A064796 -id:A064796 - cp's OEIS frontend

A035106 1, together with numbers of the form k(k+1) or k(k+2), k > 0.

1, 2, 3, 6, 8, 12, 15, 20, 24, 30, 35, 42, 48, 56, 63, 72, 80, 90, 99, 110, 120, 132, 143, 156, 168, 182, 195, 210, 224, 240, 255, 272, 288, 306, 323, 342, 360, 380, 399, 420, 440, 462, 483, 506, 528, 552, 575, 600, 624, 650, 675, 702, 728, 756, 783, 812, 840

Offset: 1

N. J. A. Sloane, revised Oct 30 2001

Largest integer m such that every permutation (p_1, ..., p_n) of (1, ..., n) satisfies p_i * p_{i+1} >= m for some i, 1 <= i <= n-1. Equivalently, smallest integer m such that there exists a permutation (p_1, ..., p_n) of (1, ..., n) satisfying p_i * p_{i+1} <= m for every i, 1 <= i <= n-1.
Also, nonsquare positive integers m such that floor(sqrt(m)) divides m. - Max Alekseyev, Nov 27 2006
Also, for n>1, a(n) is the number of non-isomorphic simple connected undirected graphs having n+1 edges and a longest path of length n. - Nathaniel Gregg, Nov 02 2021

			n=5: we must arrange the numbers 1..5 so that the max of the products of pairs of adjacent terms is minimized. The answer is 51324, with max product = 8, so a(5) = 8.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (2,0,-2,1).

Cf. A006446, A052938, A064764, A064796, A064797, A064817, A004652, A035104, A035107.
First differences give (essentially) A028242.
Bisections: A002378 (pronic numbers) and A005563.

Concatenation([1], List([2..60], n-> (2*n*(n+2) +3*((-1)^n -1))/8)); # G. C. Greubel, Jun 10 2019

import Data.List.Ordered (union)
a035106 n = a035106_list !! (n-1)
a035106_list = 1 : tail (union a002378_list a005563_list)
-- Reinhard Zumkeller, Oct 05 2015

[1] cat [(2*n*(n+2) +3*((-1)^n -1))/8: n in [2..60]]; // G. C. Greubel, Jun 10 2019

Join[{1},LinearRecurrence[{2,0,-2,1},{2,3,6,8},60]] (* or *) Join[{1}, Table[ If[EvenQ[n],(n(n+2))/4,((n-1)(n+3))/4],{n,2,60}]] (* Harvey P. Dale, May 03 2012 *)

my(x='x+O('x^60)); Vec(x*(x^4-2*x^3+x^2-1)/((x-1)^3*(x+1))) \\ Altug Alkan, Oct 23 2015

A035106(n)=!(n-1)+floor((n^2)/4+n/2); \\ R. J. Cano, Jul 24 2023

[1]+[(2*n*(n+2) +3*((-1)^n -1))/8 for n in (2..60)] # G. C. Greubel, Jun 10 2019

For n > 1, a(n) = n*(n+2)/4 if n is even and (n-1)*(n+3)/4 if n is odd. - Jud McCranie, Oct 25 2001
a(n) = a(n-1) + a(n-2) - a(n-3) + 1 = A002620(n+2) + A004526(n+2). - Henry Bottomley, Mar 08 2000
a(n+2) = (2*n^2 + 12*n + 3*(-1)^n + 13)/8, with a(1)=1, i.e., a(n+2) = (n+2)*(n+4)/4 if n is even and (n+1)*(n+5)/4 if n is odd. - Vladeta Jovovic, Oct 23 2001
From Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004: (Start)
a(n) = a(n-2) + (n-1), where a(1) = 0, a(2) = 0.
a(n) = (2*(n+1)^2 + 3*(-1)^n - 5)/8, n>=2, with a(1)=1. (End)
For n > 1, a(n) = floor((n+1)^4/(4*(n+1)^2+1)). - Gary Detlefs, Feb 11 2010
For n > 1, a(n) = n + ceiling((1/4)*(n-1)^2) - 1. - Clark Kimberling, Jan 07 2011; corrected by Arkadiusz Wesolowski, Sep 25 2012
a(1)=1, a(2)=2, a(3)=3, a(4)=6, a(5)=8; for n > 5, a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4). - Harvey P. Dale, May 03 2012
G.f.: x + x^2*(2-x) / ( (1+x)*(1-x)^3 ) = x*(x^4 - 2*x^3 + x^2 - 1)/((x-1)^3*(x+1)). - Vladeta Jovovic, Oct 23 2001; Harvey P. Dale, May 03 2012
a(n) = floor(n/2)*(1 + ceiling(n/2)), a(1) = 1. - Arkadiusz Wesolowski, Sep 25 2012
a(n) = ceiling((n-1)*(n+3)/4), n > 1. - Wesley Ivan Hurt, Jun 14 2013
a(n+1) - a(n) = A052938(n-2) for n > 1. - Reinhard Zumkeller, Oct 06 2015
E.g.f.: (8*x + 3*exp(-x) - (3-6*x-2*x^2)*exp(x))/8. - G. C. Greubel, Jun 10 2019
Sum_{n>=1} 1/a(n) = 11/4. - Amiram Eldar, Sep 24 2022

Edited by Max Alekseyev, Oct 09 2015

Definition modified to allow for the initial 1. - N. J. A. Sloane, May 17 2016

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

N. J. A. Sloane, Oct 21 2001

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

			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.

P. Erdős, R. Freud, and N. Hegyvári, Arithmetical properties of permutations of integers, Acta Mathematica Hungarica 41:1-2 (1983), pp 169-176.
D. Wasserman, Proof of terms 11-70

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

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

More terms from Vladeta Jovovic, Oct 21 2001

Further terms from David Wasserman, Aug 17 2002

A332877 Arrange the first n primes in a circle in any order. a(n) is the minimum value of the largest product of two consecutive primes out of all possible orders.

Original entry on oeis.org

6, 15, 21, 35, 55, 77, 91, 143, 187, 221, 253, 323, 391, 493, 551, 667, 713, 899, 1073, 1189, 1271, 1517, 1591, 1763, 1961, 2183, 2419, 2537, 2773, 3127, 3233, 3599, 3953, 4189, 4331, 4757, 4897, 5293, 5723, 5963, 6499, 6887, 7171, 7663, 8051, 8633, 8989, 9797, 9991, 10403, 10807, 11303

Offset: 2

Bobby Jacobs, Apr 11 2020

nonn

It might appear that all terms are either the product of two consecutive primes or two primes with a prime in between (A333747). However, 253=11*23 is the first term that is not in that sequence.

The easiest optimal permutation of n primes is probably {p_1, p_n, p_2, p_n-1, …, p_ceiling(n/2)}. - Ivan N. Ianakiev, Apr 20 2020

			Here are the different ways to arrange the first 4 primes in a circle.
  2-3
  | |  Products: 6, 21, 35, 10. Largest product: 35.
  5-7
.
  2-3
  | |  Products: 6, 15, 35, 14. Largest product: 35.
  7-5
.
  2-5
  | |  Products: 10, 15, 21, 14. Largest product: 21.
  7-3
The minimum largest product is 21, so a(4)=21.

Giovanni Resta, Examples for a(2)-a(22)

Cf. A064796, A332765, A333747.

primes[n_]:=Prime/@Range[n];
partition[n_]:=Partition[primes[n],UpTo[Ceiling[n/2]]];
riffle[n_]:=Riffle[partition[n][[1]],Reverse[partition[n][[2]]]];
a[n_]:=Max[Table[riffle[n][[i]]*riffle[n][[i+1]],{i,1,n-1}]];
a/@Range[2,60] (* Ivan N. Ianakiev, Apr 20 2020 *)

a(n) = {my(x = oo); for (k=1, (n-1)!, my(vp = Vec(numtoperm(n, k-1))); vp = apply(x->prime(x), vp); x = min(x, max(vp[1]*vp[n-1], vecmax(vector(n-1, j, vp[j]*vp[j+1]))));); x;} \\ Michel Marcus, Apr 14 2020

Probably a(n) = A332765(n+1) for n > 4.

a(12)-a(13) from Michel Marcus, Apr 14 2020
a(14) from Alois P. Heinz, Apr 15 2020
a(15)-a(22) from Giovanni Resta, Apr 19 2020
More terms from Ivan N. Ianakiev, Apr 20 2020

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

N. J. A. Sloane, Oct 21 2001

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

			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.

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

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 *)

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

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

A064817 Maximal number of squares among the n-1 numbers p_i + p_{i+1}, 1 <= i <= n-1, where (p_1, ..., p_n) is any permutation of (1, ..., n).

Original entry on oeis.org

0, 0, 1, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 16, 17, 18, 19, 20, 22, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68

Offset: 1

N. J. A. Sloane, Oct 23 2001

a(n) < n by definition, but if we counted the sum p_n + p_1, we could get a(n) = n for 32 <= n <= 49 (see A071984). - David Wasserman, Aug 20 2002
Can be formulated as a traveling salesman problem on a complete graph with node set {0, 1, ..., n} and edge cost -1 if i + j is a square, 0 otherwise. - Rob Pratt, Nov 07 2012
a(n) = n - 1 for 25 <= n <= 500, computed by solving corresponding TSP. - Rob Pratt, Nov 07 2012

			n=8: take 2,7,8,1,3,6,4,5 to get 5 squares: 2+7, 8+1, 1+3, 3+6, 4+5; a(8) = 5.
(1,8,9,7,2,14,11,5,4,12,13,3,6,10) gives 12 squares and no permutation of (1..14) gives more, so a(14)=12.

Bernardo Recamán Santos, Challenging Brainteasers, Sterling, NY, 2000, page 71, shows a(15) = 14 using 9,7,2,14,11,5,4,12,13,3,6,10,15,1,8.

Cf. A035106, A064764, A064796, A064797, A071983, A071984.

a[n_] := Which[n == 1, 0, n > 30, n - 1, True, tour = FindShortestTour[Range[n], DistanceFunction -> Function[{i, j}, If[IntegerQ[Sqrt[i + j]], -1, 0]]] // Last; cnt = 0; Do[If[IntegerQ[Sqrt[tour[[i]] + tour[[i + 1]]]], cnt++], {i, 1, n}]; cnt]; Table[an = a[n]; Print["a(", n, ") = ", an]; an, {n, 1, 69}] (* Jean-François Alcover, Nov 04 2016 *)

More terms from Vladeta Jovovic, Oct 23 2001
More terms from John W. Layman and Charles K. Layman (cklayman(AT)juno.com), Nov 07 2001
More terms from David Wasserman, Aug 20 2002
More terms from Rob Pratt, Nov 07 2012

cp's OEIS Frontend

A035106 1, together with numbers of the form k*(k+1) or k*(k+2), k > 0.

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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.

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A332877 Arrange the first n primes in a circle in any order. a(n) is the minimum value of the largest product of two consecutive primes out of all possible orders.

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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.

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A064817 Maximal number of squares among the n-1 numbers p_i + p_{i+1}, 1 <= i <= n-1, where (p_1, ..., p_n) is any permutation of (1, ..., n).

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A035106 1, together with numbers of the form k(k+1) or k(k+2), k > 0.