A238702 Sum of the smallest parts of the partitions of 4n into 4 parts.
1, 6, 21, 55, 119, 227, 396, 645, 996, 1474, 2106, 2922, 3955, 5240, 6815, 8721, 11001, 13701, 16870, 20559, 24822, 29716, 35300, 41636, 48789, 56826, 65817, 75835, 86955, 99255, 112816, 127721, 144056, 161910, 181374, 202542, 225511, 250380, 277251, 306229
Offset: 1
Examples
Add the numbers in the last column for a(n):
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 6 21 55 .. a(n)
Links
- Vincenzo Librandi, Table of n, a(n) for n = 1..100
- 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 (4,-6,5,-5,6,-4,1).
Programs
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Mathematica
CoefficientList[Series[(x + 1)*(2*x^2 + x + 1)/((1 - x)^5*(x^2 + x + 1)), {x, 0, 50}], x] (* Wesley Ivan Hurt, Jun 27 2014 *) LinearRecurrence[{4, -6, 5, -5, 6, -4, 1}, {1, 6, 21, 55, 119, 227, 396}, 50] (* Vincenzo Librandi, Aug 29 2015 *) -
PARI
Vec(-x*(x+1)*(2*x^2+x+1)/((x-1)^5*(x^2+x+1)) + O(x^100)) \\ Colin Barker, Mar 23 2014
Formula
G.f.: -x*(x+1)*(2*x^2+x+1) / ((x-1)^5*(x^2+x+1)). - Colin Barker, Mar 10 2014
a(n) = (1/9)*n^4 + (1/3)*n^3 + (5/18)*n^2 + (1/6)*n + O(1). - Ralf Stephan, May 29 2014
a(n) = Sum_{i=1..n} A238340(i). - Wesley Ivan Hurt, May 29 2014
a(n) = (1/4) * Sum_{i=1..n} A238328(i)/i. - Wesley Ivan Hurt, 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)). Then a(1) = 1, with a(n) = b(n)/(4n) + a(n-1), for n>1. - Wesley Ivan Hurt, Jun 27 2014
E.g.f.: (exp(x)*(4 + 3*x*(16 + x*(37 + 2*x*(9 + x)))) - 4*exp(-x/2)*cos(sqrt(3)*x/2))/54. - Stefano Spezia, Feb 09 2023
a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 5*a(n-4) + 6*a(n-5) - 4*a(n-6) + a(n-7). - Wesley Ivan Hurt, Jun 19 2024
Comments