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A246342
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a(0) = 12, after which, if a(n-1) = 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 k-th prime, A000040(k).
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5
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12, 23, 15, 18, 38, 35, 39, 43, 24, 68, 86, 71, 37, 21, 28, 50, 74, 62, 56, 149, 76, 104, 230, 305, 235, 186, 278, 224, 1337, 1062, 2288, 8951, 4482, 16688, 67271, 33637, 16821, 66688, 571901, 338059, 181516, 258260, 455900, 1180337, 1080207, 1817863, 1157487, 984558, 1230848, 53764115
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OFFSET
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0,1
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COMMENTS
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Iterates of A048673 starting from value 12.
All numbers 1 .. 11 are in finite cycles of A048673/A064216, thus 12 is the smallest number in this cycle, regardless of whether the cycle is infinite or finite.
This sequence soon reaches much larger values than the corresponding A246343 (iterating the same cycle in the other direction). However, with the corresponding sequences starting from 16 (A246344 & A246345), there is no such pronounced difference, and with them the bias is actually the other way.
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LINKS
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FORMULA
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a(0) = 12, and for n >= 1, a(n) = A048673(a(n-1)).
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EXAMPLE
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Start with a(0) = 12; thereafter each new term is obtained by replacing each prime factor of the previous term with the next prime, to whose product 1 is added before it is halved:
12 = 2^2 * 3 = p_1^2 * p_2 -> ((p_2^2 * p_3)+1)/2 = ((9*5)+1)/2 = 23, thus a(1) = 23.
23 = p_9 -> (p_10 + 1)/2 = (29+1)/2 = 15, thus a(2) = 15.
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PROG
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(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 = 12; for(n=0, 1001, write("b246342.txt", n, " ", k) ; k = A048673(k));
(Scheme, with memoization-macro definec)
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CROSSREFS
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A246343 gives the terms of the same cycle when going in the opposite direction from 12.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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