OFFSET
1,3
LINKS
Iain Fox, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).
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
EXAMPLE
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
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 *)
PROG
(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
(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
(Magma) [n*(2*n^2-3*n*(5+(-1)^n)/4+1)/6 : n in [1..60]]; // Wesley Ivan Hurt, Dec 03 2023
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wesley Ivan Hurt, Oct 26 2017
STATUS
approved