A238340 Number of partitions of 4n into 4 parts.
1, 5, 15, 34, 64, 108, 169, 249, 351, 478, 632, 816, 1033, 1285, 1575, 1906, 2280, 2700, 3169, 3689, 4263, 4894, 5584, 6336, 7153, 8037, 8991, 10018, 11120, 12300, 13561, 14905, 16335, 17854, 19464, 21168, 22969, 24869, 26871, 28978, 31192, 33516, 35953
Offset: 1
Examples
Count the partitions of 4*n into 4 parts:
13 + 1 + 1 + 1
12 + 2 + 1 + 1
11 + 3 + 1 + 1
10 + 4 + 1 + 1
9 + 5 + 1 + 1
8 + 6 + 1 + 1
7 + 7 + 1 + 1
11 + 2 + 2 + 1
10 + 3 + 2 + 1
9 + 4 + 2 + 1
8 + 5 + 2 + 1
7 + 6 + 2 + 1
9 + 3 + 3 + 1
8 + 4 + 3 + 1
7 + 5 + 3 + 1
6 + 6 + 3 + 1
7 + 4 + 4 + 1
6 + 5 + 4 + 1
5 + 5 + 5 + 1
9 + 1 + 1 + 1 10 + 2 + 2 + 2
8 + 2 + 1 + 1 9 + 3 + 2 + 2
7 + 3 + 1 + 1 8 + 4 + 2 + 2
6 + 4 + 1 + 1 7 + 5 + 2 + 2
5 + 5 + 1 + 1 6 + 6 + 2 + 2
7 + 2 + 2 + 1 8 + 3 + 3 + 2
6 + 3 + 2 + 1 7 + 4 + 3 + 2
5 + 4 + 2 + 1 6 + 5 + 3 + 2
5 + 3 + 3 + 1 6 + 4 + 4 + 2
4 + 4 + 3 + 1 5 + 5 + 4 + 2
5 + 1 + 1 + 1 6 + 2 + 2 + 2 7 + 3 + 3 + 3
4 + 2 + 1 + 1 5 + 3 + 2 + 2 6 + 4 + 3 + 3
3 + 3 + 1 + 1 4 + 4 + 2 + 2 5 + 5 + 3 + 3
3 + 2 + 2 + 1 4 + 3 + 3 + 2 5 + 4 + 4 + 3
1 + 1 + 1 + 1 2 + 2 + 2 + 2 3 + 3 + 3 + 3 4 + 4 + 4 + 4
4(1) 4(2) 4(3) 4(4) .. 4n
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1 5 15 34 .. a(n)
Links
- Iain Fox, Table of n, a(n) for n = 1..10000 (first 200 terms from Vincenzo Librandi)
- Antonio Osorio, A Sequential Allocation Problem: The Asymptotic Distribution of Resources, Munich Personal RePEc Archive, 2014.
- Index entries for sequences related to partitions
- Index entries for linear recurrences with constant coefficients, signature (3,-3,2,-3,3,-1).
Programs
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Magma
I:=[1,5,15,34,64,108]; [n le 6 select I[n] else 3*Self(n-1)-3*Self(n-2)+2*Self(n-3)-3*Self(n-4)+3*Self(n-5)-Self(n-6): n in [1..45]]; // Vincenzo Librandi, Aug 29 2015
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Mathematica
CoefficientList[Series[(x + 1)*(2*x^2 + x + 1)/((x - 1)^4*(x^2 + x + 1)), {x, 0, 50}], x] (* Wesley Ivan Hurt, Jun 27 2014 *) Table[2*n/9 + n^2/3 + 4*n^3/9 - Floor[n/3]/3 - Floor[(n+1)/3]/3, {n, 1, 50}] (* Vaclav Kotesovec, Jul 04 2014 *) LinearRecurrence[{3, -3, 2, -3, 3, -1}, {1, 5, 15, 34, 64, 108}, 50] (* Vincenzo Librandi, Aug 29 2015 *) -
PARI
Vec(x*(x+1)*(2*x^2+x+1)/((x-1)^4*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Mar 24 2014
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PARI
a(n)=n^2*(4*n+3)\/9 \\ Charles R Greathouse IV, Jun 29 2020
Formula
a(n) = A238328(n) / 4n.
G.f.: x*(x+1)*(2*x^2+x+1) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Mar 10 2014
a(n) = 4/9*n^3 + 1/3*n^2 + O(1). - Ralf Stephan, May 29 2014
Recurrence: Let b(1) = 4, with b(n) = (n/(n-1)) * b(n-1) + 4n * Sum_{i=0..2n} (floor((4n-2-i)/2)-i) * (floor((sign((floor((4n-2-i)/2)-i)) +2)/2)), for n>1. Then a(n) = b(n)/(4n). - Wesley Ivan Hurt, Jun 27 2014
Recurrence: (4*n^3 - 21*n^2 + 44*n - 33)*a(n) = 3*(4*n^2 - 10*n + 9)*a(n-1) + 3*(4*n^2 - 10*n + 9)*a(n-2) + (4*n^3 - 9*n^2 + 14*n - 6)*a(n-3). - Vaclav Kotesovec, Jul 04 2014
a(n) = round(((4n)^3 + 3*(4n)^2)/144). - Giacomo Guglieri, Jun 28 2020
E.g.f.: exp(-x/2)*(3*exp(3*x/2)*(1 + x*(7 + x*(15 + 4*x))) - 3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2))/27. - Stefano Spezia, Feb 09 2023
a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 3*a(n-4) + 3*a(n-5) - a(n-6). - Wesley Ivan Hurt, Jun 19 2024
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