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A016744
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a(n) = (2*n)^4.
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5
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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
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OFFSET
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0,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
Sum_{n>=1} 1/a(n) = Pi^4/1440 (conjecturally A258945).
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/11520. (End)
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MAPLE
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MATHEMATICA
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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