1,2

For each ordered pair of divisors of n, (d1,d2), a(n) can also be found using the algorithm: add d1 if d2 is prime; otherwise add 1. For example, when n = 4 the divisor pairs are: (1,1), (1,2), (1,4), (2,1), (2,2), (2,4), (4,1), (4,2), (4,4) which gives 1 + 1 + 1 + 1 + 2 + 1 + 1 + 4 + 1 = 13.

Table of n, a(n) for n=1..64.

a(n) = Sum_{d1|n} Sum_{d2|n} d1^c(d2), where c = A010051.

a(prime(n)) = prime(n) + 3.

Table[DivisorSigma[0, n] (DivisorSigma[0, n] - PrimeNu[n]) + PrimeNu[n] DivisorSigma[1, n], {n, 80}]

(PARI) a(n) = my(f=factor(n), d=numdiv(f)); d^2 + omega(f)*(sigma(f) - d); \\ Michel Marcus, Oct 05 2021

Cf. A000005 (tau), A000040 (n-th prime), A000203 (sigma), A001221 (omega), A010051 (prime characteristic).

nonn,new

Wesley Ivan Hurt, Oct 03 2021

approved

0,2

Fractional factorials are defined using the Gamma function; for example, (n/2)! := Gamma(1 + n/2).

The sequence defined by u(n) = (12*n)!*n!/((6*n)!*(4*n)!*(3*n)!) is one of the 52 sporadic integral factorial ratio sequences of height 1 found by V. I. Vasyunin (see Bober, Table 2, Entry 1). See A295431. It is known that u(n) is integral and satisfies the congruences u(n*p) == u(n) ( mod p^3 ) for prime p >= 5 and any positive integer n (Zudilin, Section 5); the o.g.f. Sum_{n >= 0} u(n)*x^n is algebraic over Q(x) (Rodriguez-Villegas).

Here we are essentially considering the sequence ( u(n/2) )n>=0. The sequence is conjectured to be integral.

Table of n, a(n) for n=0..13.

P. Bala, Some integer ratios of factorials

J. W. Bober, Factorial ratios, hypergeometric series, and a family of step functions, arXiv:0709.1977 [math.NT], 2007; J. London Math. Soc., Vol. 79, Issue 2 (2009), 422-444.

F. Rodriguez-Villegas, Integral ratios of factorials and algebraic hypergeometric functions, arXiv:math/0701362 [math.NT], 2007.

Wadim Zudilin, Congruences for q-binomial coefficients, arXiv:1901.07843 [math.NT], 2019.

a(n) = binomial(6*n,2*n)*binomial(4*n,n)/binomial(3*n/2,n).

a(2*n) = A295431(n).

a(2*n) = 24*(12*n - 1)*(12*n - 5)*(12*n - 7)*(12*n - 11)/( n*(2*n - 1)*(3*n - 1)*(3*n - 2) )*a(2*n-2);

a(2*n+1) = 96*(12*n + 1)*(12*n - 1)*(12*n + 5)*(12*n - 5)/( n*(2*n + 1)*(6*n + 1)*(6*n - 1) )*a(2*n-1).

Asymptotics: a(n) ~ 32^n/sqrt(6*Pi*n) * 3^(3*n/2) as n -> infinity.

O.g.f.: A(x) = hypergeom([1/12, 5/12, 7/12, 11/12], [1/3, 1/2, 2/3], 27648*x^2) + 40*x*hypergeom([11/12, 13/12, 7/12, 17/12], [3/2, 5/6, 7/6], 27648*x^2) is conjectured to be algebraic over Q(x).

Conjectural: a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for  prime p >= 5 and positive integers n and k.

a(11) - a(1) = 187495217080545878999040 - 40 = (2^3)*(5^3)*(11^3)*140867931690868429 == 0 (mod 11^3).

seq(binomial(6*n, 2*n)*binomial(4*n, n)/binomial(3*n/2, n), n = 0..13);

Cf. A295431, A347855 - A347858.

nonn,easy,new

Peter Bala, Sep 16 2021

approved

1,1

In this variation of A347581 the areas must be added in the order of their sizes, from 1 through n, and as each area is added the minimum possible number of line segments must be used. This forces, for example, the first three areas of size 1,2 and 3 to form a 2x3 block and thus they can never appear in any other arrangement in the final area. This is also true for n up to at least 9 due to the restriction of maximizing the usable edges for the next area. This leads to a(8) and a(10) contain one more line segment than the optimal solutions of A347581.

Table of n, a(n) for n=1..10.

Sascha Kurz, Counting polyominoes with minimum perimeter, arXiv:math/0506428 [math.CO], 2015.

Examples of n = 1 to n = 10 are given below. Note that for a(3) the configuration could also consist of the area of size 1 sitting above the area of size 2 with the area of size 3 forming an L shaped block creating the minimal 2x3 block.

.

   __

  |__|  a(1) = 4

   __ __ __

  |__|__ __|  a(2) = 9

   __ __ __

  |__|__ __|  a(3) = 14

  |__ __ __|

   __ __ __

  |__|__ __|

  |__ __ __|  a(4) = 20

  |     |

  |__ __|

   __ __ __

  |__|__ __|__

  |__ __ __|  |  a(5) = 26

  |     |     |

  |__ __|__ __|

   __ __ __

  |__|__ __|__ __ __

  |__ __ __|  |     |  a(6) = 33

  |     |     |     |

  |__ __|__ __|__ __|

         __ __ __ __

   __ __|__         |

  |__|__ __|__ __ __|

  |__ __ __|  |     |  a(7) = 40

  |     |     |     |

  |__ __|__ __|__ __|

         __ __ __ __

        |           |

        |__ __ __ __|

   __ __|__         |

  |__|__ __|__ __ __|  a(8) = 48

  |__ __ __|  |     |

  |     |     |     |

  |__ __|__ __|__ __|

   __ __ __ __ __ __ __

  |        |           |

  |        |__ __ __ __|

  |__ __ __|__         |

     |__|__ __|__ __ __|  a(9) = 55

     |__ __ __|  |     |

     |     |     |     |

     |__ __|__ __|__ __|

      __ __ __ __ __ __ __

     |        |           |

     |        |__ __ __ __|

   __|__ __ __|__         |

  |     |__|__ __|__ __ __|  a(10) = 64

  |     |__ __ __|  |     |

  |     |     |     |     |

  |     |__ __|__ __|__ __|

  |__ __|

.

Cf. A347581, A001168, A291808, A291809, A328020, A291806, A006983.

nonn,more,new

Scott R. Shannon, Oct 03 2021

approved

0,2

Table of n, a(n) for n=0..26.

a(2) = 10. Assuming the lower left quadrant is the one removed then a walk of left-down or down-left is not permitted, so the total number of 2-step walks is 4 * 3 - 2 = 10.

Cf. A001411 (four quadrants), A116903 (two quadrants), A038373 (one quadrant), A129700 (half quadrant).

nonn,walk,new

Scott R. Shannon, Sep 26 2021

approved

1,2

A number is visible from the current number if, given that it has coordinates (x,y) relative to the current number, the greatest common divisor of |x| and |y| is 1. The sequence has a(2) = 3 as that is the smallest number visible from a(1) = 1 that does not differ by 1 from a(1).

The majority of steps between terms are diagonals across the current number's nearest corner of the square spiral and thus do not approach the center of the spiral. Occasionally, however, a long diagonal step directly across the center of the spiral is taken. See the linked image.

In the first 20000 terms the longest single step is the one at n = 19534 of length sqrt(38365) units between 19743 at coordinates (-68,-70), to 19460 at coordinates (70,69). This step also yields the maximum difference between terms in the same range.

Table of n, a(n) for n=1..76.

Scott R. Shannon, Image of the path for the first 20000 terms. The colors are graduated across the spectrum to show the relative step order.

The square spiral is numbered as follows:

.

  17--16--15--14--13   .

   |               |   .

  18   5---4---3  12  29

   |   |       |   |   |

  19   6   1---2  11  28

   |   |           |   |

  20   7---8---9--10  27

   |                   |

  21--22--23--24--25--26

.

a(3) = 8 as gcd(8,3) = 1 and 8 is unvisited, visible from 3, and does not equal 2 or 4.

a(4) = 5 as gcd(5,8) = 1 and 5 is unvisited, visible from 8, and does not equal 7 or 9.

a(5) = 2 as gcd(2,5) = 1 and 2 is unvisited, visible from 5, and does not equal 4 or 6.

Cf. A348022, A331400, A335661, A063826, A332767, A347358, A335661.

nonn,new

Scott R. Shannon, Sep 25 2021

approved

1,2

A number is not visible from the current number if, given it has coordinates (x,y) relative to the current number, the greatest common divisor of |x| and |y| is greater than 1. The sequence has a(2) = 11 as that is the smallest number not visible from a(1) = 1.

In the first 10000 terms the longest single step is one at n = 6888 of length sqrt(22556) units between 22203 to 7389. The maximum difference between terms in the same range is from 3469 to 58973 at n = 9709.

Table of n, a(n) for n=1..73.

Scott R. Shannon, Image of the path for the first 10000 terms. The colors are graduated across the spectrum to show the relative step order.

The square spiral is numbered as follows:

.

  17--16--15--14--13   .

   |               |   .

  18   5---4---3  12   29

   |   |       |   |   |

  19   6   1---2  11   28

   |   |           |   |

  20   7---8---9--10   27

   |                   |

  21--22--23--24--25--26

.

a(3) = 55 as gcd(55,11) = 11 and 55 is unvisited and not visible from 11, being blocked by 29.

a(4) = 15 as gcd(15,55) = 5 and 15 is unvisited and not visible from 55, being blocked by 13, 14 and 30.

a(5) = 21 as gcd(21,15) = 3 and 21 is unvisited and not visible from 15, being blocked by 6.

Cf. A348022 (visible), A331400, A335661, A063826, A332767.

nonn,new

Scott R. Shannon, Sep 25 2021

approved

1,2

A number is visible from the current number if, given it has coordinates (x,y) relative to the current number, the greatest common divisor of |x| and |y| equals 1. See A331400 for the points visible from the starting 1 number.

In the first 10000 terms the longest single step is one at n = 9942 of length sqrt(22570) units between 31002 to 10258. The maximum difference between terms in the same range is from 5171 to 36197 at n = 9977.

Table of n, a(n) for n=1..73.

Scott R. Shannon, Image of the path for the first 10000 terms. The colors are graduated across the spectrum to show the relative step order.

The square spiral is numbered as follows:

.

  17--16--15--14--13   .

   |               |   .

  18   5---4---3  12   29

   |   |       |   |   |

  19   6   1---2  11   28

   |   |           |   |

  20   7---8---9--10   27

   |                   |

  21--22--23--24--25--26

.

a(3) = 4 as gcd(4,2) = 2 and 4 is unvisited and visible from 2.

a(4) = 6 as gcd(4,6) = 2 and 6 is unvisited and visible from 4.

a(5) = 3 as gcd(3,6) = 3 and 3 is unvisited and visible from 6.

a(6) = 12 as gcd(12,3) = 3 and 12 is unvisited and visible from 3. Note although 9 is unvisited and gcd(9,3) = 3 it is not visible from 3 due to 2.

Cf. A348025 (not visible), A331400, A335661, A063826, A332767, A347358.

nonn,new

Scott R. Shannon, Sep 25 2021

approved

1,2

This sequence uses the same rules as A336957 except with the additional restriction that a(n) must have a different number of divisors than a(n-1). This leads to the terms showing a greater variation in value. For example in the first 5000 terms the maximum is a(3915) = 228569, compared to a maximum of a(3225) = 11053 for A336957 in the same range. Like A336957 is it likely all positive integers other than the prime-powers eventually appear.

Table of n, a(n) for n=1..69.

a(4) = 45, as a(4) must share a factor with a(3) = 6, have a prime factor not in 6, have no common factor with a(2) = 2, and not have tau(6) = 4 divisors. The smallest positive integer satisfying these conditions is 45. Note that A336957(4) = 15, but 15 has four divisors thus cannot be chosen here.

Cf. A336957, A000005, A098550, A064413.

nonn,new

Scott R. Shannon, Sep 27 2021

approved

1,4

Table of n, a(n) for n=1..33.

G.f.: A(x) = (1 - sqrt(1 - 2*x - 3*x^2))/(x*(2 + x - sqrt(1 - 2*x - 3*x^2))).

If M(x) is the g.f. of A001006, then A(x) = (1 + 2*x*M(x))/(1 + 2*x + 2*x^2*M(x)).

Let M(x) be the g.f. of A001006 and F(x) = 1 + 2*x*M(x) (equivalently, x*F(x) = g.f. of A007971). Then F(-x*A(x)) = 1/F(x).

A(-x*A(x)) = 1/A(x).

a[n_] := SeriesCoefficient[(1 - Sqrt[1-2*x-3*x^2])/(x * (2 + x - Sqrt[1-2*x-3*x^2])), {x, 0, n}]; Array[a, 33, 0] (* Amiram Eldar, Oct 06 2021 *)

(PARI) my(x='x+O('x^35)); Vec((1-sqrt(1-2*x-3*x^2))/(x*(2+x-sqrt(1-2*x-3*x^2)))) \\ Michel Marcus, Oct 06 2021

Cf. A001006, A007971, A086246.

nonn,new

Alexander Burstein, Oct 06 2021

approved

1,2

Table of n, a(n) for n=1..6.

a(n) = Sum_{L partition of n} A348180(n, L) * A036038(len(L), sig(L)), where sig(L) is the partition composed by the part multiplicities of L.

Cf. A348180, A036038, A347843, A348116-A348118.

more,nonn,new

Álvar Ibeas, Oct 05 2021

approved