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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.
1
0, 36, 11628, 3744216, 1205625960, 388207814940, 125001710784756, 40250162664876528, 12960427376379457296, 4173217365031520372820, 1343763031112773180590780, 432687522800947932629858376, 139324038578874121533633806328, 44861907734874666185897455779276
OFFSET
0,2
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
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)
EXAMPLE
5*0 = 4*0;
5*36 = 4*45;
5*11628 = 4*14535;
5*3744216 = 4*4680270.
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 *)
PROG
(PARI) concat(0, Vec(36*x/((1-x)*(1-322*x+x^2)) + O(x^15))) \\ Colin Barker, Mar 02 2016
(Magma) m:=20; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(36*x/((1-x)*(1-322*x+x^2)))); // G. C. Greubel, Jul 15 2018
KEYWORD
nonn,easy
AUTHOR
Charlie Marion, Dec 20 2011
STATUS
approved