A182334 Triangular numbers that differ from a square by 1.

0, 1, 3, 10, 15, 120, 325, 528, 4095, 11026, 17955, 139128, 374545, 609960, 4726275, 12723490, 20720703, 160554240, 432224101, 703893960, 5454117903, 14682895930, 23911673955, 185279454480, 498786237505, 812293020528, 6294047334435, 16944049179226

Offset: 1

Author

Arkadiusz Wesolowski, Apr 25 2012

Keywords

• nonn
• easy

Comments

From Robert G. Wilson v, Jun 20 2015: (Start)

Actually this sequence is the union of two subsequences; the triangular numbers that are less than a square by 1 and those that are greater than a square by 1.

The first sequence by index of the triangular numbers is A072221: b(n) = 6b(n-1) - b(n-2) + 2, with b(0)=1, b(1)=4.

And obviously the second sequence by index of the triangular numbers is A006451: c(n) = 6c(n-2) - c(n-4) + 2 with c(0)=0, c(1)=2, c(2)=5, c(3)=15.

(End)

Examples

			T(2) = 3 = 2^2 - 1, T(4) = 10 = 3^2 + 1,  T(5) = 15 = 4^2 - 1, and T(15) = 120 = 11^2 - 1.
		

References

• Edward J. Barbeau, Pell's Equation (Springer 2003) at 17.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (0,0,35,0,0,-35,0,0,1).

Crossrefs

Subsequence of A000217 and of A087279.

Cf. A001110, A006451, A072221, A229131, A299921.

Programs

Magma

I:=[0,1,3,10,15,120,325,528,4095,11026,17955]; [n le 11 select I[n] else 35*Self(n-3)-35*Self(n-6)+Self(n-9): n in [1..30]]; // Vincenzo Librandi, Jun 21 2015

Mathematica

lst = {}; Do[t = n*(n + 1)/2; If[IntegerQ[(t - 1)^(1/2)] || IntegerQ[(t + 1)^(1/2)], AppendTo[lst, t]], {n, 0, 10^4}]; lst (* Arkadiusz Wesolowski, Aug 06 2012 *)
b[n_] := b[n] = 6 b[n - 1] - b[n - 2] + 2; b[0] = 1; b[1] = 4; c[n_] := c[n] = 6 c[n - 2] - c[n - 4] + 2; c[0] = 0; c[1] = 2; c[2] = 5; c[3] = 15; #(# + 1)/2 & /@ Union@ Join[ Array[b, 9, 0], Array[c, 18, 0]] (* or *)
#(# + 1)/2 & /@ Join[{0, 1}, LinearRecurrence[{1, 0, 6, -6, 0, -1, 1}, {2, 4, 5, 15, 25, 32, 90}, 35]] (* or *)
#(# + 1)/2 & /@ CoefficientList[ Series[x + x^2 (1 + x) (2 + x^2 - 3 x^3 + x^4)/((1 - x) (1 - 6 x^3 + x^6)), {x, 0, 36}], x] (* Robert G. Wilson v, Jun 20 2015 *)
a[n_] := a[n] = 35 a[n - 3] - 35 a[n - 6] + a[n - 9]; a[1] = 0; a[2] = 1; a[3] = 3; a[4] = 10; a[5] = 15; a[6] = 120; a[7] = 325; a[8] = 528; a[9] = 4095; a[10] = 11026; a[11] = 17955; Array[a, 36] (* Robert G. Wilson v after Charles R Greathouse IV, Apr 25 2012 *)
Select[Accumulate[Range[0,6*10^6]],AnyTrue[Sqrt[#+{1,-1}],IntegerQ]&] (* or *) LinearRecurrence[{0,0,35,0,0,-35,0,0,1},{0,1,3,10,15,120,325,528,4095,11026,17955},40] (* The first program uses the AnyTrue function from Mathematica version 10 *) (* Harvey P. Dale, Dec 24 2015 *)

PARI

concat(0, Vec(x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9)/((1-x)*(1+x+x^2)*(1-34*x^3+x^6)) + O(x^30))) \\ Colin Barker, Sep 17 2016

Formula

a(n) = 35*a(n-3) - 35*a(n-6) + a(n-9). - Charles R Greathouse IV, Apr 25 2012

G.f.: x^2*(1+3*x+10*x^2-20*x^3+15*x^4-25*x^5+38*x^6+x^8-x^9) / ((1-x)*(1+x+x^2)*(1-34*x^3+x^6)). - Colin Barker, Sep 17 2016