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A074347
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Smallest number requiring n steps to reach 0 when iterating the function: f(n)=abs(lpd(n)-Lpf(n)), where lpd(n) is the largest proper divisor of n and Lpf(n) is the largest prime factor of n.
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1
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1, 2, 3, 12, 13, 52, 53, 131, 271, 811, 1601, 2711, 8111, 13997, 34589, 74551, 147773, 310567, 621227, 1230343, 2627759, 4921373, 10741931, 24965191, 45887291, 111477631, 183638843, 394195667, 788380493, 1576798931
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OFFSET
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1,2
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COMMENTS
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2*10^9 < a(31) <= 2938669883. a(32) <= 7511549827. a(33) <= 11754740251. a(34) <= 30050593523. - Donovan Johnson, Dec 22 2010
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LINKS
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PROG
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(PARI) {m=25; z=11000000; v=listcreate(m); for(i=1, m, listinsert(v, -1, i)); for(n=1, z, c=1; b=1; k=n; while(b&&c<=m, d=divisors(k); i=matsize(d)[2]-1; p=if(i>0, d[i], 1); q=if(k==1, 1, vecmax(component(factor(k), 1))); a=abs(p-q); if(a==0, b=0, k=a; c++)); if(a==0, if(v[c]<0, v[c]=n; print1([c, n])))); print(); for(i=1, m, print1(v[i], ", "))}
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CROSSREFS
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KEYWORD
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more,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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