A376399 a(0) = 1, and for n > 0, a(n) is the least k such that A276075(k) = a(n-1) + A276075(a(n-1)), where A276075 is the factorial base log-function.

1, 2, 6, 30, 1050, 519090, 1466909163669353522118

Offset: 0

Author

Antti Karttunen, Nov 02 2024

Keywords

• nonn

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.

The n-th term can be computed by applying A276076 to A376403(n), i.e., to the partial sums of the preceding terms a(0) .. a(n-1) (see the examples). This follows because all terms are in A276078 by the "least k" condition of the definition (see comment in A376417).

Examples

			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.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..7
• Index entries for sequences related to factorial base representation

Crossrefs

Cf. A276075, A276076, A376400 (variant).

Cf. A376403 (= A276075(a(n)), also gives the partial sums from its second term onward).

Subsequence of A276078.

Cf. also analogous sequences A002110 (for A276085), A093502 (for A056239), A376406 (for A048675).

Programs

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];

Formula

a(n) = A276076(A376403(n)) = A276076(Sum_{i=0..n-1} a(i)).