cp's OEIS Frontend

Chebyshev's polynomials T(n,x) evaluated at x=8.

The a(n) give all (unsigned, integer) solutions of Pell equation a(n)^2 - 63*b(n)^2 = +1 with b(n)= A077412(n-1), n>=1 and b(0)=0.

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(7). - Benoit Cloitre, Feb 14 2004

a(7+14k)-1 and a(7+14k)+1 are consecutive odd powerful numbers. The first pair is 130576328+-1. See A076445. - T. D. Noe, May 04 2006

a(n)^2 - 7 * A001080(n)^2 = 1 (this property is equivalent to the second comment). - Vincenzo Librandi, Feb 17 2013

a(n+3)*a(n) - a(n+2)*a(n+1) = 16*63. - Bruno Berselli, Feb 18 2013

Entries are == 1 (mod 3). - N. J. A. Sloane, Sep 22 2007

lim(n -> Infinity, a(n)/a(n-1)) = 8 + 3*sqrt(7). - Ant King, Nov 03 2011

a(n) == 0 (modulo 4) since no integer == 3 (modulo 4) can be represented as the sum of two squares.

This sequence has as a subsequence 72, 288, 800, 1800, ... which is 8 * (triangular numbers)^2. Proof: If x = 8*(n(n+1)/2)^2 then x = (n(n+1))^2 + (n(n+1))^2, x+1 = ((n-1)(n+1))^2 + (n(n+2))^2 and x+2 = (n^2+n-1)^2 + (n^2+n+1)^2. See A254371 - Joshua Zucker, Nov 01 2002

From Altug Alkan, Apr 13 2016: (Start)

If n is in this sequence, so is n*(n+2). Proof:

If n is in this sequence, then n = a^2 + b^2, n+1 = c^2 + d^2, n+2 = e^2 + f^2 for a, b, c, d, e, f being nonzero integers.

So, n*(n+2) = (a^2 + b^2)*(e^2 + f^2) = (a*e + b*f)^2 + (a*f - b*e)^2. Note that a*f cannot be equal to b*e because of their definitions.

n*(n+2) + 1 = n^2 + 2*n + 1 = (n+1)^2. Since we know that n mod 4 = 0, then n+1 cannot be of the form 2*k^2, that is, c and d must be different. So (n+1)^2 is the sum of two nonzero squares because n+1 = c^2 + d^2.

n*(n+2) + 2 = (n+1)^2 + 1, that is obviously the sum of two nonzero squares.

So if n is in this sequence, then n*(n+2), n*(n+2) + 1 and n*(n+2) + 2 are the sums of two nonzero squares, that is n*(n+2) must also be member of this sequence.

Note that it can be produced by repeating of this result and n*(n+2)*(n*(n+2)+2)*(n*(n+2)*(n*(n+2)+2)+2)... is always a member, if n is a member. (End)

For k > 0, 25*k^2*(10*k+2)^2 and 8*A001080(k)^2 are terms. - Jinyuan Wang, Feb 23 2019

Denote as {a,b,c,d} the second-order linear recurrence a(n) = c*a(n-1) + d*a(n-2) with initial terms a, b. The following sequences and recurrence formulas are related to integer solutions of k*x^2 + 1 = y^2.

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k x y

- ----------------------- -----------------------

2 A001542 {0,2,6,-1} A001541 {1,3,6,-1}

3 A001353 {0,1,4,-1} A001075 {1,2,4,-1}

5 A060645 {0,4,18,-1} A023039 {1,9,18,-1}

6 A001078 {0,2,10,-1} A001079 {1,5,10,-1}

7 A001080 {0,3,16,-1} A001081 {1,8,16,-1}

8 A001109 {0,1,6,-1} A001541 {1,3,6,-1}

10 A084070 {0,1,38,-1} A078986 {1,19,38,-1}

11 A001084 {0,3,20,-1} A001085 {1,10,20,-1}

12 A011944 {0,2,14,-1} A011943 {1,7,14,-1}

13 A075871 {0,180,1298,-1} A114047 {1,649,1298,-1}

14 A068204 {0,4,30,-1} A069203 {1,15,30,-1}

15 A001090 {0,1,8,-1} A001091 {1,4,8,-1}

17 A121740 {0,8,66,-1} A099370 {1,33,66,-1}

18 A202299 {0,4,34,-1} A056771 {1,17,34,-1}

19 A174765 {0,39,340,-1} A114048 {1,179,340,-1}

20 a(n) {0,2,18,-1} A023039 {1,9,18,-1}

21 A174745 {0,12,110,-1} A114049 {1,55,110,-1}

22 A174766 {0,42,394,-1} A114050 {1,197,394,-1}

23 A174767 {0,5,48,-1} A114051 {1,24,48,-1}

24 A004189 {0,1,10,-1} A001079 {1,5,10,-1}

26 A174768 {0,10,102,-1} A099397 {1,51,102,-1}

The sequence of the c parameter is listed in A180495.

The corresponding solutions y1(n) are given in A307169.

The (generalized) Pell equation x^2 - 7*y^2 = 9 has two proper classes of positive solutions (x1(n), y1(n)) and (x2(n), y2(n)), for n >= 1, where x2 = A307172, and y2 = A307173.

The improper class of nonnegative solutions is given by (xi(n) = 3*X(n), yi(n) = 3*Y(n)), with the nonnegative solutions of the Pell equation X^2 - 7*Y^2 = +1, given by X(n) = A001081(n) and Y(n) = A001080(n), for n >= 0.

The proper positive solutions (x1(n), y1(n)) are given in matrix notation by -R(0)*R(2)*Auto(n)*R^{-1}(6)*(1, 0)^T (T for transposed) with the R-matrix R(t) = Matrix([[0, -1],[1, t]]) and its inverse R^{-1}(t) = Matrix([t, 1],[-1, 0]) and the automorphic matrix Auto = Matrix([2, 9],[3, 14]). The matrix power Auto^n can be given in terms of Chebyshev S-polynomials S(n, x=16) from A077412 as Auto^n = Matrix([S(n, 16) - 14*S(n-1, 16), 9*S(n-1, 16)],[3*S(n-1, 16), S(n, 16) - 2*S(n-1, 16)]).

This results from the reduced principal binary quadratic form F_p = [1, 4, -3] of the non-reduced Pell form FPell = [1, 0, -7], and the primitive representative parallel form FPara1 = [9, 8, 1] for discriminant 4*7 = 28 and the representation of 9. These forms are then connected via equivalence transformations using R(t) matrices.

This is a strong divisibility sequence, that is, GCD(a(n),a(m)) = a(GCD(n,m)) for all positive integers n and m. Consequently, this is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Sep 01 2019

Florian Luca proved that this sequence is infinite, by showing that 37*x(7*k) + 98*y(7*k) is in the sequence, where x(k) = A001081(k) and y(k) = A001080(k) are solutions of the Pell equation x^2 - 7*y^2 = 1. The sequence of these numbers is 37, 9667939010, 2524807950507510523, 659360302164952911361460078, ... - Amiram Eldar, Aug 22 2018

a(7) <= 457189690981. - Giovanni Resta, Aug 23 2018

The full set of integer solutions to this equation consists of the pairs [X(i),Y(i)] = [1+-A001081(i), Y(i)=A001080(i)]. The present generates every second one of them: a(n) = [A001081(2n)-1]/2. - R. J. Mathar, Nov 20 2007

Also, integers w >= 1 for each row n >= 1 such that z+(1/z) is an integer, where x = A000037(n), y = w*sqrt(x), and z = (y+ceiling(y))/2.

All terms of row n are positive integer multiples of T(n, 1).

Limit_{k->oo} T(n, k+1)/T(n, k) = (sqrt(b^2-4)+b)/2 where b=T(n, 2)/T(n, 1).