A051747 a(n) = n*(n+1)*(n+2)*(n^2+7*n+32)/120.
2, 10, 31, 76, 161, 308, 546, 912, 1452, 2222, 3289, 4732, 6643, 9128, 12308, 16320, 21318, 27474, 34979, 44044, 54901, 67804, 83030, 100880, 121680, 145782, 173565, 205436, 241831, 283216, 330088, 382976, 442442, 509082, 583527, 666444, 758537
Offset: 1
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..1000
- Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).
Programs
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Magma
[n*(n+1)*(n+2)*(n^2+7*n+32)/120: n in [1..40]]; // Vincenzo Librandi, Jun 15 2011
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Mathematica
Table[(1/120)*n*(n + 1)*(n + 2)*(n^2 + 7*n + 32), {n, 60}] (* Vladimir Joseph Stephan Orlovsky, Jun 14 2011 *) LinearRecurrence[{6,-15,20,-15,6,-1},{2,10,31,76,161,308},60] (* Harvey P. Dale, Oct 03 2012 *) -
PARI
conv(u,v)=local(w); w=vector(length(u),i,sum(j=1,i,u[j]*v[i+1-j])); w; t(n)=n*(n+1)/2; u=vector(10,i,t(i)); v=vector(10,i,t(i)+1); conv(u,v)
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PARI
Vec(x*(x^2-2*x+2)/(x-1)^6 + O(x^100)) \\ Colin Barker, Mar 18 2015
Formula
a(n) = binomial(n+4, n-1)+binomial(n+2, n-1).
Convolution of triangular numbers with triangular numbers + 1, i.e. [1, 3, 6, 10, 15, 21, ...] with [2, 4, 7, 11, 16, 22, ...].
a(1)=2, a(2)=10, a(3)=31, a(4)=76, a(5)=161, a(6)=308, a(n)=6*a(n-1)- 15*a(n-2)+ 20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Harvey P. Dale, Oct 03 2012
G.f.: x*(x^2-2*x+2) / (x-1)^6. - Colin Barker, Mar 18 2015