A339183 Number of partitions of n into two parts such that the smaller part is a nonzero square.
0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7
Offset: 0
Examples
a(7) = 1; The partitions of 7 into two parts are (6,1), (5,2) and (4,3). Since 1 is the only nonzero square appearing among the smallest parts, a(7) = 1. a(8) = 2; The partitions of 8 into two parts are (7,1), (6,2), (5,3) and (4,4). Since 1 and 4 are the only nonzero squares appearing among the smallest parts, a(8) = 2.
Crossrefs
Programs
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Mathematica
Table[Sum[Floor[Sqrt[i]] - Floor[Sqrt[i - 1]] , {i, Floor[n/2]}], {n, 0, 120}]
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Python
from math import isqrt def A339183(n): return isqrt(n>>1) # Chai Wah Wu, Oct 07 2024
Formula
a(n) = Sum_{i=1..floor(n/2)} c(i), where c is the square characteristic (A010052).
a(n) = floor(sqrt(n/2)), for n >= 0. [See the comment, and {4*k+2 = A016825 (k)} is the first difference sequence of the sequence b(k) = 2*k^2 = A001105(k) for which sqrt(b(k)/2) becomes an integer.] - Wolfdieter Lang, Dec 07 2020
Comments