A181433 Numbers k such that 11*k is 5 less than a square.

1, 4, 20, 29, 61, 76, 124, 145, 209, 236, 316, 349, 445, 484, 596, 641, 769, 820, 964, 1021, 1181, 1244, 1420, 1489, 1681, 1756, 1964, 2045, 2269, 2356, 2596, 2689, 2945, 3044, 3316, 3421, 3709, 3820, 4124, 4241, 4561, 4684, 5020, 5149, 5501, 5636, 6004

Offset: 1

Author

Jon Perry, Oct 20 2010

Keywords

• nonn
• easy

Comments

a(k)^k+1==0 (mod 3) for k of the form 3*(2*j+1); for other forms of k, a(k)^k-1==0 (mod 3). - Bruno Berselli, Oct 29 2010

Links

• G. C. Greubel, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Programs

Magma

&cat[ [((4+11*k)^2-5)/11, ((7+11*k)^2-5)/11] : k in [0..23] ]; // Klaus Brockhaus, Oct 20 2010

Mathematica

Sort[Flatten[ Table[{((7 + 11 k)^2 - 5)/11, ((4 + 11 k)^2 - 5)/11}, {k, 0, 20, 1}]]]
Select[Range[7000],IntegerQ[Sqrt[11#+5]]&] (* Harvey P. Dale, Nov 21 2014 *)

PARI

x='x+O('x^50); Vec(x*(1+3*x+14*x^2+3*x^3+x^4)/((1-x)^3*(1+x)^2)) \\ G. C. Greubel, Feb 25 2017

Formula

G.f.: x*(1+3*x+14*x^2+3*x^3+x^4)/((1-x)^3*(1+x)^2). - Alexander R. Povolotsky, Oct 21 2010

a(n) = (22*n*(n-1) - 5*(2*n-1)*(-1)^n + 3)/8.

a(n) = a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5) for n>5, a(1)=1, a(2)=4, a(3)=20, a(4)=29, a(5)=61.

Sum_{i=1..n} a(i) = n*(22*n^2-15*(-1)^n-13)/24.

Extensions

Formulas and more terms from Klaus Brockhaus and Bruno Berselli, Oct 20 2010

Sum added and superfluous formula removed by Bruno Berselli, Oct 22 2010 - Nov 15 2010