

A072491


Define f(1) = 0. For n>=2, let f(n) = n  p where p is the largest prime <= n. a(n) = number of iterations of f to reach 0, starting from n.


3



1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 2, 3, 2, 1, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 1, 2, 2, 2, 3, 2, 3, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2
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OFFSET

1,4


COMMENTS

a(p)=1, a(p+1)=2 and a(p+4)=3 if p is an odd prime but p+2 and p+4 are composite.


REFERENCES

S. S. Pillai, "An arithmetical function concerning primes", Annamalai University Journal (1930), pp. 159167.


LINKS

Table of n, a(n) for n=1..105.


FORMULA

On CramÃ©r's conjecture, a(n) = O(log* n).  Charles R Greathouse IV, Feb 04 2013


EXAMPLE

a(27)=3 as f(27)=2723=4, f(4)=43=1 and f(1)=0.


MATHEMATICA

f[1]=0; f[n_] := nPrime[PrimePi[n]]; a[n_] := Module[{k, x}, For[k=0; x=n, x>0, k++; x=f[x], Null]; k]


PROG

(PARI) a(n)=if(n<4, n>0, 1+a(nprecprime(n))) \\ Charles R Greathouse IV, Feb 04 2013


CROSSREFS

Cf. A072492. A066352(n) is the smallest k such that a(k)=n.
Not the same as A051034: a(122) = 3, but A051034(122) = 2.
Sequence in context: A071854 A183025 A072410 * A051034 A082477 A036430
Adjacent sequences: A072488 A072489 A072490 * A072492 A072493 A072494


KEYWORD

nonn,easy


AUTHOR

Amarnath Murthy, Jul 14 2002


EXTENSIONS

Edited by Dean Hickerson, Nov 26 2002


STATUS

approved



