Gary Yane - cp's OEIS frontend

A358787 a(1)=1; let x=gcd(a(n-1),n); for n > 1, a(n) = a(n-1) + n if x=1 or a(n-1)/x=1, otherwise a(n) = a(n-1)/x.

1, 3, 6, 3, 8, 4, 11, 19, 28, 14, 25, 37, 50, 25, 5, 21, 38, 19, 38, 19, 40, 20, 43, 67, 92, 46, 73, 101, 130, 13, 44, 11, 44, 22, 57, 19, 56, 28, 67, 107, 148, 74, 117, 161, 206, 103, 150, 25, 74, 37, 88, 22, 75, 25, 5, 61, 118, 59, 118, 59, 120, 60, 20, 5, 70, 35, 102, 3, 72, 36

Offset: 1

Gary Yane, Nov 30 2022

nonn

			For n=2, gcd(2,1)=1 so a(2) = 2+1 = 3.
For n=3, gcd(3,3)=3=a(2) so a(3) = 3+3 = 6.
For n=4, gcd(4,6)=2 so a(4) = 6/2 = 3.

Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^16, with x = gcd(a(n-1), n), showing x=1 or a(n-1)/x = 1 in red, else dark blue.

Cf. A264767, A133058, A133579, A133580, A255051, A255140, A262922.

nn = 2^16; a[1] = 1; Do[g = GCD[a[n - 1], n]; If[Or[g == 1, Set[k, a[n - 1]/g] == 1], a[n] = a[n - 1] + n, a[n] = k], {n, 2, nn}]; Array[a, nn] (* Michael De Vlieger, Dec 13 2022 *)

from math import gcd
from itertools import count, islice
def agen(): # generator of terms
    an = 1
    for n in count(2):
        yield an
        x = gcd(an, n)
        an = an + n if x == 1 or x == an else an//x
print(list(islice(agen(), 70))) # Michael S. Branicky, Dec 15 2022

A342831 a(n) is the smallest positive integer k such that the n-dimensional cube [0,k]^n contains at least as many internal lattice points as external lattice points.

Original entry on oeis.org

3, 6, 9, 12, 15, 18, 21, 24, 26, 29, 32, 35, 38, 41, 44, 47, 50, 52, 55, 58, 61, 64, 67, 70, 73, 76, 78, 81, 84, 87, 90, 93, 96, 99, 101, 104, 107, 110, 113, 116, 119, 122, 125, 127, 130, 133, 136, 139, 142, 145, 148, 151, 153, 156, 159, 162, 165, 168, 171, 174, 177, 179, 182

Offset: 1

Gary Yane, Mar 23 2021

nonn

			a(2) > 5 because the number of internal lattice points = 4^2 = 16 < 20 = 6^2 - 16 = the number of external lattice points, therefore a(2)=6 because the number of internal lattice points = 5^2 = 25 > 24 = 7^2 - 25 = number of external lattice points.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A078608.

a:= n-> ceil(1+2/(2^(1/n)-1)):
seq(a(n), n=1..65);  # Alois P. Heinz, Apr 20 2021

a[1] = 3; a[n_] := Floor[2^(1/n + 1)/(2^(1/n) - 1)]; Array[a, 100] (* Amiram Eldar, Mar 31 2021 *)

a(1) = 3 and a(n) = floor(2^(1/n+1)/(2^(1/n)-1)) for n > 1.

a(n) = A078608(n) + 1.

A343576 Number of permutations of [n] without fixed points and all cycles equal length.

Original entry on oeis.org

1, 0, 1, 2, 9, 24, 175, 720, 6405, 42560, 436401, 3628800, 48073795, 479001600, 7116730335, 88966701824, 1474541093025, 20922789888000, 400160588853025, 6402373705728000, 133991603578884051, 2457732174030848000, 55735573291977790575, 1124000727777607680000

Offset: 0

Gary Yane, Apr 20 2021

			a(4) = 9: (1,2)(3,4), (1,3)(2,4), (1,4)(2,3), (2,3,4,1), (2,4,1,3), (3,1,4,2), (3,4,2,1), (4,1,2,3), (4,3,1,2).

Alois P. Heinz, Table of n, a(n) for n = 0..450

Cf. A000040, A001358, A005225, A261431.

a:= n-> `if`(n=0, 1, add(n!/d!*(d/n)^d, d=numtheory[divisors](n) minus {n})):
seq(a(n), n=0..23);  # Alois P. Heinz, Apr 20 2021

a(n) = if (n, sumdiv(n, d, if (dMichel Marcus, Apr 21 2021

a(n) = Sum_{d|n, d0, a(0) = 1.

a(n) = A261431(n) for n in { A000040, A001358 }.

A341216 Triangle read by columns T(n,k) k > n >= 1: Last survivor positions in a modified Josephus problem for n numbers, where after each deletion the counting starts over at the lowest existing number n, rather than continuing from the current position.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 2, 1, 2, 3, 4, 5, 6, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 3, 3, 4, 5, 6, 1, 1, 2, 3, 3, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2

Offset: 1

Gary Yane, Feb 06 2021

Arrange 1,2,3,...,n clockwise in a circle. Start the count at the lowest surviving value and delete the k-th value counting clockwise around the circle. Repeat this procedure until one number remains, which is T(n,k).
Note: In the complete n X k array with n >= 1 and k >= 1, T(n,k) = T(k-1,k) for all n >= k > 1 and T(n,1)=n.
That makes the bottom triangle of the array unchanging, so it is omitted.

			n\k    2    3    4    5    6    7    8    9   10   11   12   13
_______________________________________________________________
1      1    1    1    1    1    1    1    1    1    1    1    1
2           2    1    2    1    2    1    2    1    2    1    2
3                2    3    1    3    1    2    2    3    1    3
5                     4    1    4    1    3    3    4    1    4
6                          2    5    1    3    3    5    1    5
7                               6    1    4    3    6    1    6
8                                    2    5    4    7    1    7
9                                         6    5    8    1    8
10                                             6    9    1    9
11                                                 10    1   10
12                                                       2   11
13                                                           12

Index entries for sequences related to the Josephus Problem

The last entry in each column is A128982.

T(1,k) = 1, for k > 1;
T(n,k) = T(n-1,k) if k mod n > T(n-1,k) or k mod n = 0;
T(n,k) = T(n-1,k) + 1 otherwise.

A178776 a(n) is the largest number that appears twice in an n x n multiplication table of positive integers, disregarding the pairs that are obviously produced by the commutative property.

Original entry on oeis.org

4, 4, 12, 12, 24, 36, 40, 40, 72, 72, 84, 120, 144, 144, 180, 180, 240, 252, 252, 252, 360, 400, 400, 432, 504, 504, 600, 600, 672, 672, 672, 840, 900, 900, 900, 936, 1120, 1120, 1260, 1260, 1320, 1440, 1440, 1440, 1680, 1764, 1800, 1800, 1872, 1872

Offset: 4

Gary Yane, Jun 11 2010

nonn

It appears that all terms of this sequence are multiple of 4 (checked up to 10^7). - Michel Marcus, Aug 26 2013

			The first term is 4, which appears twice in a 4 x 4 multiplication table. a(4) = 2x2 = 4x1 = 4. For all n = prime a(n) = a(n-1) so a(5) = 4. a(6) = 12 = 2x6 = 3x4, a(7) = 12, a(8) = 24 = 3x8 = 4x6, a(9) = 36 = 4x9 = 6x6.

a(n) = {skeep = Set(); mmax = 0; for (i = 1, n, for (j = i, n, v = i*j; if (! setsearch(skeep, v), skeep = setunion(skeep, Set(v)), mmax = max(mmax, v)););); mmax;} \\ Michel Marcus, Aug 26 2013

findxy(n, d) = {mins = 2*n; if (#d % 2, nd = #d\2 +1, nd = #d/2); for (i = 1, nd, if ((s = d[i]+n/d[i]) < mins, mins = s; mini = i);); x = d[mini]; y = n/x; return (x*y*(x-1)*(y-1));}
lista(nn) = {lmmx = 4; print1(lmmx, ", "); for (n=5, nn, d = divisors(n); nmmx = findxy(n, d); lmmx = max(lmmx, nmmx); print1(lmmx, ", "););} \\ Michel Marcus, Aug 26 2013

Let xy = n be the factorization of n such that x+y is a minimum. We then have a(4)= 4 and a(n)= max{a(n-1), xy(x-1)(y-1)} for all n>4.

Edited by N. J. A. Sloane, Jun 12 2010

Corrected by Michel Marcus, Aug 26 2013

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User: Gary Yane

A358787 a(1)=1; let x=gcd(a(n-1),n); for n > 1, a(n) = a(n-1) + n if x=1 or a(n-1)/x=1, otherwise a(n) = a(n-1)/x.

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A342831 a(n) is the smallest positive integer k such that the n-dimensional cube [0,k]^n contains at least as many internal lattice points as external lattice points.

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A343576 Number of permutations of [n] without fixed points and all cycles equal length.

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A341216 Triangle read by columns T(n,k) k > n >= 1: Last survivor positions in a modified Josephus problem for n numbers, where after each deletion the counting starts over at the lowest existing number n, rather than continuing from the current position.

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A178776 a(n) is the largest number that appears twice in an n x n multiplication table of positive integers, disregarding the pairs that are obviously produced by the commutative property.

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