A174768 -id:A174768 - cp's OEIS frontend

A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

0, 2, 36, 646, 11592, 208010, 3732588, 66978574, 1201881744, 21566892818, 387002188980, 6944472508822, 124613502969816, 2236098580947866, 40125160954091772, 720016798592704030

Offset: 0

Gary Detlefs, Feb 20 2012

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.
.
   k             x                        y
   -  -----------------------  -----------------------
A001542 {0,2,6,-1}       A001541 {1,3,6,-1}
A001353 {0,1,4,-1}       A001075 {1,2,4,-1}
A060645 {0,4,18,-1}      A023039 {1,9,18,-1}
A001078 {0,2,10,-1}      A001079 {1,5,10,-1}
A001080 {0,3,16,-1}      A001081 {1,8,16,-1}
A001109 {0,1,6,-1}       A001541 {1,3,6,-1}
A084070 {0,1,38,-1}      A078986 {1,19,38,-1}
A001084 {0,3,20,-1}      A001085 {1,10,20,-1}
A011944 {0,2,14,-1}      A011943 {1,7,14,-1}
A075871 {0,180,1298,-1}  A114047 {1,649,1298,-1}
A068204 {0,4,30,-1}      A069203 {1,15,30,-1}
A001090 {0,1,8,-1}       A001091 {1,4,8,-1}
A121740 {0,8,66,-1}      A099370 {1,33,66,-1}
A202299 {0,4,34,-1}      A056771 {1,17,34,-1}
A174765 {0,39,340,-1}    A114048 {1,179,340,-1}
a(n)    {0,2,18,-1}      A023039 {1,9,18,-1}
A174745 {0,12,110,-1}    A114049 {1,55,110,-1}
A174766 {0,42,394,-1}    A114050 {1,197,394,-1}
A174767 {0,5,48,-1}      A114051 {1,24,48,-1}
A004189 {0,1,10,-1}      A001079 {1,5,10,-1}
A174768 {0,10,102,-1}    A099397 {1,51,102,-1}
The sequence of the c parameter is listed in A180495.

Bruno Berselli, Table of n, a(n) for n = 0..500
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
Index entries for linear recurrences with constant coefficients, signature (18,-1).

Cf. A023039, A049660, A079962.

m:=16; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(2*x/(1-18*x+x^2))); // Bruno Berselli, Jun 19 2019

readlib(issqr):for x from 1 to 720016798592704030 do if issqr(20*x^2+1) then print(x) fi od;

LinearRecurrence[{18, -1}, {0, 2}, 16] (* Bruno Berselli, Feb 21 2012 *)
Table[2 ChebyshevU[-1 + n, 9], {n, 0, 16}]  (* Herbert Kociemba, Jun 05 2022 *)

makelist(expand(((2+sqrt(5))^(2*n)-(2-sqrt(5))^(2*n))/(4*sqrt(5))), n, 0, 15); /* Bruno Berselli, Jun 19 2019 */

a(n) = 18*a(n-1) - a(n-2).
From Bruno Berselli, Feb 21 2012: (Start)
G.f.: 2*x/(1-18*x+x^2).
a(n) = -a(-n) = 2*A049660(n) = ((2 + sqrt(5))^(2*n)-(2 - sqrt(5))^(2*n))/(4*sqrt(5)). (End)
a(n) = Fibonacci(6*n)/4. - Bruno Berselli, Jun 19 2019
For n>=1, a(n) = A079962(6n-3). - Christopher Hohl, Aug 22 2021

A217441 Numbers k such that 26*k+1 is a square.

Original entry on oeis.org

0, 24, 28, 100, 108, 228, 240, 408, 424, 640, 660, 924, 948, 1260, 1288, 1648, 1680, 2088, 2124, 2580, 2620, 3124, 3168, 3720, 3768, 4368, 4420, 5068, 5124, 5820, 5880, 6624, 6688, 7480, 7548, 8388, 8460, 9348, 9424, 10360, 10440, 11424, 11508, 12540, 12628

Offset: 1

Bruno Berselli, Nov 14 2012

Equivalently, numbers of the form m*(26*m+2), where m = 0,-1,1,-2,2,-3,3,...

Also, integer values of 2*h*(h+1)/13.

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. similar sequences listed in A219257.

Cf. A174768 (the squares A174768^2 belong to the sequence), A175886.

[n: n in [0..13000] | IsSquare(26*n+1)];

I:=[0,24,28,100,108]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A217441:=proc(q)
local n;
for n from 1 to q do if type(sqrt(26*n+1), integer) then print(n);
fi; od; end:
A217441(1000); # Paolo P. Lava, Feb 19 2013

Select[Range[0, 13000], IntegerQ[Sqrt[26 # + 1]] &]
CoefficientList[Series[4 x (6 + x + 6 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,24,28,100,108},50] (* Harvey P. Dale, Nov 03 2019 *)

a(n)=is(n)=issquare(26*n+1) \\ Charles R Greathouse IV, Oct 16 2015

G.f.: 4*x^2*(6 + x + 6*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (26*n*(n-1) + 11*(-1)^n*(2*n - 1) - 3)/4 + 3 = (26*n + 11*(-1)^n - 15)*(26*n + 11*(-1)^n - 11)/104.
26*a(2*n-1)+1 = A175886(4*n-3)^2, 26*a(2*n)+1 = A175886(4*n)^2.
Sum_{n>=2} 1/a(n) = 13/2 - cot(Pi/13)*Pi/2. - Amiram Eldar, Mar 17 2022

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A207832 Numbers x such that 20*x^2 + 1 is a perfect square.

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A217441 Numbers k such that 26*k+1 is a square.

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