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A084384
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a(1) = 2; a(n+1) = smallest k > a(n) that is divisible by at most (1/2)*[tau(k)] previous terms.
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1
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2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 46, 47, 48, 49, 51, 53, 55, 57, 58, 59, 61, 62, 65, 67, 69, 71, 72, 73, 74, 77, 79, 80, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 101, 103, 106, 107, 108
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OFFSET
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1,1
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COMMENTS
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Membership in this sequence depends only on the prime signature, see PARI script. - Charles R Greathouse IV, Oct 19 2015
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LINKS
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EXAMPLE
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12 is not a member as tau(12) = 6 and there are four terms 2,3,4 and 6 that divide 12.
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PROG
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(PARI) has(f)=f=select(n->n, f); if(#f==0, return(0)); if(#f==1 && f[1]<4, return(1)); my(t=prod(i=1, #f, f[i]+1)\2); forvec(x=vector(#f, i, [0, f[i]]), if(x==f, return(1)); t-=has(x); if(t<0, return(0)))
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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