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A294513
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Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.
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2
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2, 5, 120, 1320, 9240, 52360, 52360, 602140, 70450380, 2043061020, 16344488160, 3268897632, 62109055008, 2546471255328, 1157486934240, 54401885909280, 272009429546400, 4805499921986400, 4805499921986400, 283524495397197600, 418536159872053600
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OFFSET
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0,1
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COMMENTS
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The corresponding numerators are given by A250328(n+1), n >= 0.
Twice the positive pentagonal numbers are A049450(k+1) = (k+1)*(3*k+2), k >= 0.
For the general case V(m,r;n) = Sum_{k=0..n} 1/((k + 1)*(m*k + r)) = (1/(m - r))*Sum_{k=0..n} (m/(m*k + r) - 1/(k+1)), for r = 1, ..., m-1 and m = 2, 3, ..., and their limits see a comment in A294512. Here [m,r] = [3,2].
The limit of the series is V(3,2) = lim_{n -> oo} V(3,2;n) = (3/2)*log(3) - Pi/(2*sqrt(3)) = 0.74101875088505561179... given in A294514.
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REFERENCES
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Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, pp. 189 - 193 (with v_m(r) = ((m-r)/m)*V(m,r)).
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LINKS
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FORMULA
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a(n) = denominator(V(3,2;n)) with V(3,2;n) = Sum_{k=0..n} 1/((k + 1)*(3*k + 2)) = Sum_{k=0..n} 1/A049450(k+1) = Sum_{k=0..n} (3/(3*k + 2) - 1/(k+1)).
a(n) = 2*A250327(n+1)/(n+1) [conjecture].
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EXAMPLE
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The rationals V(3,2;n), n >= 0, begin: 1/2, 3/5, 77/120, 877/1320, 6271/9240, 36049/52360, 36423/52360, 422137/602140, 49691099/70450380, 1448086909/2043061020, ...
V(3,2;10^4) = 0.7409854223(Maple, 10 digits) to be compared with 0.7410187513 from V(3,2) given in A294514.
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MATHEMATICA
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Denominator@ Accumulate@ Array[1/(2 PolygonalNumber[5, #]) &, 21] (* Michael De Vlieger, Nov 02 2017 *)
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CROSSREFS
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KEYWORD
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nonn,frac,easy
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AUTHOR
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STATUS
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approved
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