%I #24 Oct 31 2023 12:30:05
%S 2,1,3,1,4,1,2,4,2,1,6,1,2,2,6,1,3,1,7,2,2,1,4,6,2,2,3,1,9,1,3,2,2,2,
%T 10,1,2,2,4,1,10,1,3,3,2,1,5,8,3,2,3,1,4,2,11,2,2,1,6,1,2,3,11,2,4,1,
%U 3,2,4,1,14,1,2,3,3,2,4,1,5,11,2,1,6,2,2
%N Differences of the sum of distinct values of {floor(n/k), k=1,...,n}.
%C Differences of A051201.
%C It appears that a(n)=1 if and only if n>1 and n+1 is a prime. For example, the indices where 1 occurs in {a(n)} are {2,4,6,10,12,16,...}. Adding 1 to each of these gives {3,5,7,11,13,17,...} each of which is a prime.
%H Alois P. Heinz, <a href="/A210256/b210256.txt">Table of n, a(n) for n = 1..10000</a>
%p b:= proc(n) option remember; add(i, i={seq(floor(n/k), k=1..n)}) end:
%p a:= n-> b(n+1)-b(n):
%p seq(a(n), n=1..150); # _Alois P. Heinz_, Mar 19 2012
%t b[n_] := b[n] = Total@ Union@ Table[Floor[n/k], {k, 1, n}];
%t a[n_] := b[n+1] - b[n];
%t Array[a, 150] (* _Jean-François Alcover_, Nov 20 2020, after _Alois P. Heinz_ *)
%o (Python)
%o from math import isqrt
%o def A210256(n): return ((m:=isqrt((n+1<<2)+1)+1>>1)*(m-1)>>1)+sum((n+1)//k for k in range(1,(n+1)//m+1))-((r:=isqrt((n<<2)+1)+1>>1)*(r-1)>>1)-sum(n//k for k in range(1,n//r+1)) # _Chai Wah Wu_, Oct 31 2023
%Y Cf. A006218, A051201, A055086.
%K nonn
%O 1,1
%A _John W. Layman_, Mar 19 2012