A201008 Triangular numbers, T(m), that are five-sixths of another triangular number: T(m) such that 6*T(m)=5*T(k) for some k.
0, 55, 26565, 12804330, 6171660550, 2974727580825, 1433812522297155, 691094661019647940, 333106192798948009980, 160556493834431921162475, 77387896922003387052303025, 37300805759911798127288895630
Offset: 0
Examples
6*0 = 5*0; 6*55 = 5*66; 6*26565 = 5*31878; 6*12804330 = 5*15365196.
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..229
- Index entries for linear recurrences with constant coefficients, signature (483,-483,1).
Programs
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Magma
I:=[0, 55, 26565]; [n le 3 select I[n] else 483*Self(n-1)-483*Self(n-2)+Self(n-3): n in [1..15]]; // Vincenzo Librandi, Dec 22 2011
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Mathematica
LinearRecurrence[{483,-483,1},{0,55,26565},30] (* Vincenzo Librandi, Dec 22 2011 *) -
Maxima
makelist(expand(((11-2*sqrt(30))^(2*n+1)+(11+2*sqrt(30))^(2*n+1)-22)/192), n, 0, 11); /* Bruno Berselli, Dec 21 2011 */
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PARI
concat(0,Vec(55/(1-x)/(1-482*x+x^2)+O(x^98))) \\ Charles R Greathouse IV, Dec 23 2011
Formula
For n > 1, a(n) = 482*a(n-1) - a(n-2) + 55. See A200993 for generalization.
From Bruno Berselli, Dec 21 2011: (Start)
G.f.: 55*x/((1-x)*(1-482*x+x^2)).
a(n) = a(-n-1) = 483*a(n-1)-483*a(n-2)+a(n-3).
a(n) = ((11-2*r)^(2*n+1)+(11+2*r)^(2*n+1)-22)/192, where r=sqrt(30). (End)
Extensions
a(11) corrected by Bruno Berselli, Dec 21 2011
a(6) corrected by Vincenzo Librandi, Dec 22 2011