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a(n) = n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6/(1!*2!*3!*4!*5!*6!).
1

%I #25 Sep 14 2024 02:24:56

%S 162000,26471025,1376829440,36294822144,600112800000,7031325609000,

%T 63117561830400,457937132487120,2790771598030416,14702257341646875,

%U 68449036271616000,286552568263270400,1093771338292039680,3849852478998931776,12612749124441600000

%N a(n) = n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6/(1!*2!*3!*4!*5!*6!).

%H <a href="/index/Rec#order_22">Index entries for linear recurrences with constant coefficients</a>, signature (22,-231,1540,-7315,26334,-74613,170544,-319770,497420,-646646, 705432,-646646,497420,-319770,170544,-74613,26334,-7315,1540,-231,22,-1).

%F Sum_{n>=1} 1/a(n) = 789878089*Pi^2/18000 + 64687*Pi^4/150 - 16*Pi^6/21 + 6603436*zeta(3)/25 + 80136*zeta(5) - 56698539425671/64800000. - _Amiram Eldar_, Sep 08 2022

%p [seq(n^1*(n+1)^2*(n+2)^3*(n+3)^4*(n+4)^5*(n+5)^6/(1!*2!*3!*4!*5!*6!),n=1..27)];

%t Table[(Times@@Table[(n+k)^(k+1),{k,0,5}])/Times@@(Range[6]!),{n,15}] (* _Harvey P. Dale_, Jun 07 2022 *)

%o (Sage) [binomial(n,1)*binomial(n,3)*binomial(n,5)*binomial(n,2)*binomial(n,4)*binomial(n,6) for n in range(6, 19)] # _Zerinvary Lajos_, May 17 2009

%Y Cf. A090447, A090448, A090449.

%K easy,nonn

%O 1,1

%A _Zerinvary Lajos_, Jul 05 2006

%E Offset changed from 0 to 1 by _Georg Fischer_, May 08 2021