OFFSET
0,2
COMMENTS
a(7) has 212 digits, a(8) has 10654 digits.
The lexicographically earliest infinite sequence x for which A276075(x(n)) gives the partial sums of x (shifted right once).
For any a(n), the next term a(n+1) <= a(n) * A276076(a(n)).
Conjecture: there are infinitely many variants b of this sequence, such that A276075(b(n)) = partial sums of b (shifted once right). One way to construct them: set i for some value >= 4, construct b first as here, but at point i, set b(i+1) = b(i) * A276076(b(i)), and after that, proceed as before, always finding a minimal k satisfying the condition. Unless b(i+1) = a(i+1), then b differs from this sequence but satisfies the same general condition, except that it is not the lexicographically earliest one. See also A376400.
LINKS
EXAMPLE
Starting with a(0) = 1, we take partial sums of previous terms, and apply A276076 to get the next term as:
a(1) = A276076(1) = 2,
a(2) = A276076(1+2) = 6,
a(3) = A276076(1+2+6) = 30,
a(4) = A276076(1+2+6+30) = 1050,
a(5) = A276076(1+2+6+30+1050) = 519090,
a(6) = A276076(1+2+6+30+1050+519090) = 1466909163669353522118,
etc.
PROG
(PARI)
\\ Do it hard way, by searching:
up_to = 12;
A276075(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*(primepi(f[k, 1])!)); };
A376399list(up_to) = { my(v=vector(up_to), x); v[1]=1; for(n=2, up_to, x=v[n-1]+A276075(v[n-1]); for(k=1, oo, if(A276075(k)==x, v[n]=k; break)); print1(v[n], ", ")); (v); };
v376399 = A376399list(1+up_to);
A376399(n) = v376399[1+n];
(PARI)
\\ Compute, do not search, much faster:
up_to = 8;
A276076(n) = { my(m=1, p=2, i=2); while(n, m *= (p^(n%i)); n = n\i; p = nextprime(1+p); i++); (m); };
A376399list(up_to) = { my(v=vector(up_to), s=1); v[1]=1; for(n=2, up_to, v[n] = A276076(s); s += v[n]); (v); };
v376399 = A376399list(1+up_to);
A376399(n) = v376399[1+n];
CROSSREFS
Subsequence of A276078.
KEYWORD
nonn
AUTHOR
Antti Karttunen, Nov 02 2024
STATUS
approved