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A376410
Number of integers whose arithmetic derivative (A003415) is equal to n!, the n-th factorial.
3
0, 1, 4, 13, 40, 186, 952, 5533, 38719, 346207, 3130816, 34444968, 382437431, 4637235152
OFFSET
2,3
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.
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).
PROG
(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)<pv[#pv]), 0, if(!(x%s) && isprime(x/s), (x/s), 1)); };
KEYWORD
nonn,hard,more
AUTHOR
Antti Karttunen, Nov 06 2024
STATUS
approved