A124174 - OEIS

Sophie Germain triangular numbers are one of an infinite number of triangular number sets tr where 2*tn^2*tr + tn is a triangular number: tr and tn both also being triangular numbers with tn being held constant. For the present numbers, a(n) = tr, 8*(2*tr + 1) + 1 = 16*tr + 9 is also a square, the square root of which is 2*y+1 with y being the argument of the triangular number 2*tr + 1. Now (1/2)*(y^2+y) = a^2 + a + 1 from the definition of Sophie Germain triangular numbers. Multiply both sides by 4 and subtract 3 to get 2*y^2 + 2*y - 3 = 4*a^2 + 4*a + 1 (a square). Cf. A124124: Numbers y such that 2*y^2 + 2*y - 3 is a square. The values y are the same y such that 2*y+1 = sqrt(16*tr + 9). - Kenneth J Ramsey, Jun 25 2011

Values of k such that 2*k+1 and 9*k+1 are both triangular numbers. - Colin Barker, Jun 29 2016

a(n) = 35*(a(n-2) - a(n-4)) + a(n-6).

From Peter Pein, Dec 04 2006: (Start)

a(n) = -11/32 + (-3 - 2*sqrt(2))^n/64 + (5*(3 - 2*sqrt(2))^n)/32 + (-3 - 2*sqrt(2))^n/(32*sqrt(2)) - (5*(3 - 2*sqrt(2))^n)/(32*sqrt(2)) + (-3 + 2*sqrt(2))^n/64 - (-3 + 2*sqrt(2))^n/(32*sqrt(2)) + (5*(3 + 2*sqrt(2))^n)/32 + (5*(3 + 2*sqrt(2))^n)/(32*sqrt(2));

O.g.f.: (x*(1 + 9*x + x^2))/((1 - x)*(1 - 6*x + x^2)*(1 + 6*x + x^2));

E.g.f.: (-22*exp(x) + exp(-3*x+2*x*sqrt(2))*(1-sqrt(2)) - 5*exp(3*x-2*x*sqrt(2))*(-2 + sqrt(2)) + exp(-3*x-2*x*sqrt(2))*(1+sqrt(2)) + 5*exp(3*x+2*x*sqrt(2))*(2+sqrt(2)))/64. (End)

a(n) = a(n-1) + 34*a(n-2) - 34*a(n-3) - a(n-4) + a(n-5) with a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=351. - Harvey P. Dale, Sep 28 2011

a(n) = x*(x + 1)/2 where x = A216134(n) = (2*A000129(n) + (-1)^n*(A000129(2*floor(n/2) - 1) - (-1)^n)/2). - Raphie Frank, Jan 04 2013

a(n+2) = 1/2*((3/2*sqrt(8*a(n) + 1) + sqrt(16*a(n) + 9) - 1/2)*(3/2*sqrt(8*a(n) + 1) + sqrt(16*a(n) + 9) + 1/2)); a(0) = 0, a(1) = 1. - Raphie Frank, Jan 29 2013

a:= n-> (Matrix([[10, 1, 0, 0, 1]]). Matrix(5, (i, j)-> if i=j-1 then 1 elif j=1 then [1, 34, -34, -1, 1][i] else 0 fi)^n)[1, 4]: seq(a(n), n=1..30); # Alois P. Heinz, Apr 27 2009

LinearRecurrence[{1, 34, -34, -1, 1}, {0, 1, 10, 45, 351}, 30] (* Harvey P. Dale, Sep 28 2011 *)

(Magma) I:=[0, 1, 10, 45]; [n le 4 select I[n] else 34*Self(n-2)-Self(n-4)+11: n in [1..30]]; // Vincenzo Librandi, Sep 29 2011

(PARI) a=[0, 1, 10, 45, 351]; for(n=5, 20, a=concat(a, a[#a]+34*a[#a-1]- 34*a[#a-2]-a[#a-3]+a[#a-4])); a \\ Charles R Greathouse IV, Sep 29 2011