A376410 Number of integers whose arithmetic derivative (A003415) is equal to n!, the n-th factorial.

0, 1, 4, 13, 40, 186, 952, 5533, 38719, 346207, 3130816, 34444968, 382437431, 4637235152

Offset: 2

Author

Antti Karttunen, Nov 06 2024

Keywords

• nonn
• hard
• more

Comments

For 1! = 1, there are an infinite number of integers k for which A003415(k) = 1 (namely, all the primes), therefore the starting offset is 2.

Like with A351029, also here most of the solutions seem to be squarefree semiprimes, counted by A062311.

Terms a(12)..a(15) were obtained by summing the corresponding terms of A062311 and A377986.

Links

• Index entries for sequences related to factorial numbers

Crossrefs

Cf. A000142, A002620, A003415, A062311, A099302, A377986.

Cf. A011371, A054861, A027868, A054896.

Cf. also A351029, A369239.

Programs

PARI

\\ Slow program, for computing just a few terms:
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]));
A376410(n) = { my(g=n!); sum(k=1,A002620(g),A003415(k)==g); };

PARI

A376410(n) = AntiDeriv(n!);
AntiDeriv(n,startvlen=1,solsfilename="") = { my(v = vector(startvlen,i,2), ip = #v, r, c=0); while(1, r = A003415vrl(v,n); if(0==r, ip--, if(r > 1, c++; if(solsfilename!="", write(solsfilename, r*factorback(v)))); ip = #v); if(0==ip, v = vector(1+#v,i,2); ip = #v; if(A003415vec(v) > n, return(c)), v[ip] = nextprime(1+v[ip]); for(i=1+ip, #v, v[i]=v[i-1]))); };
A003415vec(tv) = { my(n=factorback(tv), s=0, m=1, spf); for(i=1,#tv,spf = tv[i]; n /= spf; s += m*n; m *= spf); (s); }; \\ Compute Arithmetic derivative from the vector of primes.
A003415vrl(pv,lim) = { my(n=factorback(pv), x=lim-n, s=0, m=1, spf, u=n); for(i=1,#pv,spf = pv[i]; u /= spf; s += m*u; m *= spf); if(((x/s)
				

Formula

a(n) = A099302(A000142(n)).

a(n) = Sum_{k=1..A002620(n!)} [A003415(k) = n!], where [ ] is the Iverson bracket.

a(n) = A062311(n) + A377986(n).