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a(n) = A183002(n)/2.
4

%I #20 Jan 19 2024 04:57:47

%S 0,0,0,1,1,2,2,3,4,5,5,7,7,8,9,11,11,13,13,15,16,17,17,20,21,22,23,25,

%T 25,28,28,30,31,32,33,37,37,38,39,42,42,45,45,47,49,50,50,54,55,57,58,

%U 60,60,63,64,67,68,69,69,74,74,75,77,80,81,84,84,86,87,90,90,95,95,96,98,100,101,104,104,108,110,111,111,116,117,118,119,122,122,127,128,130,131,132,133,138,138,140,142,146

%N a(n) = A183002(n)/2.

%C For n >= 2, a(n) is the number of partitions of n-1 into 3 parts such that the largest part is greater than or equal to the product of the other two. For example, a(9) = 4 since the partitions for 8 would be 1+1+6 = 1+2+5 = 1+3+4 = 2+2+4, but not 2+3+3 since 2*3 > 3. - _Wesley Ivan Hurt_, Jan 03 2022

%C Conjecture: partial sums of A072670. - _Sean A. Irvine_, Jul 14 2022

%F a(n) = Sum_{k=1..floor((n-1)/3)} Sum_{i=k..floor((n-k-1)/2)} sign(floor((n-i-k-1)/(i*k))). - _Wesley Ivan Hurt_, Jan 03 2022

%F a(n) = (1/2) * Sum_{k=1..n} (tau(k)-2 + (tau(k) mod 2)), tau = A000005. - _Alois P. Heinz_, Jan 04 2022

%F a(n) ~ n * (log(n) + 2*gamma - 3) / 2, where gamma is Euler's constant (A001620). - _Amiram Eldar_, Jan 19 2024

%t Accumulate[Table[d = DivisorSigma[0, n]; If[OddQ[d], d - 1, d - 2], {n, 100}]]/2

%o (PARI) a(n) = sum(k=1, n, numdiv(k) - 2 + numdiv(k)%2)/2; \\ _Michel Marcus_, Jan 04 2022

%Y Cf. A000005, A001620, A072670, A183002.

%K nonn,easy

%O 1,6

%A _Omar E. Pol_, Jan 27 2011