%I
%S 2,36,108,284
%N a(n) is the index of the first occurrence of n in A195061.
%C In sequence A195061, element 2 through 25 are 1, which means it is possible to find a sum of p=b+c, p is the nth prime number, the prime factors of b and c traverse all primes smaller than square root p(n). From element 36, the above case does no longer stand for most of primes. However, if you pick two eligible sum p=b1+c1=b2+c2, it becomes possible that the union set of prime factors of b1, c1, b2, and c2 traverse all primes smaller than square root p(n). This stands until p(107)=587. For p(108)=593, no group of two b and cs can have the union set of their prime factors to traverse all primes smaller than square root p(n). Three groups of b and cs will be needed to do so. And starting from p(284)=1861, four groups are needed for some of the numbers.
%C The listed Mathematica program failed in finding a(5) since it exceeded the integer range as index.
%e A195061[2]=1 => a(1)=2;
%e A195061[3]=A195061[4]=...=A195061[35]=1; A195061[36]=2 => a(2)=36;
%e A195061[1..107]<=2; A195061[108]=3 => a(3)=108;
%e A195061[1..283]<=3; A195061[284]=4 => a(4)=284;
%t (* Taking the function Checks[n_] in the Mathematica program for A195061, the following program gives the first four terms: *) k = 0; i = 1; a = 0; Array[ff, 4]; Do[ff[j] = 0, {j, 1, 4}]; While[(k < 4) && (a < 4), i++; a = Checks[i]; If[(a <= 4) && (ff[a] == 0), ff[a] = i;k++]];Table[ff[m], {m, 4}]
%Y Cf. A196526, A195061.
%K nonn,hard,bref
%O 1,1
%A _Lei Zhou_, Oct 03 2011
