1,2

Conjectured to be a permutation of the natural numbers (and if so, -1 will never appear).

Table of n, a(n) for n=1..75.

Cf. A361321, A000469, A361323, A336957, A360519, A361102.

nonn,new

Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 11 2023

approved

1,2

Conjectured to be a permutation of the natural numbers (if not then there exists a k such that A000469(k) does not appear in A361321).

Table of n, a(n) for n=1..75.

Cf. A361321, A000469, A361324, A336957, A360519, A361102.

nonn,new

Scott R. Shannon, Rémy Sigrist, and N. J. A. Sloane, Mar 11 2023

approved

4,1

Table of n, a(n) for n=4..80.

Eric Weisstein's World of Mathematics, Centered Polygonal Number.

Wikipedia, Centered polygonal number.

a(16) = 5 since 16 is a centered pentagonal number, but not a centered square or centered triangular number.

seq[len_] := Module[{s = Table[0, {len}], c = 0, k = 3, n, ckn}, While[c < len, n = 2; While[(ckn = k*n*(n - 1)/2 - 2) <= len, If[s[[ckn]] == 0, c++; s[[ckn]] = k]; n++]; n = 4; k++]; s]; seq[100] (* Amiram Eldar, Mar 06 2023 *)

Cf. A101321, A176774, A333914.

nonn,new

Ilya Gutkovskiy, Feb 15 2023

approved

1,10

This sequence studies the special case of Bertrand's postulate where n is a square and p of the form k^2+1.

Conjecture: Except for n = 7, between the square of an integer n > 1 and the double of its square there is always at least one prime number of the form k^2+1.

The following table shows the first 15 cases, with p prime of the form k^2+1.

+----+-------+-------+-----------------------------+------+

| n | n^2 | 2*n^2 | p such that n^2 < p < 2*n^2 | a(n) |

+----+-------+-------+-----------------------------+------+

| 1 | 1 | 2 | --------------------------- | 0 |

| 2 | 4 | 8 | 5 | 1 |

| 3 | 9 | 18 | 17 | 1 |

| 4 | 16 | 32 | 17 | 1 |

| 5 | 25 | 50 | 37 | 1 |

| 6 | 36 | 72 | 37 | 1 |

| 7 | 49 | 98 | --------------------------- | 0 |

| 8 | 64 | 128 | 101 | 1 |

| 9 | 81 | 162 | 101 | 1 |

| 10 | 100 | 200 | 101, 197 | 2 |

| 11 | 121 | 242 | 197 | 1 |

| 12 | 144 | 288 | 197, 257 | 2 |

| 13 | 169 | 338 | 197, 257 | 2 |

| 14 | 196 | 392 | 197, 257 | 2 |

| 15 | 225 | 450 | 257, 401 | 2 |

Michel Lagneau, Table of n, a(n) for n = 1..1000

a(20) = 3 because there are 3 prime numbers of the form k^2+1 between 20^2 and 2*20^2. We obtain 400 < 401, 577, 677 < 800.

nn:=90:

for n

from 1 to nn do:

n1:=n^2:n2:=2*n^2:it:=0:

for p from n1+1 to n2-1 do:

p1:=sqrt(p-1):p2:=floor(p1):

if isprime(p) and p1=p2

then

it:=it+1:

else

fi:

od:

printf(`%d, `, it):

od:

(PARI) a(n) = my(nb=0); forprime(p=n^2+1, 2*n^2-1, if (issquare(p-1), nb++)); nb; \\ Michel Marcus, Feb 22 2023

Cf. A000290, A002496, A060715.

nonn,new

Michel Lagneau, Feb 21 2023

approved

1,1

The abundance of each term is A033880(a(n)) and s = Sum_{i=1..n} A033880(a(i)).

All imperfect numbers A132999 will appear in this sequence and the abundant numbers A005101 will appear in order.

Table of n, a(n) for n=1..67.

The sequence starts with a(1) = 12, since 12 is the first imperfect number with abundance greater than 0. Then the next term not yet in the sequence, such that s is not less than 1, is 1.

a(5) is the next abundant number 18, since any deficient number would bring s below 1.

n : 1 2 3 4 5 6 7 8 9 10

a(n): 12 1 2 4 18 3 8 20 10 24

s : 4 3 2 1 4 2 1 3 1 13

(Python)

from sympy.ntheory import abundance

from itertools import count, filterfalse

def A361206_list(nmax):

A, s = [], 0

for n in range(1, nmax+1):

A2 = set(A)

for y in filterfalse(A2.__contains__, count(1)):

ab = abundance(y)

if ab != 0 and ab + s >= 1:

A.append(y)

s += ab

break

return(A)

Cf. A005101, A033879, A033880, A132999.

nonn,easy,new

John Tyler Rascoe, Mar 04 2023

approved

1,2

First differs from A037279 at a(20).

Table of n, a(n) for n=1..50.

Divisors of 20 are 1,2,4,5,10,20, so a(20)=24501.

(PARI) a(n) = if (isprime(n) || (n==1), return (n)); my(d=divisors(n), list=List()); for (i=2, #d-1, my(dd=digits(d[i])); forstep (j=#dd, 1, -1, listput(list, dd[j]))); fromdigits(Vec(list)); \\ Michel Marcus, Mar 09 2023

Cf. A037279, A130140.

nonn,base,new

Tyler Busby, Mar 09 2023

approved

1,3

Tilings that are rotations or reflections of each other are considered distinct.

Pieces can have any combination of integer side lengths, but for the optimal sets computed so far (up to (n,k) = (7,5)), all pieces have one side of length 1.

Table of n, a(n) for n=1..26.

T(n,1) = A102462(n).

Triangle begins:

n\k| 1 2 3 4 5 6 7

---+-----------------------------------------------------

1 | 1

2 | 1 4

3 | 2 11 56

4 | 3 29 370 5752

5 | 4 94 2666 82310 2519124

6 | 6 263 19126 1232770 88117873 6126859968

7 | 12 968 134902 19119198 2835424200 ? ?

8 | 20 3416 1026667 307914196 109979838540 ? ?

A 3 X 3 square can be tiled by three 1 X 2 pieces and three 1 X 1 pieces in the following ways:

+---+---+---+ +---+---+---+ +---+---+---+

| | | | | | | | | | | |

+---+---+---+ + +---+---+ +---+ +---+

| | | | | | | | | | |

+---+---+ + +---+---+ + +---+---+ +

| | | | | | | | |

+---+---+---+ +---+---+---+ +---+---+---+

.

+---+---+---+ +---+---+---+ +---+---+---+

| | | | | | | | | |

+---+---+---+ +---+---+ + +---+---+---+

| | | | | | | | | |

+---+---+ + +---+---+---+ +---+---+---+

| | | | | | | | |

+---+---+---+ +---+---+---+ +---+---+---+

.

+---+---+---+ +---+---+---+

| | | | | |

+---+---+---+ +---+---+---+

| | | | | |

+---+---+---+ +---+---+---+

| | | | | |

+---+---+---+ +---+---+---+

The first six of these have no symmetries, so they account for 8 tilings each. The last two has a mirror symmetry, so they account for 4 tilings each. In total there are 6*8+2*4 = 56 tilings. This is the maximum for a 3 X 3 square, so T(3,3) = 56.

The following table shows the sets of pieces that give the maximum number of tilings up to (n,k) = (7,5). The solutions are unique except for (n,k) = (2,1) and (n,k) = (6,1).

\ Number of pieces of size

(n,k)\ 1 X 1 | 1 X 2 | 1 X 3 | 1 X 4

------+-------+-------+-------+------

(1,1) | 1 | 0 | 0 | 0

(2,1) | 2 | 0 | 0 | 0

(2,1) | 0 | 1 | 0 | 0

(2,2) | 2 | 1 | 0 | 0

(3,1) | 1 | 1 | 0 | 0

(3,2) | 2 | 2 | 0 | 0

(3,3) | 3 | 3 | 0 | 0

(4,1) | 2 | 1 | 0 | 0

(4,2) | 4 | 2 | 0 | 0

(4,3) | 3 | 3 | 1 | 0

(4,4) | 5 | 4 | 1 | 0

(5,1) | 3 | 1 | 0 | 0

(5,2) | 4 | 3 | 0 | 0

(5,3) | 4 | 4 | 1 | 0

(5,4) | 7 | 5 | 1 | 0

(5,5) | 7 | 6 | 2 | 0

(6,1) | 2 | 2 | 0 | 0

(6,1) | 1 | 1 | 1 | 0

(6,2) | 4 | 4 | 0 | 0

(6,3) | 7 | 4 | 1 | 0

(6,4) | 8 | 5 | 2 | 0

(6,5) | 10 | 7 | 2 | 0

(6,6) | 11 | 8 | 3 | 0

(7,1) | 2 | 1 | 1 | 0

(7,2) | 5 | 3 | 1 | 0

(7,3) | 8 | 5 | 1 | 0

(7,4) | 10 | 6 | 2 | 0

(7,5) | 11 | 7 | 2 | 1

Main diagonal: A361217.

Columns: A361218 (k = 2), A361219 (k = 3), A361220 (k = 4).

Cf. A360629, A361221.

nonn,tabl,more,new

Pontus von Brömssen, Mar 05 2023

approved

0,4

The number of all noncrossing caterpillars with n edges is given by A361356.

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

Index entries for linear recurrences with constant coefficients, signature (5,4,-34,7,63,-30,-46,31,13,-14,0,3,-1).

G.f.: (1 - 4*x - 8*x^2 + 28*x^3 + 15*x^4 - 55*x^5 - 2*x^6 + 46*x^7 - 11*x^8 - 19*x^9 + 10*x^10 + 2*x^11 - 2*x^12)/((1 - x)^2*(1 + x)^2*(1 - 5*x + 3*x^2 - x^3)*(1 - 5*x^2 + 3*x^4 - x^6)).

a(n) = 5*a(n-1) + 4*a(n-2) - 34*a(n-3) + 7*a(n-4) + 63*a(n-5) - 30*a(n-6) - 46*a(n-7) + 31*a(n-8) + 13*a(n-9) - 14*a(n-10) + 3*a(n-12) - a(n-13) for n >= 13.

(PARI)

G(x)={ my(f = x*(2 - x)/(1 - 5*x + 3*x^2 - x^3), g = 1 + x + x^2*(3 - 2*x + (4 - 3*x + x^2)*f + (1 + 2*x)*f^2)/(1 - x)^2); (intformal(g) - 3)/x/2 + x*subst((3 + 2*x*(3-x)*f)/(1-x)^2, x, x^2)/4 + subst(1/(1-x) + x*f/(1-x), x, x^2)/2}

{ Vec(G(x) + O(x^30)) }

Cf. A296533 (noncrossing trees), A361356, A361358, A361359 (up to rotation only).

nonn,easy,new

Andrew Howroyd, Mar 10 2023

approved

0,4

The number of all noncrossing caterpillars with n edges is given by A361356.

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

Index entries for linear recurrences with constant coefficients, signature (6,-3,-26,36,-2,-18,6,5,-4,1).

G.f.: (1 - 5*x - 2*x^2 + 27*x^3 - 20*x^4 - 13*x^5 + 23*x^6 - 5*x^7 - 6*x^8 + 3*x^9)/((1 - x)*(1 - 5*x + 3*x^2 - x^3)*(1 - 5*x^2 + 3*x^4 - x^6)).

a(n) = 6*a(n-1) - 3*a(n-2) - 26*a(n-3) + 36*a(n-4) - 2*a(n-5) - 18*a(n-6) + 6*a(n-7) + 5*a(n-8) - 4*a(n-9) + a(n-10) for n >= 10.

(PARI)

G(x)={ my(f = x*(2 - x)/(1 - 5*x + 3*x^2 - x^3), g = 1 + x + x^2*(3 - 2*x + (4 - 3*x + x^2)*f + (1 + 2*x)*f^2)/(1 - x)^2); (intformal(g) - 3)/x + x*subst((1 + 2*x*f)/(1-x)^2, x, x^2)/2 }

{ Vec(G(x) + O(x^30)) }

Cf. A296532 (noncrossing trees), A361356, A361358, A361360 (up to rotation and reflection).

nonn,easy,new

Andrew Howroyd, Mar 09 2023

approved

0,4

A permutation is said to be Grassmannian if it has at most one descent. A permutation is even if it has an even number of inversions.

Avoiding any of the patterns 2314 or 3412 gives the same sequence.

Table of n, a(n) for n=0..50.

Juan B. Gil and Jessica A. Tomasko, Pattern-avoiding even and odd Grassmannian permutations, arXiv:2207.12617 [math.CO], 2022.

Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

G.f.: -(x^5-x^4-4*x^3+2*x^2+x-1)/((x+1)^2*(x-1)^4).

a(n) = 1 - 5*n/24 + n^3/12 - (-1)^n * n/8. - Robert Israel, Mar 10 2023

For n=4 the a(4) = 5 permutations are 1234, 1342, 2314, 3124, 3412.

seq(1 - 5*n/24 + n^3/12 - (-1)^n * n/8, n = 0 .. 100); # Robert Israel, Mar 10 2023

Cf. A356185, A361272, A361273, A361274.

For the corresponding odd permutations, cf. A005993.

nonn,easy,new

Juan B. Gil, Mar 10 2023

approved