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Integers of the form (n^2 - 49) / 120.
1

%I #17 Sep 08 2022 08:46:02

%S 0,1,2,4,11,15,18,23,37,44,49,57,78,88,95,106,134,147,156,170,205,221,

%T 232,249,291,310,323,343,392,414,429,452,508,533,550,576,639,667,686,

%U 715,785,816,837,869,946,980,1003,1038,1122,1159,1184,1222,1313,1353

%N Integers of the form (n^2 - 49) / 120.

%H G. C. Greubel, <a href="/A214429/b214429.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,2,-2,0,0,-1,1).

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

%F a(n) = 2*a(n-4) - a(n-8) + 15 = a(-1 - n).

%t CoefficientList[Series[x*(1+x+2*x^2+7*x^3+2*x^4+x^5+x^6)/((1-x)*(1- x^4)^2), {x,0,50}], x] (* _G. C. Greubel_, Aug 10 2018 *)

%t Select[(Range[0,500]^2-49)/120,IntegerQ] (* or *) LinearRecurrence[ {1,0,0,2,-2,0,0,-1,1},{0,1,2,4,11,15,18,23,37},80] (* _Harvey P. Dale_, Oct 23 2019 *)

%o (PARI) {a(n) = (((n*3 + 1) \ 4 * 10 + 5 + 2*(-1)^n)^2 - 49) / 120 }

%o (Magma) m:=25; R<x>:=PowerSeriesRing(Integers(), m); Coefficients(R!(x*(1+x+2*x^2+7*x^3+2*x^4+x^5+x^6)/((1-x)*(1-x^4)^2))); // _G. C. Greubel_, Aug 10 2018

%Y Cf. A093722.

%K nonn,easy

%O 0,3

%A _Michael Somos_, Jul 17 2012