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A182132
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Carmichael numbers of the form C = (30n-7)*(90n-23)*(300n-79).
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1
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340561, 4335241, 153927961, 542497201, 1678569121, 2598933481, 3164207761, 25923026641, 63280622521, 88183003921, 155999871721, 209850699601, 240893092441, 274855097881, 380692027321, 733547013841, 1688729866321
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OFFSET
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1,1
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COMMENTS
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Note that in this sequence, 30n-7, 90n-23, and 300n-79 do not have to be prime.
Conjecture: The number C = (30n-7)*(90n-23)*(300n-79) is a Carmichael number if (but not only if) 30n-7, 90n-23 and 300n-79 are all three prime numbers.
The conjecture is checked for 0<n<130; we got Carmichael numbers with three prime divisors for n = 2, 9, 15, 32, 43, 48, 58, 64, 67, 78, 97, 128.
We got Carmichael numbers with more than three prime divisors for n = 1, 6, 13, 70.
Note that if n is 5 mod 7 then 90n-23 and 300n-79 are both multiples of 7 and hence C is not a Carmichael number. Otherwise, the three factors are relatively prime. - Charles R Greathouse IV, Jun 29 2017
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LINKS
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PROG
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(PARI) list(lim)=my(v=List(), C, n); while(n++ && (C=(30*n-7)*(90*n-23)*(300*n-79))<=lim, if(n%7==5, next); my(f1=factor(30*n-7), f2=factor(90*n-23), f3=factor(300*n-79), t); if(vecmax(f1[, 2])>1 || vecmax(f2[, 2])>1 || vecmax(f3[, 2])>1, next); t=concat(concat(f1[, 1], f2[, 1]), f3[, 1]); for(i=1, #t, if((C-1)%(t[i]-1), next(2))); listput(v, C)); Vec(v) \\ Charles R Greathouse IV, Jun 29 2017
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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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