OFFSET
0,2
COMMENTS
Equivalently, m such that 28*m*(m+1)+1 is a square. - Bruno Berselli, May 19 2014
LINKS
Colin Barker, Table of n, a(n) for n = 0..1000
Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021.
Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021.
Vladimir Pletser, Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers, 2022.
Index entries for linear recurrences with constant coefficients, signature (1,16,-16,-1,1).
FORMULA
G.f.: x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)).
a(n) = 16*a(n-2) - a(n-4) + 7, n >= 3. [corrected by Vladimir Pletser, Feb 29 2020]
Let b(n) be A077397 then a(n+2) = 2*a(n+1) - a(n) + b(n) with a(0)=0 a(1)=2.
a(0)=0, a(1)=2; a(n+2) = (7 + 16*a(n) + 3*sqrt((1+28*a(n)+28*a(n)^2))/2. - Herbert Kociemba, May 12 2008
a(n) = a(n-1) + 16*a(n-2) - 16*a(n-3) - a(n-4) + a(n-5). - Wesley Ivan Hurt, Jul 24 2020
MAPLE
f := gfun:-rectoproc({a(-2) = 2, a(-1) = 0, a(0) = 0, a(1) = 2, a(n) = 16*a(n - 2) - a(n - 4) + 7}, a(n), remember): map(f, [$ (0 .. 40)])[]; # Vladimir Pletser, Jul 24 2020
MATHEMATICA
LinearRecurrence[{1, 16, -16, -1, 1}, {0, 2, 5, 39, 87}, 30] (* G. C. Greubel, Jan 18 2018 *)
CoefficientList[Series[x (2+3x+2x^2)/((1-x)(1-16x^2+x^4)), {x, 0, 40}], x] (* Harvey P. Dale, Aug 19 2022 *)
PROG
(PARI) {a(n)=if(n<0, 0, polcoeff(x*(2+3*x+2*x^2)/((1-x)*(1-16*x^2+x^4)) + x*O(x^n), n))};
(Magma) I:=[0, 2, 5, 39, 87]; [n le 5 select I[n] else Self(n-1)+16*Self(n-2) -16*Self(n-3)-Self(n-4)+Self(n-5): n in [1..30]]; // G. C. Greubel, Jan 18 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Bruce Corrigan (scentman(AT)myfamily.com), Nov 05 2002
STATUS
approved