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

0, 16, 256, 1296, 4096, 10000, 20736, 38416, 65536, 104976, 160000, 234256, 331776, 456976, 614656, 810000, 1048576, 1336336, 1679616, 2085136, 2560000, 3111696, 3748096, 4477456, 5308416, 6250000, 7311616, 8503056, 9834496, 11316496, 12960000, 14776336, 16777216

Offset

0,2

Comments

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

Links

Vincenzo Librandi, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

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

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

From Amiram Eldar, Oct 10 2020: (Start)

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

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

Maple

A016744:=n->(2*n)^4: seq(A016744(n), n=0..50); # Wesley Ivan Hurt, Sep 15 2018

Mathematica

Table[(2*n)^4, {n, 0, 30}] (* G. C. Greubel, Sep 15 2018 *)

Prog

(Magma) [(2*n)^4: n in [0..40]]; // Vincenzo Librandi, Sep 05 2011
(PARI) vector(30, n, n--; (2*n)^4) \\ G. C. Greubel, Sep 15 2018

Crossrefs

Cf. A016756, A258945.

Sequence in context: A208417 A208025 A207805 * A207401 A207750 A207126

Adjacent sequences: A016741 A016742 A016743 * A016745 A016746 A016747

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved