

A246344


a(0) = 16, after which, if a(n1) = product_{k >= 1} (p_k)^(c_k), then a(n) = (1/2) * (1 + product_{k >= 1} (p_{k+1})^(c_k)), where p_k indicates the kth prime, A000040(k).


5



16, 41, 22, 20, 32, 122, 101, 52, 77, 72, 338, 434, 611, 451, 280, 1040, 4820, 7907, 3960, 30713, 15364, 22577, 12154, 9791, 4902, 8108, 9131, 5815, 4099, 2056, 3551, 2095, 1474, 1385, 984, 2903, 1455, 1768, 4361, 5869, 2940, 19058, 18845, 13227, 11053, 8707, 4357, 2182, 1640, 4064, 15917, 9432, 46238
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OFFSET

0,1


COMMENTS

Iterates of A048673 starting from value 16.
Either this sequence is actually part of the cycle containing 12 (see A246342) or 16 is the smallest member of this cycle (regardless of whether this cycle is finite or infinite), which follows because all numbers 1 .. 11 are in finite cycles, and also 13 and 14 are in closed cycles and 15 is in the cycle of 12.


LINKS



FORMULA

a(0) = 16, and for n >= 1, a(n) = A048673(a(n1)).


EXAMPLE

Start with a(0) = 16; then after each new term is obtained by replacing each prime factor of the previous term with the next prime, to whose product is added one before it is halved:
16 = 2^4 = p_1^4 > ((p_2^4)+1)/2 = (3^4 + 1)/2 = (81+1)/2 = 41, thus a(1) = 41.
41 = p_13 > ((p_14)+1)/2 = (43+1)/2 = 22, thus a(2) = 22.


PROG

(PARI)
default(primelimit, 2^30);
A003961(n) = my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); \\ Using code of Michel Marcus
k = 16; for(n=0, 1001, write("b246344.txt", n, " ", k) ; k = A048673(k));
(Scheme, with memoizationmacro definec)


CROSSREFS

A246345 gives the terms of the same cycle when going to the opposite direction from 16.


KEYWORD

nonn


AUTHOR



STATUS

approved



