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A135038
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Sums of the products of n consecutive quadruples of numbers.
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1
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0, 840, 8760, 41520, 134544, 347064, 768264, 1523424, 2780064, 4754088, 7715928, 11996688, 17994288, 26179608, 37102632, 51398592, 69794112, 93113352, 122284152, 158344176, 202447056, 255868536, 320012616, 396417696, 486762720
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OFFSET
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1,2
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LINKS
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FORMULA
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a(n) = floor(51.2*n^5 - 32*n^4 - 48*n^3 + 20*n^2 + 8.8*n).
a(1) = 0*1*2*3, a(2) = 0*1*2*3 + 4*5*6*7, ..., a(n) = 0*1*2*3 + 4*5*6*7 + 8*9*10*11 + ... + (2n-1)*(2n)*(2n+1)(2n+2).
O.g.f.: 24*x^2*(35 +155*x +65*x^2 +x^3)/(1-x)^6 . - R. J. Mathar, Feb 14 2008
a(1)=0, a(2)=840, a(3)=8760, a(4)=41520, a(5)=134544, a(6)=347064, a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Harvey P. Dale, May 09 2015
a(n) = (100*n^2 + 44*n - 240*n^3 - 160*n^4 + 256*n^5)/5.
E.g.f.: (4/5)*x^2*(525 + 1300*x + 600*x^2 + 64*x^3)*exp(x). (End)
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EXAMPLE
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For n = 3, the sum of the first 3 quadruples is 0*1*2*3 + 4*5*6*7 + 8*9*10*11 = 8760, the 3rd entry in the sequence.
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MAPLE
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MATHEMATICA
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Accumulate[Times@@@Table[4n+Range[0, 3], {n, 0, 30}]] (* or *) LinearRecurrence[ {6, -15, 20, -15, 6, -1}, {0, 840, 8760, 41520, 134544, 347064}, 30] (* Harvey P. Dale, May 09 2015 *)
Table[(100*n^2 + 44*n - 240*n^3 - 160*n^4 + 256*n^5)/5, {n, 1, 25}] (* G. C. Greubel, Sep 17 2016 *)
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PROG
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(PARI) sumprod3(n)= local(x, s=0); forstep(x=0, n, 4, s+=x*(x+1)*(x+2)*(x+3); print1(s", ") )
(Magma) [(100*n^2+44*n-240*n^3-160*n^4+256*n^5)/5: n in [1..40]]; // Vincenzo Librandi, Sep 18 2016
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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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