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.

Showing 1-5 of 5 results.

A292587 Compound filter: a(n) = P(A001221(n), A292582(n)), where P(n,k) is sequence A000027 used as a pairing function.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 4, 2, 3, 1, 5, 1, 3, 3, 7, 1, 5, 1, 5, 3, 3, 1, 8, 2, 3, 4, 5, 1, 6, 1, 11, 3, 3, 3, 23, 1, 3, 3, 8, 1, 6, 1, 5, 5, 3, 1, 12, 2, 5, 3, 5, 1, 8, 3, 8, 3, 3, 1, 9, 1, 3, 5, 22, 3, 6, 1, 5, 3, 6, 1, 38, 1, 3, 5, 5, 3, 6, 1, 12, 7, 3, 1, 9, 3, 3, 3, 8, 1, 9, 3, 5, 3, 3, 3, 17, 1, 5, 5, 23, 1, 6, 1, 8, 6
Offset: 1

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Author

Antti Karttunen, Sep 26 2017

Keywords

Comments

This is essentially also a filter constructed from the runlengths of numbers of the form 4k+0 and the runlengths of numbers of the form 4k+2 encountered in trajectories of A005940-tree. See comments in A083399 and A292586.
For all i, j: A291757(i) = A291757(j) => a(i) = a(j), that is, this filter matches to a subset of the sequences matched by filter A291757.
Moreover, for all i, j: a(i) = a(j) <=> A101296(i) = A101296(j), thus the subset is exactly the sequences matched by A101296 (A046523). This follows because the prime signature of n can be recovered from the two components as A046523(n) = A046523(A003557(n)) * A292586(n) and also vice versa as A046523(A003557(n)) = A003557(A046523(n)).

Crossrefs

Formula

a(n) = (1/2)*(2 + ((A001221(n) + A292582(n))^2) - A001221(n) - 3*A292582(n)).

A292588 Compound filter: a(n) = P(A292582(n), A292583(n)), where P(n,k) is sequence A000027 used as a pairing function.

Original entry on oeis.org

1, 1, 2, 3, 2, 2, 4, 6, 3, 2, 7, 5, 7, 4, 4, 10, 7, 3, 11, 5, 2, 7, 16, 9, 3, 7, 9, 8, 16, 4, 22, 15, 4, 7, 7, 21, 22, 11, 11, 9, 22, 2, 29, 12, 5, 16, 37, 14, 3, 3, 11, 12, 37, 9, 4, 13, 7, 16, 46, 8, 46, 22, 8, 28, 4, 4, 56, 12, 11, 7, 67, 36, 67, 22, 5, 17, 7, 11, 79, 14, 10, 22, 92, 5, 11, 29, 22, 18, 92, 5, 7, 23, 16, 37, 16, 20, 92, 3, 122, 21, 92, 11
Offset: 1

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Author

Antti Karttunen, Sep 26 2017

Keywords

Comments

A filter constructed from the runlengths of numbers of the form 4k+0 and the runlengths of numbers of the form 4k+3 encountered in trajectories of A005940-tree.

Crossrefs

Formula

a(n) = (1/2)*(2 + ((A292582(n) + A292583(n))^2) - A292582(n) - 3*A292583(n)).

A083399 Number of divisors of n that are not divisors of other divisors of n.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 2, 2, 3, 2, 3, 2, 3, 3, 2, 2, 3, 2, 3, 3, 3, 2, 3, 2, 3, 2, 3, 2, 4, 2, 2, 3, 3, 3, 3, 2, 3, 3, 3, 2, 4, 2, 3, 3, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 3, 3, 3, 2, 4, 2, 3, 3, 2, 3, 4, 2, 3, 3, 4, 2, 3, 2, 3, 3, 3, 3, 4, 2, 3, 2, 3, 2, 4, 3, 3, 3, 3, 2, 4, 3, 3, 3, 3, 3, 3, 2, 3, 3, 3, 2, 4
Offset: 1

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Author

Reinhard Zumkeller, Jun 12 2003

Keywords

Comments

a(n) <= tau(n); a(n) = tau(n) iff n is prime or n=1 (A008578, A000040); a(n)=tau(n)-1 iff n is semiprime (A001358).
Number of noncomposite divisors of n, (cf. A008578). - Jaroslav Krizek, Nov 25 2009
From Wilf A. Wilson, Jul 21 2017: (Start)
a(n) is the number of maximal subsemigroups of the annular Jones monoid of degree n.
a(n) is the number of maximal subsemigroups of the monoid of orientation-preserving mappings on a set with n elements.
a(n) + 1 is the number of maximal subsemigroups of the monoid of orientation-preserving partial mappings on a set with n elements.
(End)
This is the restricted growth sequence transform of A001221 (and thus also of A007875, A034444, A082476, A292586 and many other sequences). This follows from the formula a(n) = 1+A001221(n), and from the fact that for any n, A001221(n) <= 1+A001221(k) for all k = 1..(n-1). A067003 gives the ordinal transform of A001221. See also A292582, A292583, A292585. - Antti Karttunen, Sep 25 2017

Examples

			{1,2,3,4,6,8,12,24} are the divisors of n=24: 1, 2, 3, 4 and 6 divide not only 24, but also 8 or 12, therefore a(24) = 3.
{1,2,3,4,6,8,12,24} are the divisors of n=24: 1, 2 and 3 are noncomposites, therefore a(24) = 3. - _Jaroslav Krizek_, Nov 25 2009
		

Crossrefs

Programs

Formula

a(n) = omega(n) + 1, where omega = A001221.
a(n) = tau(n) - A055212(n) = A000005(n)-A055212(n).
a(n) = A000005(n) - A033273(n) + 1. - Jaroslav Krizek, Nov 25 2009
a(n) = A010553(A007947(n)) = A000005(A000005(A007947(n))) = tau_2(tau_2(rad(n))). - Enrique Pérez Herrero, Jun 25 2010
G.f.: x/(1 - x) + Sum_{k>=1} x^prime(k)/(1 - x^prime(k)). - Ilya Gutkovskiy, Mar 21 2017
Sum_{k=1..n} a(k) = n * (log(log(n)) + B + 1) + O(n/log(n)), where B is Mertens's constant (A077761). - Amiram Eldar, Sep 29 2024

A292583 Restricted growth sequence transform of A278222(A292383(n)); a filter related to runs of numbers of the form 4k+3 encountered on trajectories of A005940-tree.

Original entry on oeis.org

1, 1, 2, 1, 2, 2, 3, 1, 1, 2, 4, 2, 4, 3, 3, 1, 4, 1, 5, 2, 2, 4, 6, 2, 1, 4, 2, 3, 6, 3, 7, 1, 3, 4, 4, 1, 7, 5, 5, 2, 7, 2, 8, 4, 2, 6, 9, 2, 1, 1, 5, 4, 9, 2, 3, 3, 4, 6, 10, 3, 10, 7, 3, 1, 3, 3, 11, 4, 5, 4, 12, 1, 12, 7, 2, 5, 4, 5, 13, 2, 1, 7, 14, 2, 5, 8, 7, 4, 14, 2, 4, 6, 6, 9, 6, 2, 14, 1, 15, 1, 14, 5, 16, 4, 3
Offset: 1

Views

Author

Antti Karttunen, Sep 20 2017

Keywords

Comments

Term a(n) essentially records the run lengths of numbers of form 4k+3 encountered when starting from that node in binary tree A005940 which contains n, and by then traversing towards the root by iterating the map n -> A252463(n). The actual run lengths can be read from the exponents of primes in the prime factorization of A278222(A292383(m)), where m = min_{k=1..n} for which a(k) = a(n). In compound filter A292584 this is combined with similar information about the run lengths of the numbers of the form 4k+1 (A292585).
From Antti Karttunen, Sep 25 2017: (Start)
For all i, j: a(i) = a(j) => A053866(i) = A053866(j).
This follows from the interpretation of A053866 (A093709) as the characteristic function of squares and twice-squares. In binary tree A005940 each number is "born" by repeated applications of two functions: when we descend leftward we apply A003961, which shifts all prime factors of n one step towards larger primes. On the other hand, when we descend rightward the terms grow by doubling: n -> 2n (A005843). No square is ever of the form 4k+3, and for any square x, A003961(x) is also a square. Multiplying a square by 2 gives twice a square, and then multiplying by 2 again gives 4*square, which is also a square. In general, applying an even number of doubling steps in succession keeps a square as a square, while an odd number of doubling steps gives twice a square. Applying A003961 to any 2*square gives 3*(some square) which is always of the form 4k+3. Moreover, after any such "wrong turn" in A005940-tree no square nor twice a square can ever be encountered under any of the further descendants, because with this process it is impossible to find a pair for the lone prime factor now present. On the other hand, when turning left from any (2^2k)*s (where s is a square), one again gets a square of the form (3^2k)*A003961(s). All this implies that there are no numbers of the form 4k+3 in any trajectory leading to a square or twice a square in A005940-tree, while all trajectories to any other kind of number contain at least one number of the form 4k+3. Because each a(n) in this sequence contains enough information to count the 4k+3 numbers encountered on a A005940-trajectory to n (being 1 iff there are none), this filter matches A053866.
(End)

Examples

			When traversing from the root of binary tree A005940 from the node containing 7, one obtains path 7 -> 5 -> 3 -> 2 -> 1. Of these numbers, 7 and 3 are of the form 4k+3, while others are not, thus there are two separate runs of length 1: [1, 1]. On the other hand, when traversing from 15 as 15 -> 6 -> 3 -> 2 -> 1, again only two terms are of the form 4k+3: 15 and 3 and they are not next to each other, so we have the same two runs of one each: [1, 1], thus a(7) and a(15) are allotted the same value by the restricted growth sequence transform, which in this case is 3. Note that 3 occurs in this sequence for the first time at n=7, with A292383(7) = 5 and A278222(5) = 6 = 2^1 * 3^1, where those run lengths 1 and 1 are the prime exponents of 6.
		

Crossrefs

Programs

  • PARI
    allocatemem(2^30);
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ Modified from code of M. F. Hasler
    A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ This function from Charles R Greathouse IV, Aug 17 2011
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A278222(n) = A046523(A005940(1+n));
    A252463(n) = if(!(n%2),n/2,A064989(n));
    A292383(n) = if(1==n,0,(if(3==(n%4),1,0)+(2*A292383(A252463(n)))));
    write_to_bfile(1,rgs_transform(vector(16384,n,A278222(A292383(n)))),"b292583_upto16384.txt");

A292589 a(n) = A046523(A003557(n)) = A003557(A046523(n)); the least representative of the prime signature of {n divided by largest squarefree divisor of n}.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 4, 2, 1, 1, 2, 1, 1, 1, 8, 1, 2, 1, 2, 1, 1, 1, 4, 2, 1, 4, 2, 1, 1, 1, 16, 1, 1, 1, 6, 1, 1, 1, 4, 1, 1, 1, 2, 2, 1, 1, 8, 2, 2, 1, 2, 1, 4, 1, 4, 1, 1, 1, 2, 1, 1, 2, 32, 1, 1, 1, 2, 1, 1, 1, 12, 1, 1, 2, 2, 1, 1, 1, 8, 8, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 16, 1, 2, 2, 6, 1, 1, 1, 4, 1
Offset: 1

Views

Author

Antti Karttunen, Sep 27 2017

Keywords

Examples

			For n = 18 = 2 * 3^2, A003557(18) = 3^1. The least representative of the same prime signature is 2^1, thus a(18) = 2.
For n = 27 = 3^3, A003557(27) = 9 = 3^2. The least representative of the same prime signature is 2^2, thus a(27) = 4.
		

Crossrefs

Differs from A162510 for the first time at n=36 where a(36) = 6, while A162510(36) = 4.

Formula

a(n) = A046523(A003557(n)) = A003557(A046523(n)).
Showing 1-5 of 5 results.