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A182995
Sum of parts of the n-th subsection of the head of the last section of the set of partitions of any odd integer >= 2n+1.
7
3, 7, 18, 44, 82, 158, 303, 507, 873, 1470, 2354, 3756, 5923, 9065, 13815, 20824, 30853, 45365, 66210, 95415, 136696, 194414, 274057, 384136, 535219, 740559, 1019529, 1396212, 1901533, 2577918, 3479291, 4673711, 6253003, 8332767
OFFSET
1,1
COMMENTS
The last section of the set of partitions of 2n+1 contains n subsections.
Also first differences of A182737. - Omar E. Pol, Mar 03 2011
FORMULA
a(n) = A138880(2n+1) - A138880(2n-1).
EXAMPLE
a(5)=82 because the 5th subsection of the head of the last section of any odd integer >= 11 looks like this:
(11 . . . . . . . . . . )
( 6 . . . . . 5 . . . . )
( 7 . . . . . . 4 . . . )
( 8 . . . . . . . 3 . . )
( 4 . . . 4 . . . 3 . . )
( 5 . . . . 3 . . 3 . . )
. (2 . )
. (2 . )
. (2 . )
. (2 . )
. (2 . )
. (2 . )
. (2 . )
. (2 . )
There are 21 parts whose sum is 11+6+5+7+4+8+3+4+4+3+5+3+3+2+2+2+2+2+2+2+2 = 11*6 + 2*8 = 82, so a(5) = 82.
KEYWORD
nonn
AUTHOR
Omar E. Pol, Feb 06 2011
EXTENSIONS
a(17) corrected and more terms from Omar E. Pol, Mar 03 2011.
a(12) corrected by Georg Fischer, Aug 31 2020
STATUS
approved