A125650 Numerator of n(n+3)/(4(n+1)(n+2)) = sum(k=1..n, 1/(k(k+1)(k+2)) ).

0, 1, 5, 9, 7, 5, 27, 35, 11, 27, 65, 77, 45, 26, 119, 135, 38, 85, 189, 209, 115, 63, 275, 299, 81, 175, 377, 405, 217, 116, 495, 527, 140, 297, 629, 665, 351, 185, 779, 819, 215, 451, 945, 989, 517, 270, 1127, 1175, 306, 637, 1325, 1377, 715, 371, 1539, 1595, 413, 855

Offset: 0

Author

Alexander Adamchuk, Nov 29 2006

Keywords

• nonn
• frac
• easy

Comments

3^2 divides a(3k). p divides a(p) for an odd prime p. p divides a(p-3) for prime p>3. p^k divides a(p^k) for an odd prime p. a(n) = m^2 is a perfect square for n = {1,3,24,147,864,5043,29400,171363,...} = A125651(n). Corresponding numbers m such that m^2 = a[ A125651(n) ] are listed in A125652(n) = {1,3,9,105,306,3567,10395,121173,...}.

Examples

			The rationals n(n+3)/(4(n+1)(n+2)) = a(n)/A230328(n) begin:
0, 1/6, 5/24, 9/40, 7/30, 5/21, 27/112, 35/144, 11/45, 27/110, 65/264, 77/312, 45/182, 26/105, 119/480, ... - _Wolfdieter Lang_, Mar 08 2018
		

References

• L. B.W. Jolley, Summation of Series, Second revised ed., Dover, 1961, p.38, (201).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A125651, A125652. A160050, A230328 (denominators).

Programs

Magma

[Numerator(n*(n+3)/(4*(n+1)*(n+2))): n in [0..60]]; // Vincenzo Librandi, May 21 2012

Mathematica

Table[Numerator[n(n+3)/(4(n+1)(n+2))],{n,0,100}]

PARI

a(n)=n*(n+3)/2^min(3,valuation(n*(n+3),2)); \\ Max Alekseyev, Jan 11 2007

Formula

a(n) = Sum_{k=1..n} 1/(k(k+1)(k+2)).

a(n) = n*(n+3)/2^min(3,valuation(n*(n+3),2)). a(n)=n*(n+3)/4 for n=1 or 4 (mod 8); a(n)=n*(n+3)/8 for n=0 or 5 (mod 8); a(n) = n*(n+3)/2 for n=2, 3, 6, or 7 (mod 8). - Max Alekseyev, Jan 11 2007

a(n) = A106609(n)*A106609(n+3). - Paul Curtz, Jan 13 2011

G.f.: x*(x^19 -2*x^18 +3*x^17 -5*x^16 +3*x^15 -6*x^14 +7*x^13 -11*x^12 -12*x^11 +24*x^10 -36*x^9 +24*x^8 -38*x^7 +28*x^6 -18*x^5 -3*x^4 -2*x -1) / ((x-1)^3*(x^2+1)^3*(x^4+1)^3). - Colin Barker, Feb 21 2013

G.f. for rationals r(n) = a(n)/A230328(n): (1/4)*(1 - hypergeometric([1, 2], [3], -x/(1-x)))/(1-x) = (- 2*x + 3*x^2 + 2*(2*x - (1 + x^2))*log(1-x))/(4*(1-x)*x^2). For the r(n) formula see Jolley's general remark (201) on p.38. Thanks to Gary Detlefs for pointing to this remark. - Wolfdieter Lang, Mar 08 2018