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%I #32 Aug 21 2022 04:18:38
%S 1,1,9,17,33,49,73,97,129,161,201,241,289,337,393,449,513,577,649,721,
%T 801,881,969,1057,1153,1249,1353,1457,1569,1681,1801,1921,2049,2177,
%U 2313,2449,2593,2737,2889,3041,3201,3361,3529,3697,3873,4049,4233,4417,4609
%N A081585 and A069129 interleaved.
%C a(A000129(n)) is a square.
%C (n^2)*a(n) = A275496(n) which is a triangular number.
%C (A000129(n)^2)*a(A000129(n)) = A275496(A000129(n)) = A001110(n) which is a square triangular number.
%C a(2n+1)/a(2n) is convergent to 1.
%H Daniel Poveda Parrilla, <a href="/A275543/b275543.txt">Table of n, a(n) for n = 0..500000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1).
%F a(0) = 1; a(n) = A275496(n)/(n^2) for n > 0.
%F From _Colin Barker_, Aug 01 2016: (Start)
%F a(n) = (2*n^2 + (-1)^n).
%F a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) for n > 3.
%F G.f.: (1 -x +7*x^2 +x^3) / ((1 - x)^3*(1 + x)).
%F (End)
%F From _Daniel Poveda Parrilla_, Aug 18 2016: (Start)
%F a(2n) = A077221(2n) + 1.
%F a(2n + 1) = A077221(2n + 1). (End)
%F Sum_{n>=0} 1/a(n) = (1 + (tan(c) + coth(c))*c)/2, where c = Pi/(2*sqrt(2)) is A093954. - _Amiram Eldar_, Aug 21 2022
%e a(1) = A275496(1) = 1.
%e a(5) = A275496(5)/25 = 1225/25 = 49.
%e a(7) = A275496(7)/49 = 4753/49 = 97.
%e a(12) = A275496(12)/144 = 41616/144 = 289.
%t CoefficientList[Series[(1 - x + 7 x^2 + x^3)/((1 - x)^3 (1 + x)), {x, 0, 48}], x] (* or as defined *)
%t Riffle[LinearRecurrence[{3, -3, 1}, {1, 9, 33}, #], FoldList[#1 + #2 &, 1, 16 Range@ #]] &@ 25 (* _Michael De Vlieger_, Aug 01 2016, after _Vincenzo Librandi_ at A081585 and _Robert G. Wilson v_ at A069129 *)
%o (PARI) a(n)=(-1)^n + 2*n^2 \\ _Charles R Greathouse IV_, Aug 03 2016
%o (PARI) Vec((1-x+7*x^2+x^3)/((1-x)^3*(1+x)) + O(x^100)) \\ _Colin Barker_, Aug 21 2016
%Y Cf. A081585(n) = a(2n), A069129(n) = a(2n + 1).
%Y Cf. A000129, A000217, A000290, A001110, A077221, A093954, A275496.
%K nonn,easy
%O 0,3
%A _Daniel Poveda Parrilla_, Aug 01 2016