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Triangular numbers that are 1/6 of another triangular number.
12

%I #34 Oct 01 2021 11:26:50

%S 0,1,6,105,595,10296,58311,1008910,5713890,98862891,559902916,

%T 9687554415,54864771885,949281469786,5376187741821,93019896484620,

%U 526811533926580,9115000574022981,51622154137063026,893177036357767525,5058444293898249975,87522234562487194476

%N Triangular numbers that are 1/6 of another triangular number.

%C The triangular numbers multiplied by 6 are in A077290.

%H Colin Barker, <a href="/A077289/b077289.txt">Table of n, a(n) for n = 0..1000</a>

%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.13494">Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations</a>, arXiv:2102.13494 [math.NT], 2021.

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,98,-98,-1,1).

%F Let b(n) be A077288. Then a(n)=b(n)*(b(n)+1)/2.

%F G.f.: -x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)). - _Colin Barker_, Jul 02 2013

%F a(n) = 98*a(n-2) - a(n-4) + 7. - _Vladimir Pletser_, Feb 19 2021

%F 96*a(n) = 9*A072256(n+1) -2*(-1)^n*A054320(n) -7. - _R. J. Mathar_, Oct 01 2021

%e b(3)=14 so a(3) = 14*15/2 = 105, etc.

%p f := gfun:-rectoproc({a(-2) = 1, a(-1) = 0, a(0) = 0, a(1) = 1, a(n) = 98*a(n-2)-a(n-4)+7}, a(n), remember); map(f, [`$`(0 .. 1000)])[]; # _Vladimir Pletser_, Feb 19 2021

%t tr6Q[n_]:= IntegerQ[1/2 (Sqrt[1+48n]-1)]; Select[Accumulate[ Range[0,1380000]],tr6Q] (* _Harvey P. Dale_, Apr 21 2011 *)

%o (PARI)

%o T(n)=n*(n+1)\2;

%o istriang(n)=issquare(8*n+1);

%o for(n=0, 10^10, t=T(n); if ( t%6==0 && istriang(t\6), print1(t\6, ", ") ) );

%o \\ _Joerg Arndt_, Jul 03 2013

%o (PARI) concat(0, Vec(-x*(x^2+5*x+1) / ((x-1)*(x^2-10*x+1)*(x^2+10*x+1)) + O(x^100))) \\ _Colin Barker_, May 15 2015

%Y Cf. A077288, A077290, A077291.

%K easy,nonn

%O 0,3

%A Bruce Corrigan (scentman(AT)myfamily.com), Nov 03 2002