A219259 - OEIS

%I #42 Dec 20 2024 09:24:54

%S 0,23,27,96,104,219,231,392,408,615,635,888,912,1211,1239,1584,1616,

%T 2007,2043,2480,2520,3003,3047,3576,3624,4199,4251,4872,4928,5595,

%U 5655,6368,6432,7191,7259,8064,8136,8987,9063,9960,10040,10983,11067,12056,12144

%N Numbers k such that 25*k+1 is a square.

%C Equivalently, numbers of the form m*(25*m+2), where m = 0,-1,1,-2,2,-3,3,...

%C Also, integer values of h*(h+2)/25.

%C Exponents in the expansion of Product_{n >= 1} (1 - q^(50*n))*(1 - q^(50*n-23))*(1 - q^(50*n-27)) = 1 - q^23 - q^27 + q^96 + q^104 - q^219 - q^231 + + - - .... - _Peter Bala_, Dec 18 2024

%H Bruno Berselli, <a href="/A219259/b219259.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-2,-1,1).

%F G.f.: x^2*(23 + 4*x + 23*x^2)/((1 + x)^2*(1 - x)^3).

%F a(n) = a(-n+1) = (50*n*(n-1) + 21*(-1)^n*(2*n - 1) + 5)/8 + 2.

%F 25*a(n)+1 = A047209(A197652(n+1))^2.

%F Sum_{n>=2} 1/a(n) = 25/4 - cot(2*Pi/25)*Pi/2. - _Amiram Eldar_, Mar 17 2022

%p A219259:=proc(q)

%p local n;

%p for n from 1 to q do if type(sqrt(25*n+1), integer) then print(n);

%p fi; od; end:

%p A219259(1000); # _Paolo P. Lava_, Feb 19 2013

%t Select[Range[0, 13000], IntegerQ[Sqrt[25 # + 1]] &]

%t CoefficientList[Series[x (23 + 4 x + 23 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* _Vincenzo Librandi_, Aug 18 2013 *)

%o (Magma) [n: n in [0..13000] | IsSquare(25*n+1)];

%o (Magma) I:=[0,23,27,96,104]; [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

%Y Cf. similar sequences listed in A219257.

%Y Cf. A047209, A197652.

%K nonn,easy

%O 1,2

%A _Bruno Berselli_, Nov 19 2012