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A016744 a(n) = (2*n)^4.

%I A016744 #31 Sep 08 2022 08:44:41
%S A016744 0,16,256,1296,4096,10000,20736,38416,65536,104976,160000,234256,
%T A016744 331776,456976,614656,810000,1048576,1336336,1679616,2085136,2560000,
%U A016744 3111696,3748096,4477456,5308416,6250000,7311616,8503056,9834496,11316496,12960000,14776336,16777216
%N A016744 a(n) = (2*n)^4.
%C A016744 Suppose the vertices of a triangle are (S(n), S(n+j)), (S(n+2*j), S(n+3*j)) and (S(n+4*j), S(n+5*j)) where S(n) is the n-th square number, A000290(n). Then the area of this triangle will be a(j). A generalization follows: let P(k,n) = the n-th k-gonal number and suppose the vertices of a triangle are (P(k,n), P(k,n+j)), (P(k,n+2*j), P(k,n+3*j)) and (P(k,n+4*j), P(k,n+5*j)). Then the area of this triangle will be (2*k-4)^2*j^4. See also A141046 for k = 3. - _Charlie Marion_, Mar 26 2021
%H A016744 Vincenzo Librandi, <a href="/A016744/b016744.txt">Table of n, a(n) for n = 0..10000</a>
%H A016744 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A016744 G.f.: 16*x*(x+1)*(x^2+10*x+1)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009
%F A016744 E.g.f.: 16*x*(1 + 7*x + 6*x^2 + x^3)*exp(x). - _G. C. Greubel_, Sep 15 2018
%F A016744 From _Amiram Eldar_, Oct 10 2020: (Start)
%F A016744 Sum_{n>=1} 1/a(n) = Pi^4/1440 (conjecturally A258945).
%F A016744 Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/11520. (End)
%p A016744 A016744:=n->(2*n)^4: seq(A016744(n), n=0..50); # _Wesley Ivan Hurt_, Sep 15 2018
%t A016744 Table[(2*n)^4, {n,0,30}] (* _G. C. Greubel_, Sep 15 2018 *)
%o A016744 (Magma) [(2*n)^4: n in [0..40]]; // _Vincenzo Librandi_, Sep 05 2011
%o A016744 (PARI) vector(30, n, n--; (2*n)^4) \\ _G. C. Greubel_, Sep 15 2018
%Y A016744 Cf. A016756, A258945.
%K A016744 nonn,easy
%O A016744 0,2
%A A016744 _N. J. A. Sloane_