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

%I

%S 0,16,256,1296,4096,10000,20736,38416,65536,104976,160000,234256,

%T 331776,456976,614656,810000,1048576,1336336,1679616,2085136,2560000,

%U 3111696,3748096,4477456,5308416,6250000,7311616,8503056,9834496,11316496,12960000,14776336,16777216

%N a(n) = (2*n)^4.

%C 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 Vincenzo Librandi, <a href="/A016744/b016744.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F G.f.: 16*x*(x+1)*(x^2+10*x+1)/(1-x)^5. - Maksym Voznyy (voznyy(AT)mail.ru), Aug 11 2009

%F E.g.f.: 16*x*(1 + 7*x + 6*x^2 + x^3)*exp(x). - _G. C. Greubel_, Sep 15 2018

%F From _Amiram Eldar_, Oct 10 2020: (Start)

%F Sum_{n>=1} 1/a(n) = Pi^4/1440 (conjecturally A258945).

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/11520. (End)

%p A016744:=n->(2*n)^4: seq(A016744(n), n=0..50); # _Wesley Ivan Hurt_, Sep 15 2018

%t Table[(2*n)^4, {n,0,30}] (* _G. C. Greubel_, Sep 15 2018 *)

%o (MAGMA) [(2*n)^4: n in [0..40]]; // _Vincenzo Librandi_, Sep 05 2011

%o (PARI) vector(30, n, n--; (2*n)^4) \\ _G. C. Greubel_, Sep 15 2018

%Y Cf. A016756, A258945.

%K nonn,easy,changed

%O 0,2

%A _N. J. A. Sloane_

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Last modified April 15 03:00 EDT 2021. Contains 342974 sequences. (Running on oeis4.)