1,2

We solve the equation P(p)=P(r)+P(r+1) with unknowns p and r, equivalent to (6*p-1)^2=2*(6*r+2)^2+17. The diophantine equation X^2=2*Y^2+17 whose solutions give p and r are obtained by (x(n), y(n)) such that:

x(1)=5, x(2)=215, x(3)=4517, x(4)=248087 and the same recurrence relation on the odd and even indices x(n+2)=1154*x(n+1)-x(n)

y(1)=2, y(2)=152, y(3)=3194, y(4)=175424 and the same recurrence relation on the odd and even indices y(n+2)=1154*y(n+1)-y(n)

The solutions (p,r) are given by the (u(n),v(n)) such that

u(1)=1, u(2)=36, u(3)=753, u(4)=41348 and the same recurrence relation on the odd and even indices u(n+2)=1154*u(n+1)-u(n) -192 or u(n+1)=577*u(n)-96+68*(72*u(n)^2-24*u(n)-32)^0.5

v(1)=0, v(2)=25, v(3)=532, v(4)=29237 and the same recurrence relation on the odd and even indices v(n+2)=1154*v(n+1)-v(n) +384 or v(n+1)=577*v(n)+192+68*(72*u(n)^2+48*u(n)+15)^0.5

For odd and even indices respectively : a(n+2)=1331714*a(n+1)-a(n)-416160. on the odd and the even indices respectively we have also : a(n+1)=665857*a(n)-208080+19618*(1152*a(n)^2-720*a(n)-32)^0.5. the g.f fonction h such that h(z)=a(1)*z+a(2)*z^2+... is given by h(z)=((z+1925*z^2-483503*z^3+65395*z^4-22*z^5)/((1-z)*(1-1331714*z^2+z^4))

with P(m)=m*(3*m-1)/2, a(1)=1 because a(1)=P(1)=P(0)+P(1); a(2)=1926 because P(36)=1926=P(25)+P(26)=925+1001 ; a(3)=850137 because P(753)=850137=P(532)+P(533)=424270+425867 ...

(from Emeric Deutsch) a:=proc(m) if type (sqrt(18*m^2-6*m-8)/6-1/3), integer=true then m*(3*m-1)/2 else fi end : seq(a(m), m=1..100000)od;

Sequence in context: A035768 A107564 A135648 this_sequence A099482 A141593 A031766

Adjacent sequences: A133298 A133299 A133300 this_sequence A133302 A133303 A133304

nonn

Richard Choulet (richardchoulet(AT)yahoo.fr), Dec 20 2007