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some small corrections
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Michael Hardy
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This is Sloane's A121559, which essentially iterates A064722. The behavior is controlled by the (appropriately weighted) distribution of prime gaps below N.$N.$

Heuristically, with $\ell=1-1/\log N$, you'd expect something like $(1-\ell)\left(1+\ell^3+\ell^5+\ell^7+\ell^{11}+\cdots\right)$ 1s$1$s below N,$N,$ where the exponents are 1$1$ less than the locations of the 1s$1$s. You can stop the sum around $\log^2 N$.

This is Sloane's A121559, which essentially iterates A064722. The behavior is controlled by the (appropriately weighted) distribution of prime gaps below N.

Heuristically, with $\ell=1-1/\log N$, you'd expect something like $(1-\ell)\left(1+\ell^3+\ell^5+\ell^7+\ell^{11}+\cdots\right)$ 1s below N, where the exponents are 1 less than the locations of the 1s. You can stop the sum around $\log^2 N$.

This is Sloane's A121559, which essentially iterates A064722. The behavior is controlled by the (appropriately weighted) distribution of prime gaps below $N.$

Heuristically, with $\ell=1-1/\log N$, you'd expect something like $(1-\ell)\left(1+\ell^3+\ell^5+\ell^7+\ell^{11}+\cdots\right)$ $1$s below $N,$ where the exponents are $1$ less than the locations of the $1$s. You can stop the sum around $\log^2 N$.

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Charles
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This is Sloane's A121559, which essentially iterates A064722. The behavior is controlled by the (appropriately weighted) distribution of prime gaps below N.

Heuristically, with $\ell=1-1/\log N$, you'd expect something like $(1-\ell)\left(1+\ell^3+\ell^5+\ell^7+\ell^{11}+\cdots\right)$ 1s below N, where the exponents are 1 less than the locations of the 1s. You can stop the sum around $\log^2 N$.