OFFSET
1,2
COMMENTS
For the generic case x^2 + (x + p^2)^2 = y^2 with p = m^2 - 2 a (prime) number in A028871, m>=7 (means p>=47), the first five consecutive solutions are: (0; p^2), (2*m^3+2*m^2-4*m-4; m^4+2*m^3-4*m-4), (4*m^3+8*m^2+8*m; m^4+4*m^3+12*m^2+8*m+4), (3*m^4-20*m^3+44*m^2-40*m+12; 5*m^4-28*m^3+60*m^2-56*m+20), (3*m^4-10*m^3+2*m^2+20*m-16; 5*m^4-14*m^3+28*m-20) and the other solutions are defined by: (X(n); Y(n))= (3*X(n-5)+2*Y(n-5)+p^2; 4*X(n-5)+3*Y(n-5)+2*p^2).
X(n) = 6*X(n-5) - X(n-10) + 2*p^2, and Y(n) = 6*Y(n-5) - Y(n-10) (can be easily proved using X(n) = 3*X(n-5) + 2*Y(n-5) + p^2, and Y(n) = 4*X(n-5) + 3*Y(n-5) + 2*p^2).
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,6,-6,0,0,0,-1,1).
FORMULA
a(n) = 6*a(n-5) - a(n-10) + 4418 for n >= 11; a(1)=0, a(2)=752, a(3)=1820, a(4)=2231, a(5)=3995, a(6)=6627, a(7)=10575, a(8)=16511, a(9)=18840, a(10)=28952.
From Colin Barker, Feb 04 2020: (Start)
G.f.: x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)).
a(n) = a(n-1) + 6*a(n-5) - 6*a(n-6) - a(n-10) + a(n-11) for n>11.
(End)
EXAMPLE
For p=47 (m=7) the first five (5) consecutive solutions are (0, 2209), (752, 3055), (1820, 4421), (2231, 4969), (3995, 7379).
MATHEMATICA
LinearRecurrence[{1, 0, 0, 0, 6, -6, 0, 0, 0, -1, 1}, {0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952, 44180}, 40] (* Jean-François Alcover, Feb 08 2020 *)
PROG
(Magma) I:=[0, 752, 1820, 2231, 3995, 6627, 10575, 16511, 18840, 28952]; [n le 10 select I[n] else 6*Self(n-5) - Self(n-10)+4418: n in [1..100]];
(PARI) concat(0, Vec(x^2*(752 + 1068*x + 411*x^2 + 1764*x^3 + 2632*x^4 - 564*x^5 - 472*x^6 - 137*x^7 - 472*x^8 - 564*x^9) / ((1 - x)*(1 - 6*x^5 + x^10)) + O(x^40))) \\ Colin Barker, Feb 04 2020
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mohamed Bouhamida, Feb 04 2020
STATUS
approved