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A328243
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Numbers whose arithmetic derivative (A003415) is larger than 1 and one of the terms of A143293 (partial sums of primorials).
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10
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14, 45, 74, 198, 5114, 10295, 65174, 1086194, 20485574, 40354813, 465779078, 12101385979, 15237604243, 18046312939, 29501083259, 52467636437, 65794608773, 86725630997, 87741700037, 131833085077, 168380217557, 176203950283, 177332276971, 226152989747, 292546582253
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OFFSET
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1,1
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COMMENTS
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Let k' be the arithmetic derivative of k. Then to find terms of the form k = p * q where p, q are prime, we could see that k' = p + q. Then as one of them needs to be two, say p, needs to be 2, we have q = A143293(m) - 2 a prime. This would give terms 2 * q.
If terms are of the form k = p * q * r where p, q, r are distinct primes then k' = p*q + p*r + q*r. For m we like, we could solve p*q + p*r + q*r = A143293(m). checking p * q below some bound, we can solve for r and get r = (A143293(m) - p*q) / (p + q). With some extra constraints and searching different prime signatures, one might confirm terms found are all below some chosen upper bound. (End)
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LINKS
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FORMULA
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PROG
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(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A143293(n) = if(n==0, 1, my(P=1, s=1); forprime(p=2, prime(n), s+=P*=p); (s)); \\ From A143293.
A276150(n) = { my(s=0, p=2, d); while(n, d = (n%p); s += d; n = (n-d)/p; p = nextprime(1+p)); (s); };
A276086(n) = { my(i=0, m=1, pr=1, nextpr); while((n>0), i=i+1; nextpr = prime(i)*pr; if((n%nextpr), m*=(prime(i)^((n%nextpr)/pr)); n-=(n%nextpr)); pr=nextpr); m; };
k=0; for(n=1, A002620(A143293(6)), if(isA328243(n), k++; print1(n, ", ")));
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CROSSREFS
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Sequence A369240 sorted into ascending order.
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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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