cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A376418 a(n) = n - A276086(A276085(n)), where A276085 and A276086 are primorial base log and exp-functions.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0, 7, 0, 0, 0, 5, 0, 0, 0, 6, 0, 0, 22, 7, 0, 0, 0, 14, 0, 0, 0, 31, 0, 0, 0, 10, 0, 0, 0, 11, 0, 0, 0, 43, 0, 0, 0, 13, 0, 44, 0, 14, 0, 0, 0, 15, 0, 0, 0, 59, 0, 0, 0, 17, 0, 0, 0, 62, 0, 0, 0, 19, 0, 0, 0, 35, 66, 0, 0, 21, 0, 0, 0, 22, 0, 0, 0, 23, 0, 0, 0, 86, 0, 0, 0, 25
Offset: 1

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Author

Antti Karttunen, Nov 03 2024

Keywords

Comments

All terms are nonnegative because for all n, x = A276086(A276085(n)) <= n, as any factor prime(i)^k || n (with k >= prime(i)) will propagate carries (in the image of fully additive A276085) towards more significant digit positions, which A276086 will convert back to the exponents of larger primes, but for each new instance of such larger prime present in x, enough instances of smaller primes in n have been eliminated (by the carry process) so that the net change of magnitude is negative, unless there are no such factors present at all in n (i.e., when n is a term of A048103), then A276086(A276085(n)) = n, and a(n) = 0.
This implies also that the least k for which A276085(k) = n is k = A276086(n).
There are several conspicuous patterns among the terms. For example, for n = 392, 3992, 39992, 399992, 3999992, ..., a(n) = 98, 998, 9998, 99998, 999998, ..., = n/4, but this holds only if n/8 is not in A100716, as generally, for all terms x that are in the intersection of A051062 and A168183 and x/8 is in A048103, it follows that a(x) = x/4. There are many other similar identities.
Differs from similar A376417 for the first time at n=625, 1250, 1875, 2500, 3125, 3375, 3750, 4375, 4500, 5000, 5625, ...

Examples

			a(4) = 1, as 4 = prime(1)^2, thus A276085(4) = 2 * A002110(1-1) = 2, and A276086(2) = prime(2) = 3, and 4-3 = 1.
a(625) = 0, as 625 = prime(3)^4, thus A276085(625) = 4 * A002110(3-1) = 4*6 = 24, and A276086(24) = prime(3)^4 [because A049345(24) = 400] = 625, and 625-625 = 0.
a(2500) = 625, as 2500 = 2^2 * 5^4 = prime(1)^2 * prime(3)^4, thus A276085(2500) = 2 * A002110(1-1) + 4 * A002110(3-1) = 2*1 + 4*6 = 26, but on the other hand, A276086(26) = prime(2) * prime(3)^4 [because A049345(26) = 410] = 3 * 5^4 = 1875, and 2500 - 1875 = 625.
a(3999999992) = 999999998, as 3999999992 = 2^3 * 691 * 723589 = prime(1)^3 * prime(125) * prime(58312), thus x = A276085(3999999992) = A002110(1-1) + A002110(2-1) + A002110(125-1) + A002110(58312-1), so A276086(x) = prime(1) * prime(2) * prime(125) * prime(58312), therefore a(3999999992) = (8-6)*prime(125)*prime(58312) = 3999999992/4 = 999999998. Note that A049345(8) = "110", as 8 = 6+2.
		

Crossrefs

Cf. A049345, A048103 (indices of 0's), A100716 (of terms > 0), A276085, A276086, A376417.

Programs

  • PARI
    A276085(n) = { my(f=factor(n)); sum(k=1, #f~, f[k, 2]*prod(i=1,primepi(f[k, 1]-1),prime(i))); };
    A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
    A376418(n) = (n - A276086(A276085(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.

Original entry on oeis.org

1, 2, 6, 30, 1050, 519090, 1466909163669353522118
Offset: 0

Views

Author

Antti Karttunen, Nov 02 2024

Keywords

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.
		

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)).

A378518 a(n) = n - A019565(A048675(n)), where A019565 and A048675 are base-2 exp and log-functions.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 2, 4, 0, 0, 7, 0, 0, 0, 11, 0, 8, 0, 5, 0, 0, 0, 14, 18, 0, 12, 7, 0, 0, 0, 22, 0, 0, 0, 21, 0, 0, 0, 10, 0, 0, 0, 11, 38, 0, 0, 33, 38, 36, 0, 13, 0, 24, 0, 14, 0, 0, 0, 53, 0, 0, 28, 49, 0, 0, 0, 17, 0, 0, 0, 42, 0, 0, 54, 19, 0, 0, 0, 73, 74, 0, 0, 49, 0, 0, 0, 22, 0, 76, 0, 23, 0, 0, 0, 66, 0, 76
Offset: 1

Views

Author

Antti Karttunen, Nov 29 2024

Keywords

Comments

No negative terms because A097246(n) <= n for all n.

Crossrefs

Cf. A005117 (positions of 0's), A019565, A048675, A097246, A097248, A376406.
Cf. also A376417, A376418 (analogous sequences for other bases than base-2).

Programs

  • PARI
    A097246(n) = { my(f=factor(n)); prod(i=1, #f~, (nextprime(f[i, 1]+1)^(f[i, 2]\2))*((f[i, 1])^(f[i, 2]%2))); };
    A097248(n) = { my(k=A097246(n)); while(k<>n, n = k; k = A097246(k)); k; };
    A378518(n) = (n - A097248(n));

Formula

a(n) = n - A097248(n).
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