A294286 Sum of the squares of the parts in the partitions of n into two distinct parts.
0, 0, 5, 10, 30, 46, 91, 124, 204, 260, 385, 470, 650, 770, 1015, 1176, 1496, 1704, 2109, 2370, 2870, 3190, 3795, 4180, 4900, 5356, 6201, 6734, 7714, 8330, 9455, 10160, 11440, 12240, 13685, 14586, 16206, 17214, 19019, 20140, 22140, 23380, 25585, 26950, 29370
Offset: 1
Examples
For n = 6, there are two ways of partitioning 6 into two distinct parts: 6 = 1+5 and 6 = 2+4. So a(6) = 1^2 + 5^2 + 2^2 + 4^2 = 46. For n = 7, there are three ways of partitioning 7 into two distinct parts: 7 = 1+6, 7 = 2+5, and 7 = 3+4. So a(7) = 1^2 + 6^2 + 2^2 + 5^2 + 3^2 + 4^2 = 91. - _Michael B. Porter_, Nov 05 2017
Links
- Iain Fox, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)
- Index entries for sequences related to partitions
- Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).
Crossrefs
Cf. A000330.
Programs
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Magma
[n*(2*n^2-3*n*(5+(-1)^n)/4+1)/6 : n in [1..60]]; // Wesley Ivan Hurt, Dec 03 2023
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Mathematica
Table[Sum[i^2 + (n - i)^2, {i, Floor[(n-1)/2]}], {n, 40}] Table[Total[Flatten[Select[IntegerPartitions[n,{2}],#[[1]]!=#[[2]]&]]^2],{n,50}] (* Harvey P. Dale, Dec 02 2022 *) -
PARI
first(n) = my(res = vector(n, i, i^3 / 3 - i^2 / 2 + i / 6)); forstep(i = 2, n, 2, res[i] -= i^2 >> 2); res \\ David A. Corneth, Oct 27 2017
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PARI
concat(vector(2), Vec(x^3*(5 + 5*x + 5*x^2 + x^3) / ((1 - x)^4*(1 + x)^3) + O(x^60))) \\ Colin Barker, Nov 04 2017
Formula
a(n) = Sum_{i=1..floor((n-1)/2)} i^2 + (n-i)^2.
From David A. Corneth, Oct 27 2017: (Start)
For odd n, a(n) = n^3/3 - n^2/2 + n/6 = A000330(n + 1).
For even n, a(n) = n^3/3 - 3*n^2/4 + n/6.
(End)
From Colin Barker, Nov 04 2017: (Start)
G.f.: x^3*(5 + 5*x + 5*x^2 + x^3) / ((1 - x)^4*(1 + x)^3).
a(n) = a(n-1) + 3*a(n-2) - 3*a(n-3) - 3*a(n-4) + 3*a(n-5) + a(n-6) - a(n-7) for n > 7.
(End)
a(n) = n*(2*n^2-3*n*(5+(-1)^n)/4+1)/6. - Wesley Ivan Hurt, Dec 03 2023