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

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

1,2

Comments

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.

Links

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).

Formula

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

Maple

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

Mathematica

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 *)

Prog

(Magma) [n: n in [0..13000] | IsSquare(26*n+1)];
(Magma) 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
(PARI) a(n)=is(n)=issquare(26*n+1) \\ Charles R Greathouse IV, Oct 16 2015

Crossrefs

Cf. similar sequences listed in A219257.

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

Sequence in context: A030500 A107406 A206261 * A045668 A045659 A319901

Adjacent sequences: A217438 A217439 A217440 * A217442 A217443 A217444

Keyword

nonn,easy

Author

Bruno Berselli, Nov 14 2012

Status

approved