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a(n) is the index of the first occurrence of n in A195061.
0

%I #18 Feb 14 2023 08:56:47

%S 2,36,108,284

%N a(n) is the index of the first occurrence of n in A195061.

%C In sequence A195061, the 2nd through 25th terms are 1's, which means it is possible to find a sum of p=b+c, p=prime(n), the prime factors of b and c traverse all primes smaller than sqrt(prime(n)). From the 36th term, the above case no longer holds for most primes. However, if you choose two eligible sums p=b1+c1=b2+c2, it becomes possible that the union set of prime factors of b1, c1, b2, and c2 traverses all primes smaller than sqrt(prime(n)). This stands until prime(107)=587. For prime(108)=593, no group of two b's and c's can have the union set of their prime factors to traverse all primes smaller than sqrt(prime(n)). Three groups of b's and c's will be needed to do so. And starting from prime(284)=1861, four groups are needed for some of the numbers.

%C The listed Mathematica program fails to find a(5) since it exceeds 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: *)

%t k = 0; i = 1; a = 0;

%t Array[ff, 4];

%t 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++]];

%t Table[ff[m], {m, 4}]

%Y Cf. A196526, A195061.

%K nonn,hard,bref,uned

%O 1,1

%A _Lei Zhou_, Oct 03 2011