A154145 Indices k such that 15 plus the k-th triangular number is a perfect square.

1, 4, 6, 11, 20, 33, 43, 70, 121, 196, 254, 411, 708, 1145, 1483, 2398, 4129, 6676, 8646, 13979, 24068, 38913, 50395, 81478, 140281, 226804, 293726, 474891, 817620, 1321913, 1711963, 2767870, 4765441, 7704676, 9978054, 16132331, 27775028

Offset

1,2

Comments

a(1..4)=(1,4,6,11); a(n>4)=6*a(n-2)-a(n-4)+2. [From Ctibor O. Zizka, Nov 13 2009]

Links

Table of n, a(n) for n=1..37.

F. T. Adams-Watters, SeqFan Discussion, Oct 2009

Formula

{k: 15+k*(k+1)/2 in A000290}.

Conjectures: (Start)

a(n)= +a(n-1) +6*a(n-4) -6*a(n-5) -a(n-8) +a(n-9).

G.f.: x*(-1-3*x-2*x^2-5*x^3-3*x^4+5*x^5+2*x^6+3*x^7+2*x^8)/((x-1) * (x^4+2*x^2-1) * (x^4-2*x^2-1)).

G.f.: ( 4 + (7+4*x+16*x^2+11*x^3)/(x^4-2*x^2-1) + 1/(x-1) + (-4-7*x-3*x^2-2*x^3)/(x^4+2*x^2-1) )/2. (End)

a(1..4) = (1,4,6,11); a(n) = 6*a(n-2) - a(n-4) + 2, for n>4. - Ctibor O. Zizka, Nov 13 2009

Example

1*(1+1)/2+15 = 4^2. 4*(4+1)/2+15 = 5^2. 6*(6+1)/2+15 = 6^2. 11*(11+1)/2+15 = 9^2.

Mathematica

Flatten[Position[Accumulate[Range[28000000]], _?(IntegerQ[Sqrt[#+15]]&)]] (* This program will take a long time to run. *) (* Harvey P. Dale, Jun 09 2014 *)
Join[{1, 4, 6}, Select[Range[0, 1000], ( Ceiling[Sqrt[#*(# + 1)/2]] )^2 - #*(# + 1)/2 == 15 &]] (* G. C. Greubel, Sep 03 2016 *)

Crossrefs

Cf. A000217, A000290, A006451.

Sequence in context: A291916 A047811 A244010 * A327553 A302428 A336142

Adjacent sequences: A154142 A154143 A154144 * A154146 A154147 A154148

Keyword

nonn

Author

R. J. Mathar, Oct 18 2009

Extensions

a(32)-a(37) from Donovan Johnson, Nov 01 2010

Status

approved