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.

A064764 Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies lcm(p_i, p_{i+1}) >= m for some i, 1 <= i <= n-1.

Original entry on oeis.org

1, 2, 3, 4, 6, 6, 12, 12, 12, 12, 18, 18, 24, 24, 24, 24, 35, 35, 44, 44, 44, 44, 55, 55, 55, 55, 55, 55, 68, 68, 85, 85, 85, 85, 85, 85, 102, 102, 102, 102, 119, 119, 145, 145, 145, 145, 174, 174, 174, 174, 174, 174, 203, 203, 203, 203, 203, 203, 232, 232, 261, 261, 261
Offset: 1

Views

Author

N. J. A. Sloane, Oct 21 2001

Keywords

Comments

For n >= 4, a(n) >= A073818(pi(n)), with equality for 19 <= n <= 70. - David Wasserman, Aug 17 2002

Examples

			n=6: we must arrange the numbers 1..6 so that the max of the lcm of pairs of adjacent terms is minimized. The answer is 632415, with max lcm = 6, so a(6) = 6.

Links

Crossrefs

Cf. A035106, A064736, A064796, A064797, A000720, A073818.

Formula

a(n) = (1+o(1))n^2/(4 log n) as n -> infinity.

Extensions

More terms from Vladeta Jovovic, Oct 21 2001
Further terms from David Wasserman, Aug 17 2002

A342091 a(n) is the least number k such that k! has n distinct exponents in its prime factorization.

Original entry on oeis.org

1, 2, 4, 6, 10, 15, 22, 33, 44, 55, 68, 85, 102, 119, 145, 174, 203, 232, 261, 296, 333, 370, 410, 451, 492, 533, 590, 656, 708, 767, 826, 885, 944, 1005, 1072, 1143, 1207, 1278, 1422, 1455, 1562, 1652, 1778, 1917, 2032, 2134, 2235, 2328, 2425, 2540, 2682, 2831, 2929
Offset: 0

Views

Author

Amiram Eldar, Feb 27 2021

Keywords

Comments

After n=0, first differs from A073818 at n = 27.
a(n) is the least k such that A071625(k!) = A071626(k) = n.
Is this sequence strictly increasing?

Examples

			a(1) = 2 since 2! = 2^1 is the least factorial with a single exponent (1) in its prime factorization.
a(2) = 4 since 4! = 24 = 2^3 * 3^1 is the least factorial with 2 distinct exponents (1 and 3) in its prime factorization.
a(3) = 6 since 6! = 720 = 2^4 * 3^2 * 5^1 is the least factorial with 3 distinct exponents (1, 2 and 4) in its prime factorization.

Links

Crossrefs

Cf. A000142, A071625, A071626, A073818, A133924, A336616, A336619, A336867, A336868.

Programs

  • Mathematica
    f[1] = 0; f[n_] := Length @ Union[FactorInteger[n!][[;; , 2]]]; seq[max_] := Module[{s = Table[0, {max}], n = 1, c = 0}, While[c < max, i = f[n] + 1; If[i <= max && s[[i]] == 0, c++; s[[i]] = n]; n++]; s]; seq[50]

A064797 Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies lcm(p_i, p_{i+1}) >= m for some i, 1 <= i <= n, where p_{n+1} = p_1.

Original entry on oeis.org

1, 2, 6, 6, 12, 12, 15, 15, 18, 18, 24, 24, 35, 35, 35, 35, 44, 44, 55, 55, 55, 55, 68, 68, 68, 68, 68, 68, 85, 85, 102, 102, 102, 102, 102, 102, 119, 119, 119, 119, 145, 145, 174, 174, 174, 174, 203, 203, 203, 203, 203, 203, 232, 232, 232, 232, 232, 232, 261, 261
Offset: 1

Views

Author

N. J. A. Sloane, Oct 21 2001

Keywords

Comments

Testing a trial value of a(n) is equivalent to searching for a Hamiltonian cycle in the appropriate graph. - Martin Fuller, Jul 30 2006

Examples

			n=4: we must arrange the numbers 1..4 in a circle so that the max of the lcm of pairs of adjacent terms is minimized. The answer is 1423, with max lcm = 6, so a(4) = 6.

Crossrefs

Cf. A064764, A035106, A064796, A000720, A073818.

Programs

  • Mathematica
    Table[Min[Max[LCM@@@Partition[#,2,1,1]]&/@Permutations[Range[n]]], {n,10}] (* Harvey P. Dale, Oct 05 2011 *) (* The program takes a long time to run and uses a great deal of memory *)

Formula

For n >= 3, a(n) >= A073818(pi(n)+1), with equality for 17 <= n <= 250 - Martin Fuller, Jul 30 2006

Extensions

More terms from Vladeta Jovovic, Oct 22 2001
a(11)-a(24) from Charles R Greathouse IV, Jul 23 2006
More terms from Martin Fuller, Jul 30 2006

A073819 a(n) = prime(i) such that prime(i)*(n+1-i) is maximized (1 <= i <= n).

Original entry on oeis.org

2, 2, 2, 5, 5, 11, 11, 11, 11, 17, 17, 17, 17, 29, 29, 29, 29, 29, 37, 37, 37, 41, 41, 41, 41, 59, 59, 59, 59, 59, 59, 59, 67, 67, 67, 71, 71, 97, 97, 97, 97, 97, 97, 97, 97, 97, 97, 97, 97, 127, 127, 127, 127, 127, 127, 127, 127, 149, 149, 149, 149, 149, 149, 149, 149
Offset: 1

Views

Author

David Wasserman, Aug 13 2002

Keywords

Comments

3 is the only n for which the maximum is not unique; a(3) could also be given as 3.

Examples

			For n = 5, we take the first 5 primes in ascending order and multiply them by the numbers from 5 to 1 in descending order: 2*5 = 10 3*4 = 12 5*3 = 15 7*2 = 14 11*1 = 11 The largest product is 15, so a(5) = 5.

Crossrefs

Cf. A073818, A073820.

Formula

a(n) = A073818(n)/A073820(n).

A073820 a(n) = n+1-i such that prime(i)*(n+1-i) is maximized (1 <= i <= n).

Original entry on oeis.org

1, 2, 3, 2, 3, 2, 3, 4, 5, 4, 5, 6, 7, 5, 6, 7, 8, 9, 8, 9, 10, 10, 11, 12, 13, 10, 11, 12, 13, 14, 15, 16, 15, 16, 17, 17, 18, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 20, 21, 22, 23, 24, 25, 26, 27, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 32, 33, 31, 30, 31, 32, 33, 34
Offset: 1

Views

Author

David Wasserman, Aug 13 2002

Keywords

Comments

3 is the only n for which the maximum is not unique; a(3) could also be given as 2.

Examples

			For n = 5, we take the first 5 primes in ascending order and multiply them by the numbers from 5 to 1 in descending order: 2*5 = 10 3*4 = 12 5*3 = 15 7*2 = 14 11*1 = 11 The largest product is 15, so a(5) = 3.

Crossrefs

Cf. A073818, A073819.

Formula

a(n) = A073818(n)/A073819(n).
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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