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A246362
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Numbers n such that if 2n-1 = product_{k >= 1} (p_k)^(c_k), then n < product_{k >= 1} (p_{k-1})^(c_k), where p_k indicates the k-th prime, A000040(k).
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11
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4, 6, 7, 9, 10, 12, 15, 16, 19, 20, 21, 22, 24, 27, 29, 30, 31, 34, 35, 36, 37, 40, 42, 44, 45, 46, 47, 48, 49, 51, 52, 54, 55, 56, 57, 60, 62, 64, 65, 66, 67, 69, 70, 71, 72, 75, 76, 78, 79, 80, 81, 82, 84, 85, 87, 89, 90, 91, 92, 96, 97, 99, 100, 101, 102, 103, 105, 106, 107, 108, 109, 110, 111, 112, 114, 115
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OFFSET
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1,1
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COMMENTS
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Numbers n such that A064216(n) > n.
Numbers n such that A064989(2n-1) > n.
The sequence grows as:
a(100) = 148
a(1000) = 1449
a(10000) = 14264
a(100000) = 141259
a(1000000) = 1418197
and the powers of 10 occur at:
a(5) = 10
a(63) = 100
a(701) = 1000
a(6973) = 10000
a(70845) = 100000
a(705313) = 1000000
suggesting that the ratio a(n)/n is converging to a constant and an arbitrary natural number is more than twice as likely to be here than in the complement A246361. Compare this to the ratio present in the "inverse" case A246282.
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LINKS
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EXAMPLE
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4 is present, as 2*4 - 1 = 7 = p_4, and p_{4-1} = p_3 = 5 > 4.
5 is not present, as 2*5 - 1 = 9 = p_2 * p_2, and p_1 * p_1 = 4, with 4 < 5.
6 is present, as 2*6 - 1 = 11 = p_5, and p_{5-1} = p_4 = 7 > 6.
35 is present, as 2*35 - 1 = 69 = 3*23 = p_2 * p_9, and p_1 * p_8 = 2*19 = 38 > 35.
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PROG
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(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)};
n = 0; i = 0; while(i < 10000, n++; if(isA246362(n), i++; write("b246362.txt", i, " ", n)));
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CROSSREFS
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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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