cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A294513 Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.

Original entry on oeis.org

2, 5, 120, 1320, 9240, 52360, 52360, 602140, 70450380, 2043061020, 16344488160, 3268897632, 62109055008, 2546471255328, 1157486934240, 54401885909280, 272009429546400, 4805499921986400, 4805499921986400, 283524495397197600, 418536159872053600
Offset: 0

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Author

Wolfdieter Lang, Nov 02 2017

Keywords

Comments

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.

Examples

			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.
Conjecture tests: a(0) = 2 =  A250327(1)/1, 2* a(1) = 5 = 2*A250327(2)/2 = A250327(2), a(2) = 120 = 2*A250327(2)/3 = 2*180/3, ...
		

References

  • Max Koecher, Klassische elementare Analysis, Birkhäuser, Basel, Boston, 1987, pp. 189 - 193 (with v_m(r) = ((m-r)/m)*V(m,r)).

Crossrefs

Cf. A049450, A250327(n+1), A250328(n+1), A294512.

Programs

  • Mathematica
    Denominator@ Accumulate@ Array[1/(2 PolygonalNumber[5, #]) &, 21] (* Michael De Vlieger, Nov 02 2017 *)

Formula

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].