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.

A334870 If n is a square, a(n) = A000196(n), and for nonsquare n, let p be the smallest prime dividing the squarefree part of n. Divide n by p and multiply by the product of all smaller primes.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 30, 4, 3, 5, 210, 8, 2310, 7, 10, 4, 30030, 9, 510510, 24, 14, 11, 9699690, 12, 5, 13, 18, 120, 223092870, 15, 6469693230, 16, 22, 17, 42, 6, 200560490130, 19, 26, 20, 7420738134810, 21, 304250263527210, 840, 54, 23, 13082761331670030, 32, 7, 25, 34, 9240, 614889782588491410, 27, 66, 28, 38, 29
Offset: 1

Views

Author

Antti Karttunen, Jun 08 2020

Keywords

Comments

Each natural numbers occurs exactly twice in this sequence.
In binary trees like A334860 and A334866, for n > 2, a(n) gives the parent node of node n.
For nonsquare numbers, n, with squarefree part A019565(k) and square part m, a(n) is the number with squarefree part A019565(k-1) and square part m. - Peter Munn, Jul 14 2020

Crossrefs

Programs

  • Mathematica
    Array[If[IntegerQ[#2], #2, #1/#2*Product[Prime@i, {i, PrimePi@#2 - 1}] & @@ {#1, FactorInteger[#2 /. (c_ : 1)*a_^(b_ : 0) :> (c*a^b)^2][[1, 1]]}] & @@ {#, Sqrt[#]} &, 58] (* Michael De Vlieger, Jun 26 2020 *)
  • PARI
    A334870(n) = if(issquare(n),sqrtint(n),my(c=core(n), m=n); forprime(p=2, , if(!(c % p), m/=p; break, m*=p)); (m));

Formula

a(A334747(n)) = n.
a(A000040(n)) = A002110(n-1).
a(n^2) = n.
a(n) = A225546(A252463(A225546(n))). - Peter Munn, Jun 08 2020

A334871 Number of steps needed to reach 1 when starting from n and iterating with A334870.

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 8, 3, 3, 5, 16, 4, 32, 9, 6, 3, 64, 4, 128, 6, 10, 17, 256, 5, 5, 33, 5, 10, 512, 7, 1024, 4, 18, 65, 12, 4, 2048, 129, 34, 7, 4096, 11, 8192, 18, 7, 257, 16384, 5, 9, 6, 66, 34, 32768, 6, 20, 11, 130, 513, 65536, 8, 131072, 1025, 11, 4, 36, 19, 262144, 66, 258, 13, 524288, 5, 1048576, 2049, 7, 130
Offset: 1

Views

Author

Antti Karttunen, Jun 08 2020

Keywords

Comments

Distance of n from the root (1) in binary trees like A334860 and A334866.
Each n > 0 occurs 2^(n-1) times.
a(n) is the size of the inner lining of the integer partition with Heinz number A225546(n), which is also the size of the largest hook of the same partition. (After Gus Wiseman's Apr 02 2019 comment in A252464).

Crossrefs

Programs

  • PARI
    A334870(n) = if(issquare(n),sqrtint(n),my(c=core(n), m=n); forprime(p=2, , if(!(c % p), m/=p; break, m*=p)); (m));
    A334871(n) = { my(s=0); while(n>1,s++; n = A334870(n)); (s); };
    
  • PARI
    \\ Much faster:
    A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
    A334871(n) = { my(s=0); while(n>1, if(issquare(n), s++; n = sqrtint(n), s += A048675(core(n)); n /= core(n))); (s); };

Formula

a(1) = 0; for n > 1, a(n) = 1 + a(A334870(n)).
a(n) = A252464(A225546(n)).
a(n) = A048675(A007913(n)) + a(A008833(n)).
For n > 1, a(n) = 1 + A048675(A007913(n)) + a(A000188(n)).
For n > 1, a(n) = A070939(A334859(n)) = A070939(A334865(n)).
For all n >= 1, a(n) >= A299090(n).
For all n >= 1, a(n) >= A334872(n).

A334868 Lexicographically earliest infinite sequence such that a(i) = a(j) => f(i) = f(j) for all i, j >= 1, where f(1) = 0 and for n > 1, f(n) = -1 if n is in A050376, and f(n) = A334870(n) otherwise.

Original entry on oeis.org

1, 2, 2, 2, 2, 3, 2, 4, 2, 5, 2, 6, 2, 7, 8, 2, 2, 9, 2, 10, 11, 12, 2, 13, 2, 14, 15, 16, 2, 17, 2, 18, 19, 20, 21, 22, 2, 23, 24, 25, 2, 26, 2, 27, 28, 29, 2, 30, 2, 31, 32, 33, 2, 34, 35, 36, 37, 38, 2, 39, 2, 40, 41, 6, 42, 43, 2, 44, 45, 46, 2, 47, 2, 48, 49, 50, 51, 52, 2, 53, 2, 54, 2, 55, 56, 57, 58, 59, 2, 60, 61, 62, 63, 64, 65, 66, 2, 67, 68, 8, 2, 69, 2, 70, 71
Offset: 1

Views

Author

Antti Karttunen, Jun 08 2020

Keywords

Comments

For all i, j: A305979(i) = A305979(j) => a(i) = a(j) => A334872(i) = A334872(j).

Crossrefs

Cf. A050376 (positions of 2's), A305979, A334869, A334870, A334872.

Programs

  • PARI
    up_to = 65537;
    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; };
    ispow2(n) = (n && !bitand(n,n-1));
    A302777(n) = ispow2(isprimepower(n));
    A334870(n) = if(issquare(n),sqrtint(n),my(c=core(n), m=n); forprime(p=2, , if(!(c % p), m/=p; break, m*=p)); (m));
    A334868aux(n) = if(1==n,0,if(A302777(n),-1,A334870(n)));
    v334868 = rgs_transform(vector(up_to,n,A334868aux(n)));
    A334868(n) = v334868[n];

A335428 Prime exponent of the first Fermi-Dirac factor (number of form p^(2^k), A050376) reached, when starting from n and iterating with A334870, with a(1) = 0.

Original entry on oeis.org

0, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 1, 1, 4, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 4, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 4, 4, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 4, 1, 2, 2, 1, 1, 1, 1, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Jun 26 2020

Keywords

Examples

			For n=27, when iterating with A334870, we obtain the path 27 -> 18 -> 9, with that 9 being the first prime power encountered that has an exponent of the form 2^k, as 9 = 3^2, thus a(27) = 2. See the binary tree A334860 or A334866 for how such paths go.
For n=900, when iterating with A334870 we obtain the path 900 -> 30 -> 15 -> 10 -> 5, and 5^1 is finally a prime power with an exponent that is two's power, thus a(900) = 1. Note that 900 is the first such position of 1 in this sequence that is not listed in A333634.
		

Crossrefs

Programs

  • PARI
    A209229(n) = (n && !bitand(n,n-1));
    A302777(n) = A209229(isprimepower(n));
    A334870(n) = if(issquare(n),sqrtint(n),my(c=core(n), m=n); forprime(p=2, , if(!(c % p), m/=p; break, m*=p)); (m));
    A335428(n) = { while(n>1 && !A302777(n), n = A334870(n)); isprimepower(n); };
    
  • PARI
    \\ Faster, A209229 and A302777 like in above:
    A335428(n) = if(1==n,0, while(!A302777(n), if(issquarefree(n), return(1)); if(issquare(n), n = sqrtint(n), n /= core(n))); isprimepower(n));

A335427 a(1) = 0; for k >= 2, a(prime(k)) = 0, a(k^2) = 2 * a(k); otherwise a(n) = a(A334870(n)) + 1.

Original entry on oeis.org

0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 4, 0, 1, 2, 4, 0, 1, 0, 6, 2, 1, 0, 5, 0, 1, 2, 10, 0, 3, 0, 5, 2, 1, 4, 2, 0, 1, 2, 7, 0, 3, 0, 18, 4, 1, 0, 6, 0, 1, 2, 34, 0, 3, 4, 11, 2, 1, 0, 8, 0, 1, 8, 6, 4, 3, 0, 66, 2, 5, 0, 3, 0, 1, 2, 130, 8, 3, 0, 8, 0, 1, 0, 12, 4, 1, 2, 19, 0, 5, 8, 258, 2, 1, 4, 7, 0, 1, 16, 2, 0, 3, 0, 35, 6
Offset: 1

Views

Author

Antti Karttunen and Peter Munn, Jun 15 2020

Keywords

Crossrefs

A052126, A225546, A334870, A335426 are used in formulas defining this sequence.
Related fully additive sequence: A048675.
Cf. A062090 (indices of zeros), A003159 (indices of even values), A036554 (indices of odd values).
A003961, A019565 are used to express relationship between terms of this sequence.

Programs

  • PARI
    A334870(n) = if(issquare(n),sqrtint(n),my(c=core(n), m=n); forprime(p=2, , if(!(c % p), m/=p; break, m*=p)); (m));
    A335427(n) = if(n<=2,n-1, if(isprime(n), 0, if(issquare(n), 2*A335427(sqrtint(n)), 1+A335427(A334870(n)))));
    
  • PARI
    A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; };
    A052126(n) = if(1==n,n,(n/vecmax(factor(n)[, 1])));
    A335427(n) = if(n<=2,n-1, if(issquarefree(n), A048675(A052126(n)), my(k=core(n)); A048675(k) + 2*A335427(sqrtint(n/k))));

Formula

Alternative definition: (Start)
a(1) = 0, a(2) = 1; otherwise for n = k * m^2, k squarefree:
if m = 1, a(n) = A048675(A052126(k));
if m > 1, a(n) = A048675(k) + 2 * a(m).
(End)
For n = 4 * A122132(k), a(n) = A048675(n).
More generally, a(n) = A048675(n) if and only if n is in A335738.
a(n) = A335426(A225546(n)).
a(A003961(2k+1)) = 2 * a(2k+1).
If n is in A036554, a(n) = a(n/2) + 1; otherwise for n <> 3, a(n) = 2 * a(A019565(k/2) * m^2) - a(m^2), where n = A019565(k) * m^2.
Showing 1-5 of 5 results.