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A359260
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Numbers m such that the arithmetic mean of the first k divisors of m is an integer for all k in 1..d(m), where d(m) = A000005(m).
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3
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1, 3, 5, 7, 11, 13, 15, 17, 19, 23, 29, 31, 33, 37, 41, 43, 47, 49, 51, 53, 59, 61, 67, 69, 71, 73, 79, 83, 87, 89, 91, 97, 101, 103, 107, 109, 113, 123, 127, 131, 133, 137, 139, 141, 149, 151, 157, 159, 163, 167, 169, 173, 177, 179, 181, 191, 193, 197, 199, 211
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OFFSET
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1,2
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COMMENTS
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All the terms are arithmetic numbers (A003601).
All the terms are odd numbers.
All the odd primes are terms.
There are infinitely many composite numbers in this sequence. For example, if p is a prime of the form 6*k-1 (A007528), then 3*p is a term. Also, if p is a prime of the form 6*k + 1 (A002476), then p^2 is a term.
prime(n)^k is a term for k = 0..A359262(n).
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LINKS
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EXAMPLE
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15 is a term since its divisors are {1, 3, 5, 15}, 1/1 =1, (1 + 3)/2 = 2, (1 + 3 + 5)/3 = 3, and (1 + 3 + 5 + 15)/4 = 6 are all integers.
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MATHEMATICA
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q[n_] := AllTrue[Accumulate[(d = Divisors[n])]/Range[Length[d]], IntegerQ]; Select[Range[1, 200, 2], q]
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PROG
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(PARI) is(n) = {my(s = k = 0); fordiv(n, d, k++; s += d; if(s%k, return(0))); 1; }
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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