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.

A368900 LCM-transform of Doudna sequence.

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 3, 2, 7, 1, 1, 1, 5, 1, 3, 2, 11, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 5, 1, 3, 2, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 5, 1, 3, 2, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

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Author

Keywords

Comments

Let's define "property S" for sequences as follows: If s is any sequence of positive natural numbers, normalized to begin with offset 1, then it satisfies the S-property if LCM-transform(s) is equal to the sequence obtained by applying A014963 to sequence s, or in other words, when for all n >= 1, lcm {s(1)..s(n)} / lcm {s(1)..s(n-1)} = A014963(s(n)). This holds if and only if, for all n >= 1, when, either (case A): s(n) is of the form p^k, p prime, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to p^(k-1), or (case B): when s(n) is not a prime power, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to s(n). Together the cases (A) and (B) reduce to the condition that each prime power should appear in s before any of its multiples do.
Clearly the Doudna-sequence satisfies the property by the way of its construction, as do many of its variants like A356867 (see A369060).
Also, for any base-2 related permutation b that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., if for all n >= 1, A000523(b(n)) = A000523(n), then the above property is automatically satisfied.
Furthermore, because in Doudna-sequence no multiple of any term is located on the same row as the term itself (see the tree-illustration in A005940), it follows that any composition of A005940 with any such base-2 related permutation as mentioned above also automatically satisfies the S-property, for example, the permutations A163511, A243353, A253563, A253565, A366260, A366263 and A366275.
Note: Like A005940 itself, also this sequence might be more logical with the starting offset 0 instead of 1, to better align with the underlying mapping from the binary expansion of n to the prime factorization. - Antti Karttunen, Jan 24 2024

Crossrefs

List of LCM-transforms of permutations (permutation given in parentheses):
Cf. A265576 (A064413; note that the EKG sequence permutation does not satisfy the S-property).
In all following cases, the permutation satisfies the S-property:
Cf. A369041 (A003188), A369042 (A006068), A369043 (A193231), A369044 (A057889), A369041 (A054429). [Base-2 related permutations]
Other permutations that have the same property: A303767, (and when used as an offset=1 sequence): A052330.

Programs

  • Mathematica
    nn = 120; Array[Set[{s[#], a[#]}, {#, #}] &, 2]; j = 2;
    Do[If[EvenQ[n],
      Set[s[n], 2 s[n/2]],
      Set[s[n],
        Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &,
          FactorInteger[s[(n + 1)/2]]]]];
      k = LCM[j, s[n]]; a[n] = k/j; j = k, {n, 3, nn}];
    Array[a, nn] (* Michael De Vlieger, Mar 24 2024 *)
  • PARI
    up_to = 16384;
    LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t) };
    v368900 = LCMtransform(vector(up_to,i,A005940(i)));
    A368900(n) = v368900[n];
    
  • PARI
    A000265(n) = (n>>valuation(n,2));
    A209229(n) = (n && !bitand(n,n-1));
    A368900(n)  = if(1==n, 1, my(x=A000265(n-1)); if(A209229(1+x), prime(1+valuation(n-1,2)), 1));

Formula

a(n) = A368901(n) / A368901(n-1) = lcm {1..A005940(n)} / lcm {1..A005940(n-1)}.
a(n) = A005940(n) / gcd(A005940(n), A368901(n-1)).
a(n) = A014963(A005940(n)). [Because A005940 satisfies the property given in the comments]
For n >= 1, Product_{d|n} a(A005941(d)) = n. [Implied by above]
For n >= 1, a(n) = A369030(1+A054429(n-1)).
For n > 1, if n-1 is a number of the form 2^i - 2^j with i >= j, then a(n) = prime(1+j), otherwise a(n) = 1.

A369044 LCM-transform of bijective bit reverse (A057889).

Original entry on oeis.org

1, 2, 3, 2, 5, 1, 7, 2, 3, 1, 13, 1, 11, 1, 1, 2, 17, 1, 5, 1, 1, 1, 29, 1, 19, 1, 3, 1, 23, 1, 31, 2, 1, 1, 7, 1, 41, 1, 1, 1, 37, 1, 53, 1, 1, 1, 61, 1, 1, 1, 1, 1, 43, 1, 59, 1, 1, 1, 1, 1, 47, 1, 1, 2, 1, 1, 97, 1, 3, 1, 113, 1, 73, 1, 1, 1, 89, 1, 11, 1, 1, 1, 101, 1, 1, 1, 1, 1, 1, 1, 109, 1, 1, 1, 5, 1, 67
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2024

Keywords

Comments

Bijective bit reverse, A057889, is a permutation related to the binary expansion of n that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., for all n >= 1, A000523(A057889(n)) = A000523(n), from which it immediately follows that A057889 has the property S mentioned in the comments of A368900, and therefore this sequence is equal to A014963(A057889(n)), for n >= 1.

Crossrefs

Programs

  • PARI
    up_to = 65537; \\ Checked up to 2^17;
    LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
    A030101(n) = if(n<1,0,subst(Polrev(binary(n)),x,2));
    A057889(n) = if(!n,n,A030101(n/(2^valuation(n,2))) * (2^valuation(n, 2)));
    v369044 = LCMtransform(vector(up_to,i,A057889(i)));
    A369044(n) = v369044[n];
    A014963(n) = { ispower(n, , &n); if(isprime(n), n, 1); };

Formula

a(n) = lcm {1..A057889(n)} / lcm {1..A057889(n-1)}.
a(n) = A014963(A057889(n)). [See comments.]
For n >= 1, Product_{d|n} a(A057889(d)) = n. [Implied by above.]

A369041 LCM-transform of binary Gray code (A003188).

Original entry on oeis.org

1, 3, 2, 1, 7, 5, 2, 1, 13, 1, 1, 1, 11, 3, 2, 1, 5, 3, 1, 1, 31, 29, 1, 1, 1, 23, 1, 1, 19, 17, 2, 1, 7, 1, 1, 1, 1, 53, 1, 1, 61, 1, 1, 1, 59, 1, 1, 1, 41, 43, 1, 1, 47, 1, 1, 1, 37, 1, 1, 1, 1, 1, 2, 1, 97, 1, 1, 1, 103, 101, 1, 1, 109, 1, 1, 1, 107, 1, 1, 1, 11, 1, 1, 1, 127, 5, 1, 1, 1, 1, 1, 1, 1, 113, 1, 1, 3
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2024

Keywords

Comments

Binary Gray code, A003188, is a permutation related to the binary expansion of n that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., for all n >= 1, A000523(A003188(n)) = A000523(n), from which it immediately follows that A003188 has the property S mentioned in the comments of A368900, and therefore this sequence is equal to A014963(A003188(n)), for n >= 1.

Crossrefs

Programs

  • Mathematica
    nn = 120; a[1] = s[1] = 1; Do[s[n] = LCM[s[n - 1], BitXor[n, Floor[n/2]] ]; a[n] = s[n]/s[n - 1], {n, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Mar 24 2024 *)
  • PARI
    up_to = 65537; \\ Checked up to 2^17;
    LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
    A003188(n) = bitxor(n, n>>1);
    v369041 = LCMtransform(vector(up_to,i,A003188(i)));
    A369041(n) = v369041[n];
    A014963(n) = { ispower(n, , &n); if(isprime(n), n, 1); };

Formula

a(n) = lcm {1..A003188(n)} / lcm {1..A003188(n-1)}.
a(n) = A014963(A003188(n)). [See comments.]

A369042 LCM-transform of the inverse of binary Gray code (A006068).

Original entry on oeis.org

1, 3, 2, 7, 1, 2, 5, 1, 1, 1, 13, 2, 3, 11, 1, 31, 1, 1, 29, 1, 5, 3, 1, 2, 17, 19, 1, 23, 1, 1, 1, 1, 1, 1, 61, 1, 1, 59, 1, 1, 7, 1, 1, 1, 1, 1, 53, 2, 1, 1, 1, 1, 1, 1, 37, 47, 1, 1, 1, 1, 41, 43, 1, 127, 1, 1, 5, 1, 11, 1, 1, 1, 113, 1, 1, 1, 1, 1, 1, 1, 97, 1, 1, 103, 1, 1, 101, 1, 1, 1, 109, 1, 1, 107, 1, 2, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2024

Keywords

Comments

Inverse of Binary Gray code, A006068, is a permutation related to the binary expansion of n that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., for all n >= 1, A000523(A006068(n)) = A000523(n), from which it immediately follows that A006068 has the property S mentioned in the comments of A368900, and therefore this sequence is equal to A014963(A006068(n)), for n >= 1.

Crossrefs

Programs

  • PARI
    up_to = 65537; \\ Checked up to 2^17;
    LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
    A006068(n)= { my(s=1, ns); while(1, ns = n >> s; if(0==ns, break()); n = bitxor(n, ns); s <<= 1; ); return (n); } \\ From A006068
    v369042 = LCMtransform(vector(up_to,i,A006068(i)));
    A369042(n) = v369042[n];
    A014963(n) = { ispower(n, , &n); if(isprime(n), n, 1); };

Formula

a(n) = lcm {1..A006068(n)} / lcm {1..A006068(n-1)}.
a(n) = A014963(A006068(n)). [See comments.]

A369045 LCM-transform of binary invert permutation (A054429).

Original entry on oeis.org

1, 3, 2, 7, 1, 5, 2, 1, 1, 13, 1, 11, 1, 3, 2, 31, 1, 29, 1, 3, 1, 5, 1, 23, 1, 1, 1, 19, 1, 17, 2, 1, 1, 61, 1, 59, 1, 1, 1, 1, 1, 53, 1, 1, 1, 7, 1, 47, 1, 1, 1, 43, 1, 41, 1, 1, 1, 37, 1, 1, 1, 1, 2, 127, 1, 5, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 113, 1, 1, 1, 109, 1, 107, 1, 1, 1, 103, 1, 101, 1, 1, 1, 97, 1, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 12 2024

Keywords

Comments

Binary invert permutation, A054429, is a self-inverse permutation related to the binary expansion of n that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., for all n >= 1, A000523(A054429(n)) = A000523(n), from which it immediately follows that A054429 has the property S mentioned in the comments of A368900, and therefore this sequence is equal to A014963(A054429(n)), for n >= 1.

Crossrefs

Programs

  • PARI
    up_to = 65537;
    LCMtransform(v) = { my(len = length(v), b = vector(len), g = vector(len)); b[1] = g[1] = 1; for(n=2,len, g[n] = lcm(g[n-1],v[n]); b[n] = g[n]/g[n-1]); (b); };
    A054429(n) = ((3<<#binary(n\2))-n-1);
    v369045 = LCMtransform(vector(up_to,i,A054429(i)));
    A369045(n) = v369045[n];
    
  • PARI
    A014963(n) = { ispower(n, , &n); if(isprime(n), n, 1); };
    A054429(n) = ((3<<#binary(n\2))-n-1);
    A369045(n) = A014963(A054429(n));

Formula

a(n) = lcm {1..A054429(n)} / lcm {1..A054429(n-1)}.
a(n) = A014963(A054429(n)). [See comments.]
For n >= 1, Product_{d|n} a(A054429(d)) = n. [Implied by above.]
Showing 1-5 of 5 results.