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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.

%I A376399 #37 Nov 04 2024 12:32:54
%S A376399 1,2,6,30,1050,519090,1466909163669353522118
%N 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.
%C A376399 a(7) has 212 digits, a(8) has 10654 digits.
%C A376399 The lexicographically earliest infinite sequence x for which A276075(x(n)) gives the partial sums of x (shifted right once).
%C A376399 For any a(n), the next term a(n+1) <= a(n) * A276076(a(n)).
%C A376399 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.
%C A376399 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).
%H A376399 Antti Karttunen, <a href="/A376399/b376399.txt">Table of n, a(n) for n = 0..7</a>
%H A376399 <a href="/index/Fa#facbase">Index entries for sequences related to factorial base representation</a>
%F A376399 a(n) = A276076(A376403(n)) = A276076(Sum_{i=0..n-1} a(i)).
%e A376399 Starting with a(0) = 1, we take partial sums of previous terms, and apply A276076 to get the next term as:
%e A376399 a(1) = A276076(1) = 2,
%e A376399 a(2) = A276076(1+2) = 6,
%e A376399 a(3) = A276076(1+2+6) = 30,
%e A376399 a(4) = A276076(1+2+6+30) = 1050,
%e A376399 a(5) = A276076(1+2+6+30+1050) = 519090,
%e A376399 a(6) = A276076(1+2+6+30+1050+519090) = 1466909163669353522118,
%e A376399 etc.
%o A376399 (PARI)
%o A376399 \\ Do it hard way, by searching:
%o A376399 up_to = 12;
%o A376399 A276075(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*(primepi(f[k, 1])!)); };
%o A376399 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); };
%o A376399 v376399 = A376399list(1+up_to);
%o A376399 A376399(n) = v376399[1+n];
%o A376399 (PARI)
%o A376399 \\ Compute, do not search, much faster:
%o A376399 up_to = 8;
%o A376399 A276076(n) = { my(m=1, p=2, i=2); while(n, m *= (p^(n%i)); n = n\i; p = nextprime(1+p); i++); (m); };
%o A376399 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); };
%o A376399 v376399 = A376399list(1+up_to);
%o A376399 A376399(n) = v376399[1+n];
%Y A376399 Cf. A276075, A276076, A376400 (variant).
%Y A376399 Cf. A376403 (= A276075(a(n)), also gives the partial sums from its second term onward).
%Y A376399 Subsequence of A276078.
%Y A376399 Cf. also analogous sequences A002110 (for A276085), A093502 (for A056239), A376406 (for A048675).
%K A376399 nonn
%O A376399 0,2
%A A376399 _Antti Karttunen_, Nov 02 2024