). The characteristic polynomial of M is x^3 - 5x^2 + 5x - 1. a(n)/a(n-1) tends to 2 + sqrt(3) = 3.732.... (see A019973) (a root of the polynomial and an eigenvalue of the matrix).
Numbers k such that the RootMeanSquare([1..6*k-5]) is an integer. - Ctibor O. Zizka, Dec 17 2008
Place a(n) blue and b(n) red balls in an urn. Draw 3 balls without replacement. Then Probability(3 red balls) = Probability(1 red and 2 blue balls); binomial(b(n),3) = binomial(b(n),1)*binomial(a(n),2); b(n) = A179167(n). - Paul Weisenhorn, Jul 01 2010
Conjecture: consecutive terms of this sequence and consecutive terms of A032908 provide all the positive integer solutions of (a+b)*(a+b+1) == 0 (mod (a*b)). - Robert Israel, Aug 26 2015
Conjecture is true: see StackExchange link. - Robert Israel, Sep 06 2015
Values of a unitary Y-frieze pattern associated to the linearly oriented quiver K3 (i.e., the quiver whose underlying graph is the complete graph on the vertices {1,2,3}, oriented such that i -> j whenever i < j). - Antoine de Saint Germain, Dec 30 2024

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
R. Israel, W. Jagy et al., Diophantine equation (x+y)(x+y+1)-kxy=0, Math StackExchange, Sep 1 2015.
Giovanni Lucca, Circle Chains Inscribed in Symmetrical Lenses and Integer Sequences, Forum Geometricorum, Volume 16 (2016) 419-427.
Index entries for linear recurrences with constant coefficients, signature (5,-5,1).

Cf. A001075, A001353, A001835, A005246, A019973, A032908, A061278, A211955.

a:=[1,2,6];; for n in [4..20] do a[n]:=5a[n-1]-5*a[n-2]+a[n-3]; od; a; #
G. C. Greubel, Dec 23 2019

a101265 n = a101265_list !! (n-1)
a101265_list = 1 : 2 : 6 : zipWith (+) a101265_list
    (map (* 5) $ tail $ zipWith (-) (tail a101265_list) a101265_list)
-- Reinhard Zumkeller, May 18 2014

I:=[1,2,6]; [n le 3 select I[n] else 5*Self(n-1) - 5*Self(n-2) + Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 07 2015

r:=sqrt(3): for n from 1 to 100 do a[n]:=(6+(3+r)*(2+r)^(n-1)+(3-r)*(2-r)^(n-1))/12: end do: # Paul Weisenhorn, Jul 01 2010
r:=sqrt(3): a[n]:=round((6+(3+r)*(2+r)^(n-1))/12): # Paul Weisenhorn, Jul 01 2010
f:= proc(n)
  option remember; local x;
  x:= procname(n-1);
  2*x + (sqrt(12*x^2 - 12*x + 1) - 1)/2
end proc:
f(1):= 1:
map(f, [$1..30]); # Robert Israel, Aug 26 2015
seq( simplify((ChebyshevU(n,2) - Chebyshev(n-1,2) + 1)/2), n=0..20); # G. C. Greubel, Dec 23 2019

LinearRecurrence[{5,-5,1},{1,2,6},25] (* Ray Chandler, Jan 27 2014 *)
CoefficientList[Series[(1-3x+x^2)/((1-x)(1-4x+x^2)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 07 2015 *)
Table[(ChebyshevU[n, 2] - ChebyshevU[n-1, 2] + 1)/2, {n, 0, 20}] (* G. C. Greubel, Dec 23 2019 *)

M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1,30,print1(([1,0,0]*M^i)[1],","))

{a(n)=polcoeff(x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)+x*O(x^n)),n)}

{a(n)=if(n==0,1,if(n==1,1,a(n-1)*(a(n-1)+1)/a(n-2)))} /* Paul D. Hanna, Apr 08 2012 */

vector(21, n, (polchebyshev(n, 2, 2) - polchebyshev(n-1, 2, 2) + 1)/2 ) \\ G. C. Greubel, Dec 23 2019

[(chebyshev_U(n,2) - chebyshev_U(n-1,2) + 1)/2 for n in (0..20)] # G. C. Greubel, Dec 23 2019

a(n) = A005246(n)*A005246(n+1). a(n+1) = a(n)*(a(n)+1)/a(n-1). - Franklin T. Adams-Watters, Apr 24 2006
a(n) = (A001835(n) + 1) / 2. - Ralf Stephan, May 16 2007
O.g.f.: x*(1-3*x+x^2)/((1-x)*(1-4*x+x^2)). - R. J. Mathar, Aug 22 2008
a(n) = 1 + A061278(n). - Ctibor O. Zizka, Dec 17 2008
a(n) = 4*a(n-1) - a(n-2) - 1. - N. Sato, Jan 21 2010
a(n) = (6+(3+r)*(2+r)^(n-1) + (3-r)*(2-r)^(n-1))/12; r=sqrt(3). - Paul Weisenhorn, Jul 01 2010
a(n+1) = a(n) * (a(n) + 1) / a(n-1) for n>1 with a(0)=1, a(1)=1. - Paul D. Hanna, Apr 08 2012
From Peter Bala, May 01 2012: (Start)
a(n+1) = 1 + Sum {k = 1..n} 2^(k-1)*binomial(n+k,2*k).
Row sums of A211955.
a(n) = T(n,u)*T(n+1,u)/u with u = sqrt(3) and T(n,x) denotes the Chebyshev polynomial of the first kind.
Sum_{n >= 0} 1/a(n) = sqrt(3). In fact, 3 - (Sum_{n = 0..2*N} 1/a(n))^2 = 2/(A001835(N+1))^2 and 3 - (Sum_{n = 0..2*N+1} 1/a(n))^2 = 3/(A001075(N+1))^2. (End)
From Robert Israel, Aug 26 2015: (Start)
(a(n) + a(n+1))*(a(n) + a(n+1) + 1) = 6 * a(n) * a(n+1).
a(n+1) = 2*a(n) + (sqrt(12*a(n)^2 - 12*a(n) + 1) - 1)/2. (End)
a(n) = (ChebyshevU(n, 2) - ChebyshevU(n-1, 2) + 1)/2 = (ChebyshevT(n, 2) + ChebyshevU(n, 2) + 2)/4. - G. C. Greubel, Dec 23 2019
a(n) = (1+a(n-1))*(1+a(n-2))/a(n-3) for n > 3. - Antoine de Saint Germain, Dec 30 2024

a(26)-a(27) from Vincenzo Librandi, Sep 07 2015

A201730 Triangle T(n,k), read by rows, given by (2,1/2,3/2,0,0,0,0,0,0,0,...) DELTA (0,1/2,-1/2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 2, 0, 5, 1, 0, 14, 6, 0, 0, 41, 26, 1, 0, 0, 122, 100, 10, 0, 0, 0, 365, 363, 63, 1, 0, 0, 0, 1094, 1274, 322, 14, 0, 0, 0, 0, 3281, 4372, 1462, 116, 1, 0, 0, 0, 0, 9842, 14760, 6156, 744, 18, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 04 2011

Riordan array ((1-2x)/(1-4x+3x^2),x^2/(1-4x+3x^2)).

A007318*A201701 as lower triangular matrices.

			Triangle begins:
1
2, 0
5, 1, 0
14, 6, 0, 0
41, 26, 1, 0, 0
122, 100, 10, 0, 0, 0
365, 363, 63, 1, 0, 0, 0

Cf. A007051 (1st column), A261064 (2nd column).

A201730 := proc(n,k)
    (1-2*x)/(1-4*x+(3-y)*x^2) ;
    coeftayl(%,y=0,k) ;
    coeftayl(%,x=0,n) ;
end proc:
seq(seq(A201730(n,k),k=0..n),n=0..12) ; # R. J. Mathar, Dec 06 2011

m = 13;
(* DELTA is defined in A084938 *)
DELTA[Join[{2, 1/2, 3/2}, Table[0, {m}]], Join[{0, 1/2, -1/2}, Table[0, {m}]], m] // Flatten (* Jean-François Alcover, Feb 19 2020 *)

G.f.: (1-2x)/(1-4x+(3-y)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A139011(n), A000079(n), A007051(n), A006012(n), A001075(n), A081294(n), A001077(n), A084059(n), A108851(n), A084128(n), A081340(n), A084132(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k, k>+0} T(n+k,k) = A081704(n) .
T(n,k) = 3*T(n-1,k)+ Sum_{j>0} T(n-1-j,k-1).
T(n,k) = 4*T(n-1,k)+ T(n-2,k-1) - 3*T(n-2,k) with T(0,0)=1, T(1,0)= 2, T(1,1) = 0 and T(n,k) = 0 if k<0 or if n

A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

Original entry on oeis.org

1, 17, 577, 19601, 665857, 22619537, 768398401, 26102926097, 886731088897, 30122754096401, 1023286908188737, 34761632124320657, 1180872205318713601, 40114893348711941777, 1362725501650887306817, 46292552162781456490001

Offset: 0

Henry Bottomley, Aug 16 2000

The sequence satisfies the Pell equation a(n)^2 - 18 * A202299(n+1)^2 = 1. - Vincenzo Librandi, Dec 19 2011
Also numbers n such that n - 1 and 2*n + 2 are squares. - Arkadiusz Wesolowski, Mar 15 2015
And they, n - 1 and 2*n + 2, are the squares of A005319 and A003499. - Michel Marcus, Mar 15 2015
This sequence {a(n)} gives all the nonnegative integer solutions of the Pell equation a(n)^2 - 32*(3*A091761(n))^2 = +1. - Wolfdieter Lang, Mar 09 2019

			G.f. = 1 + 17*x + 577*x^2 + 19601*x^3 + 665857*x^4 + 22619537*x^5 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..600 (terms 0..200 from Vincenzo Librandi)
Tanya Khovanova, Recursive Sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (34,-1).

Cf. A001075, A001541, A001091, A001079, A023038, A011943, A001081, A023039, A001085 and note relationship with square triangular number sequences A001110 and A001109. A091761.

Row 3 of array A188644.

I:=[1, 17]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Dec 18 2011

LinearRecurrence[{34,-1},{1,17},30] (* Vincenzo Librandi, Dec 18 2011 *)
a[ n_] := ChebyshevT[ 2 n, 3]; (* Michael Somos, May 28 2014 *)

makelist(expand(((17+sqrt(288))^n+(17-sqrt(288))^n))/2, n, 0, 15); /* Vincenzo Librandi, Dec 18 2011 */

{a(n) = polchebyshev( n, 1, 17)}; /* Michael Somos, Apr 05 2019 */

[lucas_number2(n,34,1)/2 for n in range(0,15)] # Zerinvary Lajos, Jun 27 2008

a(n) = (r^n + 1/r^n)/2 with r = 17 + sqrt(17^2-1).
a(n) = 16*A001110(n) + 1 = A001541(2n) = (4*A001109(n))^2 + 1 = 3*A001109(2n-1) - A001109(2n-2) = A001109(2n) - 3*A001109(2n-1).
a(n) = T(n, 17) = T(2*n, 3) with T(n, x) Chebyshev's polynomials of the first kind. See A053120. T(n, 3)= A001541(n).
G.f.: (1-17*x)/(1-34*x+x^2).
G.f.: (1 - 17*x / (1 - 288*x / (17 - x))). - Michael Somos, Apr 05 2019
a(n) = cosh(2n*arcsinh(sqrt(8))). - Herbert Kociemba, Apr 24 2008
a(n) = (a^n + b^n)/2 where a = 17 + 12*sqrt(2) and b = 17 - 12*sqrt(2); sqrt(a(n)-1)/4 = A001109(n). - James R. Buddenhagen, Dec 09 2011
a(-n) = a(n). - Michael Somos, May 28 2014
a(n) = sqrt(1 + 32*9*A091761(n)^2), n >= 0. See one of the Pell comments above. - Wolfdieter Lang, Mar 09 2019

More terms from James Sellers, Sep 07 2000

Chebyshev comments from Wolfdieter Lang, Nov 29 2002

A083881 a(n) = 6a(n-1) - 6a(n-2), with a(0)=1, a(1)=3.

Original entry on oeis.org

1, 3, 12, 54, 252, 1188, 5616, 26568, 125712, 594864, 2814912, 13320288, 63032256, 298271808, 1411437312, 6678993024, 31605334272, 149558047488, 707716279296, 3348949390848, 15847398669312, 74990695670784, 354859782008832

Offset: 0

Paul Barry, May 08 2003

Binomial transform of A001075.

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

Cf. A083882.

Cf. A030192.

a:=[1,3];; for n in [3..30] do a[n]:=6*a[n-1]-6*a[n-2]; od; a; # G. C. Greubel, Aug 01 2019

I:=[1,3]; [n le 2 select I[n] else 6*Self(n-1) -6*Self(n-2): n in [1..30]]; // G. C. Greubel, Aug 01 2019

f[n_]:= Simplify[(3 + Sqrt@3)^n + (3 - Sqrt@3)^n]/2; Array[f, 30, 0] (* Robert G. Wilson v, Oct 31 2010 *)
LinearRecurrence[{6,-6}, {1,3}, 30] (* G. C. Greubel, Aug 01 2019 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-6*x+6*x^2)) \\ G. C. Greubel, Aug 01 2019

((1-3*x)/(1-6*x+6*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Aug 01 2019

a(n) = ((3-sqrt(3))^n + (3+sqrt(3))^n)/2.
a(n) = Sum_{k=0..floor(n/2)} binomial(n, 2*k)*3^(n-k).
G.f.: (1-3*x)/(1-6*x+6*x^2).
E.g.f.: exp(3*x) * cosh(x*sqrt(3)).
a(n) = right and left terms in M^n * [1 1 1] where M = the 3X3 matrix [1 1 1 / 1 4 1 / 1 1 1]. M^n * [1 1 1] = [a(n) A030192(n) a(n)]. E.g. a(3) = 54 since M^3 * [1 1 1] = [54 144 54] = [a(3) A030192(3) a(3)]. - Gary W. Adamson, Dec 18 2004
a(n) = Sum_{k, 0<=k<=n}3^k*A098158(n,k). - Philippe Deléham, Dec 04 2006
G.f.: A(x) = G(0) where G(k) =  1 + 3*x/((1-3*x) - x*(1-3*x)/(x + (1-3*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 29 2012.

A084917 Positive numbers of the form 3*y^2 - x^2.

Original entry on oeis.org

2, 3, 8, 11, 12, 18, 23, 26, 27, 32, 39, 44, 47, 48, 50, 59, 66, 71, 72, 74, 75, 83, 92, 98, 99, 104, 107, 108, 111, 122, 128, 131, 138, 143, 146, 147, 156, 162, 167, 176, 179, 183, 188, 191, 192, 194, 200, 207, 218, 219, 227, 234, 236, 239, 242, 243, 251, 263, 264, 275, 282, 284

Offset: 1

Roger Cuculière, Jul 14 2003

Positive integers k such that x^2 - 4xy + y^2 + k = 0 has integer solutions. (See the CROSSREFS section for sequences relating to solutions for particular k.)
Comments on method used, from Colin Barker, Jun 06 2014: (Start)
In general, we want to find the values of f, from 1 to 400 say, for which x^2 + bxy + y^2 + f = 0 has integer solutions for a given b.
In order to solve x^2 + bxy + y^2 + f = 0 we can solve the Pellian equation x^2 - Dy^2 = N, where D = b*b - 4 and N = 4*(b*b - 4)*f.
But since sqrt(D) < N, the classical method of solving x^2 - Dy^2 = N does not work. So I implemented the method described in the 1998 sci.math reference, which says:
"There are several methods for solving the Pellian equation when |N| > sqrt(d). One is to use a brute-force search. If N < 0 then search on y = sqrt(abs(n/d)) to sqrt((abs(n)(x1 + 1))/(2d)) and if N > 0 search on y = 0 to sqrt((n(x1 - 1))/(2d)) where (x1, y1) is the minimum positive solution (x, y) to x^2 - dy^2 = 1. If N < 0, for each positive (x, y) found by the search, also take (-x, y). If N > 0, also take (x, -y). In either case, all positive solutions are generated from these using (x1, y1) in the standard way."
Incidentally all my Pell code is written in B-Prolog, and is somewhat voluminous. (End)
Also, positive integers of the form -x^2 + 2xy + 2y^2 of discriminant 12. - N. J. A. Sloane, May 31 2014 [Corrected by Klaus Purath, May 07 2023]
The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.
The equivalent sequence for x^2 - 5xy + y^2 + k = 0 is A237351.
A positive k does not appear in this sequence if and only if there is no integer solution of x^2 - 3*y^2 = -k with (i) 0 < y^2 <= k/2 and (ii) 0 <= x^2 <= k/2. See the Nagell reference Theorems 108 a and 109, pp. 206-7, with D = 3, N = k and (x_1,y_1) = (2,1). - Wolfdieter Lang, Jan 09 2015
From Klaus Purath, May 07 2023: (Start)
There are no squares in this sequence. Products of an odd number of terms as well as products of an odd number of terms and any terms of A014209 belong to the sequence.
Products of an even number of terms are terms of A014209. The union of this sequence and A014209 is closed under multiplication.
A positive number belongs to this sequence if and only if it contains an odd number of prime factors congruent to {2, 3, 11} modulo 12. If it contains prime factors congruent to {5, 7} modulo 12, these occur only with even exponents. (End)
From Klaus Purath, Jul 09 2023: (Start)
Any term of the sequence raised to an odd power also belongs to the sequence. Proof: t^(2n+1) = t*t^2n = (3*x^2 - y^2)*t^2n = 3*(x*t^n)^2 - (y*t^n)^2. It seems that t^(2n+1) occurs only if t also is in the sequence.
Joerg Arndt has proved that there are no squares in this sequence: Assume s^2 = 3*y^2 - x^2, then s^2 + x^2 = 3 * y^2, but the sum of two squares cannot be 3 * y^2, qed. (End)
That products of any 3 terms belong to the sequence can be proved by the following identity:  (na^2 - b^2) (nc^2 - d^2) (ne^2 - f^2) = n[a(nce + df) + b(cf + de)]^2 - [na(cf + de) + b(nce + df)]^2. This can be verified by expanding both sides of the equation. - Klaus Purath, Jul 14 2023

			11 is in the sequence because 3 * 3^2 - 4^2 = 27 - 16 = 11.
12 is in the sequence because 3 * 4^2 - 6^2 = 48 - 36 = 12.
13 is not in the sequence because there is no solution in integers to 3y^2 - x^2 = 13.
From _Wolfdieter Lang_, Jan 09 2015: (Start)
Referring to the Jan 09 2015 comment above.
k = 1 is out because there is no integer solution of (i) 0 < y^2 <= 1/2.
For k = 4, 5, 6, and 7 one has y = 1, x = 0, 1 (and the negative of this). But x^2 - 3 is not -k for these k and x values. Therefore, these k values are missing.
For k = 8 .. 16 one has y = 1, 2 and x = 0, 1, 2. Only y = 2 has a chance and only for k = 8, 11 and 12 the x value 2, 1 and 0, respectively, solves x^2 - 12 = -k. Therefore 9, 10, 13, 14, 15, 16 are missing.
... (End)

T. Nagell, Introduction to Number Theory, Chelsea Publishing Company, New York, 1964.

Sci.math, General Pell equation: x^2 - N*y^2 = D, 1998
Sci.math, General Pell equation: x^2 - N*y^2 = D, 1998 (Edited and cached copy)
N. J. A. Sloane et al., Binary Quadratic Forms and OEIS (Index to related sequences, programs, references)

With respect to solutions of the equation in the early comment, see comments etc. in: A001835 (k = 2), A001075 (k = 3), A237250 (k = 11), A003500 (k = 12), A082841 (k = 18), A077238 (k = 39).
Cf. A031363, A237351.
A141123 gives the primes.
For a list of sequences giving numbers and/or primes represented by binary quadratic forms, see the "Binary Quadratic Forms and OEIS" link.

r[n_] := Reduce[n == 3*y^2 - x^2 && x > 0 && y > 0, {x, y}, Integers]; Reap[For[n = 1, n <= 1000, n++, rn = r[n]; If[rn =!= False, Print["n = ", n, ", ", rn /. C[1] -> 1 // Simplify]; Sow[n]]]][[2, 1]] (* Jean-François Alcover, Jan 21 2016 *)
Select[Range[300],Length[FindInstance[3y^2-x^2==#,{x,y},Integers]]>0&] (* Harvey P. Dale, Apr 23 2023 *)

Terms 26 and beyond from Colin Barker, Feb 06 2014

A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 208, 1552, 11584, 86464, 645376, 4817152, 35955712, 268377088, 2003193856, 14952042496, 111603564544, 833020346368, 6217748512768, 46409906716672, 346408259682304, 2585626450591744, 19299378566004736

Offset: 0

Philippe Deléham, Feb 29 2004

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (8,-4).

Cf. A001075.

Sum_{k>=0} A086645(n,k)*m^k for m = 0, 1, 2, 4 gives A000007, A081294, A001541, A083884.

a:=[1,4];; for n in [3..20] do a[n]:=8*a[n-1]-4*a[n-2]; od; a; # G. C. Greubel, Feb 03 2019

m:=20; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( (1-4*x)/(1-8*x+4*x^2) )); // G. C. Greubel, Feb 03 2019

LinearRecurrence[{8,-4}, {1,4}, 20] (* G. C. Greubel, Feb 03 2019 *)

my(x='x+O('x^20)); Vec((1-4*x)/(1-8*x+4*x^2)) \\ G. C. Greubel, Feb 03 2019

[lucas_number2(n,8,4)/2 for n in range(0,21)] # Zerinvary Lajos, Jul 08 2008

a(n) = Sum_{k>=0} binomial(2*n, 2*k)*3^k = Sum_{k>=0} A086645(n, k)*3^k.
a(n) = 2^n*A001075(n).
G.f.: (1-4*x)/(1-8*x+4*x^2). - Philippe Deléham, Sep 07 2009
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x*(3*k-4)/(x*(3*k-1) - 1/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 28 2013
From Peter Bala, Feb 19 2022: (Start)
a(n) = Sum_{k = 0..floor(n/2)} 4^(n-2*k)*12^k*binomial(n,2*k).
a(n) = [x^n] (4*x + sqrt(1 + 12*x^2))^n.
G.f.: A(x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 8*x + 4*x^2) is the g.f. of A069835.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. (End)

Corrected by T. D. Noe, Nov 07 2006

A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.

Original entry on oeis.org

0, 1, 16, 225, 3136, 43681, 608400, 8473921, 118026496, 1643897025, 22896531856, 318907548961, 4441809153600, 61866420601441, 861688079266576, 12001766689130625, 167163045568562176, 2328280871270739841, 32428769152221795600, 451674487259834398561

Offset: 0

Wolfdieter Lang, Oct 18 2004

Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - Ctibor O. Zizka, Oct 15 2010

Consequently A049451(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011

Colin Barker, Table of n, a(n) for n = 0..874
S. Barbero, U. Cerruti, and N. Murru, On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A001075, A001353, A049451, A092184.

LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* Michael De Vlieger, Feb 23 2021 *)

concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ Colin Barker, Jun 15 2015

a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12.
a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2).
a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.
G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).
Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - Creighton Dement, Nov 01 2004
a(n) = (A001075(n)^2-1)/3. - Parker Grootenhuis, Nov 28 2017

A011916 a(n) = ((b(n)-1)+sqrt(3b(n)^2-4b(n)+1))/2, where b(n) is A011922(n).

Original entry on oeis.org

0, 3, 44, 615, 8568, 119339, 1662180, 23151183, 322454384, 4491210195, 62554488348, 871271626679, 12135248285160, 169022204365563, 2354175612832724, 32789436375292575, 456697933641263328, 6360981634602394019

Offset: 0

Mario Velucchi (mathchess(AT)velucchi.it)

Integers k such that k^2 = Sum_{i=1..x} (k+i) for some value of x. 3 is a term because 3^2=9 and 4+5=9; 44 is a term because 44^2=1936 and the sum of (45,46,47,...,76) = 1936. - Gil Broussard, Dec 23 2008
Also the index of the first of two consecutive octagonal numbers whose sum is equal to the sum of two consecutive squares. - Colin Barker, Dec 20 2014
Also the index of a triangular number included in A239071. - Ivan Neretin, May 31 2015

Mario Velucchi, "Seeing couples" in Recreational and Educational Computing, to appear 1997. [apparently never materialized, Colin Barker, Dec 23 2014]

Colin Barker, Table of n, a(n) for n = 0..874
Index entries for linear recurrences with constant coefficients, signature (15,-15,1).

Cf. A011918, A109437.

RecurrenceTable[{a[n] == 15 a[n - 1] - 15 a[n - 2] + a[n - 3], a[0] == 0, a[1] == 3, a[2] == 44}, a, {n, 0, 17}] (* Michael De Vlieger, Jul 02 2015 *)
LinearRecurrence[{15,-15,1},{0,3,44},30] (* Harvey P. Dale, Jul 26 2018 *)

{a(n) = if( n<0, n = -n; polcoeff( x*(1 - 3*x) / ((x-1) * (x^2 - 14*x + 1)) + x * O(x^n), n), polcoeff( x*(x - 3) / ((x-1) * (x^2 - 14*x + 1)) + x * O(x^n), n))} /* Michael Somos, Jul 27 2012 */

concat(0, Vec(x*(-3+x)/((x-1)*(x^2-14*x+1)) + O(x^100))) \\ Colin Barker, Dec 20 2014

From R. J. Mathar, Apr 15 2010: (Start)
a(n) = +15*a(n-1) -15*a(n-2) +a(n-3).
G.f.: x*(-3 + x) / ((x - 1)*(x^2 - 14*x + 1)). (End)
From Michael Somos, Jul 27 2012: (Start)
a(n) = A109437(2*n).
a(-1 - n) = -A109437(2*n + 1). (End)
a(n) = (A001353(n+1)^2 - A001075(n)^2)/4. - Richard R. Forberg, Aug 26 2013
a(n) = (-2-(7-4*sqrt(3))^n*(-1+sqrt(3))+(1+sqrt(3))*(7+4*sqrt(3))^n)/12. - Colin Barker, Mar 05 2016

More terms from R. J. Mathar, Apr 15 2010

Added a(0)=0, Michael Somos, Jul 27 2012

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A007654 Numbers k such that the standard deviation of 1,...,k is an integer.

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A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5*a(n-1) - 5*a(n-2) + a(n-3) for n > 3.

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A201730 Triangle T(n,k), read by rows, given by (2,1/2,3/2,0,0,0,0,0,0,0,...) DELTA (0,1/2,-1/2,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

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A056771 a(n) = a(-n) = 34*a(n-1) - a(n-2), and a(0)=1, a(1)=17.

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A083881 a(n) = 6*a(n-1) - 6*a(n-2), with a(0)=1, a(1)=3.

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A101265 a(1) = 1, a(2) = 2, a(3) = 6; a(n) = 5a(n-1) - 5a(n-2) + a(n-3) for n > 3.

A083881 a(n) = 6a(n-1) - 6a(n-2), with a(0)=1, a(1)=3.

A090965 a(n) = 8a(n-1) - 4a(n-2), where a(0) = 1, a(1) = 4.

A011916 a(n) = ((b(n)-1)+sqrt(3b(n)^2-4b(n)+1))/2, where b(n) is A011922(n).