A393402 Sum of the smallest parts in the partitions of n into 3 parts whose middle part is odd.

0, 0, 0, 1, 1, 1, 1, 2, 4, 7, 7, 8, 10, 13, 17, 23, 25, 28, 32, 38, 46, 56, 60, 66, 74, 84, 96, 111, 119, 129, 141, 156, 174, 195, 207, 222, 240, 261, 285, 313, 331, 352, 376, 404, 436, 472, 496, 524, 556, 592, 632, 677, 709, 745, 785, 830, 880, 935, 975, 1020, 1070, 1125, 1185, 1251, 1301, 1356, 1416, 1482, 1554, 1632, 1692

Offset

0,8

Links

Table of n, a(n) for n=0..70.

Index entries for sequences related to partitions

Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-1,0,2,-4,4,-4,2,0,-1,2,-2,2,-1).

Formula

a(n) = Sum_{j=1..floor(n/3)} Sum_{k=j..floor((n-j)/2)} j * (k mod 2).

a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - a(n-4) + 2*a(n-6) - 4*a(n-7) + 4*a(n-8) - 4*a(n-9) + 2*a(n-10) - a(n-12) + 2*a(n-13) - 2*a(n-14) + 2*a(n-15) - a(n-16).

a(n) + A309688(n) + A393412(n) = n*A309687(n). - Wesley Ivan Hurt, Feb 17 2026

a(n) ~ n^3/216. - Charles R Greathouse IV, May 31 2026

Example

a(10) = 7; There are 4 partitions of 10 into 3 parts whose middle part is odd: (8,1,1), (6,3,1), (5,3,2) and (4,3,3).  The sum of the smallest parts of these partitions is 1+1+2+3 = 7.

Mathematica

Table[Sum[Sum[j*Mod[k, 2], {k, j, Floor[(n - j)/2]}], {j, Floor[n/3]}], {n, 0, 100}]
LinearRecurrence[{2, -2, 2, -1, 0, 2, -4, 4, -4, 2, 0, -1, 2, -2, 2, -1}, {0, 0, 0, 1, 1, 1, 1, 2, 4, 7, 7, 8, 10, 13, 17, 23}, 100]

Prog

(PARI) a(n)=(2*n^3+9*n^2+12*[-3, -2, 1, 6, 1, -2, -3, -2, 1, 6, 1, -2][n%12+1]*n+[0, 13, -76, 81, 112, 77, -108, -95, 32, 189, 4, -31][n%12+1])/432 \\ Charles R Greathouse IV, May 31 2026

Crossrefs

Cf. A309687, A309688, A393412.

Sequence in context: A161211 A276807 A363166 * A161368 A396009 A023978

Adjacent sequences: A393399 A393400 A393401 * A393403 A393404 A393405

Keyword

nonn,easy

Author

Wesley Ivan Hurt, Feb 13 2026

Status

approved