OFFSET
1,2
COMMENTS
This sequence is the answer to the problem A1872 proposed on French mathematical site Diophante (see link).
Equivalent to the two Diophantine equations: m^2 = 10*k^2 + z and m-k = q*z for some q >= 1.
There exist solutions iff z = 1 or z = 9.
When (m, k, z) is a solution, then (19m+60k, 6m+19k, z) is another solution.
There is only one solution such that z = m-k: (13, 4, 9), see 1st example.
There exist two distinct families of solutions corresponding to z = 1 and z = 9, odd indices correspond to z = 1 and even indices to z = 9.
-> For z = 1, all solutions m of Pell equation m^2 - 10*k^2 = 1 are terms because z = 1 divides every m-k.
First few solutions (m, k) are (1, 0), (19, 6), (721, 228), (27379, 86568), ... with m = A078986(q) and corresponding k = 6*A078987(q).
-> For z = 9, solutions m must satisfy m^2 - 10*k^2 = 9 with 9 divides m-k. Among the 3 fundamental solutions (3, 0), (7, 2), (13, 4) of Pell equation m^2 -10*k^2 = 9, only (13, 4) gives solutions where 9 divides m-k.
First few solutions (m, k) are (13, 4), (487, 154), (18493, 5848), ... with m = A228209(3q).
LINKS
Diophante, A1872, Carrément magiques (in French).
Index entries for linear recurrences with constant coefficients, signature (0,38,0,-1).
FORMULA
a(2n+1) = A078986(n) for n >= 0.
a(2n) = A228209(3n) for n >= 1.
a(n+4) = 38*a(n+2) - a(n), a(1) = 1, a(2) = 13, a(3)= 19, a(4) = 487.
G.f.: x*(1 + 13*x - 19*x^2 - 7*x^3)/(1 - 38*x^2 + x^4). - Stefano Spezia, Apr 03 2021
EXAMPLE
For m = 13, 13^2 = 169, 4^2 = 16, 13^2 - 10*4^2 = 9 and 9 = 13-4 divides 13-4.
For m = 19, 19^2 = 361, 6^2 = 36, 19^2 - 10*6^2 = 1 and 1 divides 19-6 = 13.
For m = 487, 487^2 = 237169, 154^2 = 23716, 487^2 - 10*154^2 = 9 and 9 divides 487-154 = 333 = 9*37.
MATHEMATICA
LinearRecurrence[{0, 38, 0, -1}, {1, 13, 19, 487}, 24] (* Amiram Eldar, Apr 03 2021 *)
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Bernard Schott, Apr 03 2021
STATUS
approved