A201003 Triangular numbers, T(m), that are four-fifths of another triangular number: T(m) such that 5*T(m) = 4*T(k) for some k.

0, 36, 11628, 3744216, 1205625960, 388207814940, 125001710784756, 40250162664876528, 12960427376379457296, 4173217365031520372820, 1343763031112773180590780, 432687522800947932629858376, 139324038578874121533633806328, 44861907734874666185897455779276

Offset: 0

Author

Charlie Marion, Dec 20 2011

Keywords

• nonn
• easy

Comments

Also, numbers m such that 8*m+1 and 10*m+1 are squares. Example: 8*1205625960+1 = 98209^2 and 12056259601 = 109801^2. - Bruno Berselli, Mar 03 2016

Examples

			5*0 = 4*0;
5*36 = 4*45;
5*11628 = 4*14535;
5*3744216 = 4*4680270.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (323,-323,1).

Crossrefs

Cf. A001652, A029549, A053141, A075528, A200993-A201008, A298271.

Programs

Magma

m:=20; R:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(36*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018

Mathematica

triNums = Table[(n^2 + n)/2, {n, 0, 4999}]; Select[triNums, MemberQ[triNums, (5/4)#] &] (* Alonso del Arte, Dec 20 2011 *)
CoefficientList[Series[-36 x/((x - 1) (x^2 - 322 x + 1)), {x, 0, 30}], x] (* Vincenzo Librandi, Aug 11 2014 *)
LinearRecurrence[{323,-323,1},{0,36,11628},20] (* Harvey P. Dale, Dec 21 2015 *)

PARI

concat(0, Vec(36*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016

Formula

For n>1, a(n) = 322*a(n-1) - a(n-2) + 36. See A200993 for generalization.

G.f.: 36*x / ((1-x)*(x^2-322*x+1)). - R. J. Mathar, Aug 10 2014

From Colin Barker, Mar 02 2016: (Start)

a(n) = (-18+(9-4*sqrt(5))*(161+72*sqrt(5))^(-n)+(9+4*sqrt(5))*(161+72*sqrt(5))^n)/160.

a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n>2. (End)

a(n) = 36*A298271(n). - Amiram Eldar, Dec 01 2024