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Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.
2

%I #10 Nov 12 2017 08:07:29

%S 2,5,120,1320,9240,52360,52360,602140,70450380,2043061020,16344488160,

%T 3268897632,62109055008,2546471255328,1157486934240,54401885909280,

%U 272009429546400,4805499921986400,4805499921986400,283524495397197600,418536159872053600

%N Denominators of the partial sums of the reciprocals of twice the pentagonal numbers.

%C The corresponding numerators are given by A250328(n+1), n >= 0.

%C Twice the positive pentagonal numbers are A049450(k+1) = (k+1)*(3*k+2), k >= 0.

%C 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].

%C 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.

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

%F 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)).

%F a(n) = 2*A250327(n+1)/(n+1) [conjecture].

%e 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, ...

%e V(3,2;10^4) = 0.7409854223(Maple, 10 digits) to be compared with 0.7410187513 from V(3,2) given in A294514.

%e 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, ...

%t Denominator@ Accumulate@ Array[1/(2 PolygonalNumber[5, #]) &, 21] (* _Michael De Vlieger_, Nov 02 2017 *)

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

%K nonn,frac,easy

%O 0,1

%A _Wolfdieter Lang_, Nov 02 2017