|
%I #19 Sep 28 2017 19:49:59
%S 3,25,49,761,7381,86021,1171733,2436559,14274301,55835135,19093197,
%T 1347822955,34395742267,315404588903,9304682830147,586061125622639,
%U 54062195834749,54801925434709,2053580969474233,2078178381193813
%N Numerator of the polynomial A_l(x) = sum_{d=1..l-1} x^(l-d)/d for index l=2n+1 evaluated at x=1.
%C The polynomials A_{2n+1}(x) = sum_{d=1..2n} x^(2n+1-d)/d for small n look as follows:
%C n=1, index = 3: A_3(x) = x/2 + x^2.
%C n=2, index = 5: A_5(x) = x/4 + x^2/3 + x^3/2 + x^4.
%C n=3, index = 7: A_7(x) = x/6 + x^2/5 + x^3/4 + x^4/3 + x^5/2 + x^6.
%C n=4, index = 9: A_9(x) = x/8 + x^2/7 + x^3/6 + x^4/5 + x^5/4 + x^6/3 + x^7/2 + x^8.
%H Vincenzo Librandi, <a href="/A145609/b145609.txt">Table of n, a(n) for n = 1..200</a>
%F (1/(2*n+1))*2F1(1, 2*n+1; 2*n+2; 1/m) = Sum_{x>=0} m^(-x)/(x+2n+1) = m^(2n)*arctanh((2m-1)/(2m^2-2m+1)) - A_{2n+1}(m) = m^(2n)*log(m/(m-1)) - A_{2n+1}(m). - _Artur Jasinski_, Oct 14 2008
%F It appears that A145609(n)/A145610(n) = H(2*n+2), the harmonic number of order 2*n+2. - _Groux Roland_, Jan 08 2011
%F Yes, A145609(n)/A145610(n) = H(2*n+2), as A_l(x) = sum_{d=1..l-1} x^(l-d)/d at x=1 is just sum_{d=1..l-1}1/d = H(l-1), the harmonic number of order (l-1). - _Zak Seidov_, Jan 09 2014
%F a(n) = numerator of Integral_{x=0..1} ((1 - x^(2*n))/(1 - x). - _Peter Luschny_, Sep 28 2017
%p A := proc(l,x) add(x^(l-d)/d,d=1..l-1) ; end: A145609 := proc(n) numer( A(2*n+1,1)) ; end: seq(A145609(n),n=1..20) ; # _R. J. Mathar_, Aug 21 2009
%t m = 1; aa = {}; Do[k = 0; Do[k = k + m^(2 r + 1 - d)/d, {d, 1, 2 r}]; AppendTo[aa, Numerator[k]], {r, 1, 25}]; aa (* _Artur Jasinski_ *)
%Y For denominators see A145610.
%Y Cf. A145611-A145640.
%K frac,nonn
%O 1,1
%A _Artur Jasinski_, Oct 14 2008
%E Edited, parentheses in front of Gauss. Hypg. Fct. added by _R. J. Mathar_, Aug 21 2009
| 