A246343 a(0) = 12, after which, if (2*a(n-1)) - 1 = product_{k >= 1} (p_k)^(c_k) then a(n) = product_{k >= 1} (p_{k-1})^(c_k), where p_k indicates the k-th prime, A000040(k).

12, 19, 31, 59, 44, 46, 55, 107, 134, 166, 317, 398, 282, 557, 470, 622, 763, 531, 1051, 1267, 1807, 3607, 7211, 4522, 9041, 3700, 3725, 3982, 7951, 15889, 30053, 24018, 24189, 34535, 14630, 12916, 21769, 27599, 24524, 32678, 26094, 43073, 34446, 68881, 116479, 143359, 275221, 550439, 667462, 1051489

Offset: 0

Author

Antti Karttunen, Aug 24 2014

Keywords

• nonn

Comments

Iterates of A064216 starting from value 12.

All numbers from 1 to 11 are in finite cycles of A048673/A064216, thus 12 is the smallest number in this cycle, regardless of whether it is infinite or finite.

Examples

			Start with a(0) = 12; then after each new term is obtained by doubling the previous term, from which one is subtracted, after which each prime factor is replaced with the previous prime:
12 -> ((2*12)-1) = 23 = p_9, and p_8 = 19, thus a(1) = 19.
19 -> ((2*19)-1) = 37 = p_12, and p_11 = 31, thus a(2) = 31.
31 -> ((2*31)-1) = 61 = p_18, and p_17 = 59, thus a(3) = 59.
59 -> ((2*59)-1) = 117 = 3*3*13 = p_2 * p_2 * p_6, and p_1 * p_1 * p_5 = 2*2*11 = 44, thus a(4) = 44.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..1001

Crossrefs

A246342 gives the terms of the same cycle when going to the opposite direction from 12.

Cf. A048673, A064216, A246344, A246345.

Cf. also A246281, A246282, A246351, A246352, A246361, A246362, A246371, A246372.

Programs

PARI

default(primelimit, 2^30);
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A064216(n) = A064989((2*n)-1);
k = 12; for(n=0, 1001, write("b246343.txt", n, " ", k); k = A064216(k));
(Scheme, with memoization-macro definec)
(definec (A246343 n) (if (zero? n) 12 (A064216 (A246343 (- n 1)))))

Formula

a(0) = 12, a(n) = A064216(a(n-1)).