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Divide all prime indices by 2, round down, and take the number with those prime indices, assuming prime(0) = 1.
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%I #14 Feb 12 2023 17:29:44

%S 1,1,2,1,2,2,3,1,4,2,3,2,5,3,4,1,5,4,7,2,6,3,7,2,4,5,8,3,11,4,11,1,6,

%T 5,6,4,13,7,10,2,13,6,17,3,8,7,17,2,9,4,10,5,19,8,6,3,14,11,19,4,23,

%U 11,12,1,10,6,23,5,14,6,29,4,29,13,8,7,9,10,31

%N Divide all prime indices by 2, round down, and take the number with those prime indices, assuming prime(0) = 1.

%C A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

%C Also the Heinz number of the part-wise half (rounded down) of the partition with Heinz number n, where the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

%C Each n appears A000005(n) times at odd positions (infinitely many at even). To see this, note that our transformation does not distinguish between A066207 and A066208.

%F Completely multiplicative with a(prime(2k)) = prime(k) and a(prime(2k+1)) = prime(k). Cf. A297002.

%F a(prime(n)) = A109763(n-1).

%e The prime indices of n = 1501500 are {1,1,2,3,3,3,4,5,6}, so the prime indices of a(n) are {1,1,1,1,2,2,3}; hence we have a(1501500) = 720.

%e The 6 odd positions of 2124 are: 63, 99, 105, 165, 175, 275, with prime indices:

%e 63: {2,2,4}

%e 99: {2,2,5}

%e 105: {2,3,4}

%e 165: {2,3,5}

%e 175: {3,3,4}

%e 275: {3,3,5}

%t Table[Times@@(If[#1<=2,1,Prime[Floor[PrimePi[#1]/2]]^#2]&@@@FactorInteger[n]),{n,100}]

%Y Positions of 1's are A000079.

%Y Positions of 2's are 3 and A164095.

%Y Positions of first appearances are A297002, sorted A066207.

%Y A004526 is floor(n/2), with an extra first zero.

%Y A056239 adds up prime indices, row-sums of A112798.

%Y A109763 lists primes of index floor(n/2).

%Y Cf. A003961, A064988, A064989, A076610, A215366, A248601, A357980.

%K nonn

%O 1,3

%A _Gus Wiseman_, Oct 23 2022