|
%I #24 Apr 04 2019 13:44:23
%S 0,1,10,1540,11935,1777555,13773376,2051297326,15894464365,
%T 2367195337045,18342198104230,2731741367653000,21166880717817451,
%U 3152427171076225351,24426562006163234620,3637898223680596402450,28188231388231654934425,4198131397700237172202345
%N Integers that are simultaneously triangular (A000217) and decagonal (A001107).
%C Positive terms are of the form (m^2-9)/16 where m runs over the elements of A077443 that are congruent to 5 modulo 8. Correspondingly, for n>1, sqrt(16*a(n)+9) form a subsequence of A077443, while sqrt(8*a(n)+1) form a subsequence of A077442 with indices congruent to 2,3 modulo 4. [_Max Alekseyev_]
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1, 1154, -1154, -1, 1).
%F a(n) = A000217(A133218(n)) = A001107(A133217(n)).
%F For n>5, a(n) = 1154*a(n-2) - a(n-4) + 396.
%F For n>6, a(n) = a(n-1) + 1154*a(n-2) - 1154*a(n-3) - a(n-4) + a(n-5).
%F For n>1, a(n) = 1/64 * ( (9 + 4* sqrt(2)*(-1)^n)*(1+sqrt(2))^(4*n-6) + (9 - 4* sqrt(2)*(-1)^n)*(1-sqrt(2))^(4*n-6) - 22).
%F a(n) = floor ( 1/64 * (9 + 4*sqrt(2)*(-1)^n) * (1+sqrt(2))^(4*n-6) ).
%F G.f.: (x^5 + 9*x^4 + 376*x^3 + 9*x^2 + x)/((1 - x)*(x^2 - 34*x + 1)*(x^2 + 34*x + 1)). [corrected by _Peter Luschny_, Apr 04 2019]
%F Lim (n -> Infinity, a(2n+1)/a(2n)) = (1/49)*(3649+2580*sqrt(2)).
%F Lim (n -> Infinity, a(2n)/a(2n-1)) = (1/49)*(193+132*sqrt(2)).
%e The initial terms of the sequences of triangular (A000217) and decagonal (A001107) numbers are 0, 1, 3, 6, 10, 15, ... and 0, 1, 10, 27, ... respectively. As the third number which is common to both sequences is 10, we have a(3) = 10.
%t LinearRecurrence[{1, 1154, -1154, -1, 1} , {0, 1, 10, 1540, 11935, 1777555}, 17] (* first term 0 corrected by _Georg Fischer_, Apr 02 2019 *)
%Y Cf. A000217, A001107, A133217, A133218, A077443, A077442.
%K nonn
%O 1,3
%A _Richard Choulet_, Oct 11 2007; _Ant King_, Nov 04 2011
%E Entry revised by _N. J. A. Sloane_, Nov 06 2011
%E Term 0 prepended and entry revised accordingly by _Max Alekseyev_, Nov 06 2011
|
