A294070 a(n) = (1/4)*(n^2 - 2*n)^2 + (9/4)*(n^2 - 2*n) + 6.
4, 6, 15, 40, 96, 204, 391, 690, 1140, 1786, 2679, 3876, 5440, 7440, 9951, 13054, 16836, 21390, 26815, 33216, 40704, 49396, 59415, 70890, 83956, 98754, 115431, 134140, 155040, 178296, 204079, 232566, 263940, 298390, 336111, 377304, 422176, 470940, 523815
Offset: 1
Examples
2*2, 2*3, 3*5, 5*8, 8*12, 12*17, 17*23, 23*30, 30*38, ...
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
- SESC NSU Correspondence School, First assignments for 2018/2019 (in Russian), Kvant, 2018, No. 7, p. 42, Mathematics section, 6th grade, exercise no. 2. "Calculate and show in a reduced fraction form the following sum: 1/(2*3) + 2/(3*5) + 3/(5*8) + 4/(8*12) + 5/(12*17)."
- Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
Programs
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GAP
List([1..40],n->(n^2-3*n+6)*(n^2-n+4)/4); # Muniru A Asiru, Aug 16 2018
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Magma
[(n^2-3*n+6)*(n^2-n+4)/4: n in [1..40]]; // Vincenzo Librandi, Aug 30 2018
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Maple
b:=n->(n^2-3*n+6)/2: seq(b(n)*b(n+1),n=1..40); # Muniru A Asiru, Aug 16 2018
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Mathematica
Times@@@Partition[Array[(#^2 -3# +6)/2 &, 40], 2, 1] (* Michael De Vlieger, Sep 24 2018 *) LinearRecurrence[{5,-10,10,-5,1}, {4,6,15,40,96}, 40] (* G. C. Greubel, Feb 10 2019 *)
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PARI
Vec(x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1-x)^5 + O(x^40)) \\ Colin Barker, Nov 26 2018
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Sage
[(n^2-3*n+6)*(n^2-n+4)/4 for n in (1..40)] # G. C. Greubel, Feb 10 2019
Formula
From Muniru A Asiru, Aug 16 2018: (Start)
a(n) = (n^2 - 3*n + 6)*(n^2 - n + 4)/4.
From Colin Barker, Nov 26 2018: (Start)
G.f.: x*(4 - 14*x + 25*x^2 - 15*x^3 + 6*x^4)/(1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n > 5.
a(n) = (24 - 18*n + 13*n^2 - 4*n^3 + n^4)/4. (End)
E.g.f.: (1/4)*exp(x)*(16 + 8*x + 14*x^2 + 6*x^3 + x^4). - Stefano Spezia, Nov 30 2018