A328243 Numbers whose arithmetic derivative (A003415) is larger than 1 and one of the terms of A143293 (partial sums of primorials).

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

Offset: 1

Author

Antti Karttunen, Oct 10 2019

Keywords

• nonn

Comments

From David A. Corneth, Oct 12 2019: (Start)

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)

See sequences A369239 and A369240 for more observations and insights about the terms of this sequence. - Antti Karttunen, Jan 22 2024

Links

• Giovanni Resta, Table of n, a(n) for n = 1..39 (terms < 10^13)

Crossrefs

Sequence A369240 sorted into ascending order.

Cf. A003415, A143293, A327969, A327978, A328313, A369239.

Programs

PARI

A002620(n) = ((n^2)>>2);
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; };
isA328243(n) = { my(u=A003415(n)); ((u>1)&&(1==A276150(A276086(u)))); }; \\ This is very slow program!
k=0; for(n=1,A002620(A143293(6)),if(isA328243(n), k++; print1(n,", ")));

Formula

A327969(a(n)) <= 5 for all n.

Extensions

a(12)-a(25) from David A. Corneth and Giovanni Resta, Oct 12 2019