A330805 -id:A330805 - cp's OEIS frontend

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

0, 5, 27, 85, 205, 420, 770, 1302, 2070, 3135, 4565, 6435, 8827, 11830, 15540, 20060, 25500, 31977, 39615, 48545, 58905, 70840, 84502, 100050, 117650, 137475, 159705, 184527, 212135, 242730, 276520

Offset: 0

Luce ETIENNE, Nov 18 2020

			a(1) = 2*3-1 = 5, a(2) = 2*16-5 = 27, a(3) = 2*50-15 = 85, a(4) = 2*120-35 = 205, a(5) = 2*245-70 = 420, a(6) = 2*448-126 = 770.

Luce ETIENNE, Illustration of a(1), a(2), a(3) and a(4)
Wikipedia, Aztec diamond.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217, A000332, A002417, A004320, A045943, A258440, A330805.

CoefficientList[Series[x (2 x + 5)/(1 - x)^5, {x, 0, 30}], x] (* Michael De Vlieger, Dec 12 2020 *)

G.f.: x*(2*x + 5)/(1 - x)^5.
E.g.f.: exp(x)*x*(120 + 204*x + 76*x^2 + 7*x^3)/24. - Stefano Spezia, Nov 18 2020
a(n) = 5*(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
a(n) = n*(n + 1)*(n + 2)*(7*n + 13)/24.
a(n) = 2*A004320(n) - A000332(n+3).
a(n) = 2*A000332(n+2) + 5*A000332(n+3).

A356356 Triangle of number of rectangles in the interior of the rectangle with vertices (k,0), (0,k), (n,n+k) and (n+k,n), read by rows.

Original entry on oeis.org

0, 1, 9, 2, 19, 51, 3, 29, 86, 166, 4, 39, 121, 250, 410, 5, 49, 156, 334, 575, 855, 6, 59, 191, 418, 740, 1141, 1589, 7, 69, 226, 502, 905, 1427, 2044, 2716, 8, 79, 261, 586, 1070, 1713, 2499, 3396, 4356, 9, 89, 296, 670, 1235, 1999, 2954, 4076, 5325, 6645

Offset: 1

Evan Robinson, Oct 15 2022

The function of the triangle T(n,k), where n,k > 0, is equal to (n-k+1)*A330805(k-1) - (n-k)*T(k,k-1) + k*(n-k). This is equivalent to saying that this function is (n-k+1) Aztec diamonds (A330805(k-1)) minus the overlaps of those diamonds (two Aztec diamonds of size k-1 overlapped, hence f(k,k-1)) plus (n-k) copies of k extra rectangles. For this last part, the rectangles are of sizes 1 X (2k-1), 3 X (2k-3), 5 X (2k-5), ..., (2k-3) X 3, (2k-1) X 1 and there are (n-k) copies per overlap.
T(n,n) = A330805(n-1).
If n or k <= 0, T(n,k) = 0.
T(n,k) = T(k,k) + (n-k)*A000447(k). That is, incrementing n for fixed k adds a fixed number of new rectangles, equal to A000447(k).
This sequence was prompted by the codegolf.se question linked below, where the problem was to find T(n,k) plus the number of squares and rectangles in an n X k rectangular lattice with diagonals (lines y+a=+-x).

			Triangle T(n,k) begins:
  n\k  1    2    3    4    5    6    7    8    9   10
   1   0
   2   1    9
   3   2   19   51
   4   3   29   86  166
   5   4   39  121  250  410
   6   5   49  156  334  575  855
   7   6   59  191  418  740 1141 1589
   8   7   69  226  502  905 1427 2044 2716
   9   8   79  261  586 1070 1713 2499 3396 4356
  10   9   89  296  670 1235 1999 2954 4076 5325 6645
For n = 7, k = 3, T(n,k) = (7-3+1)*A330805(3-1) - (7-3)*f(3,2) + 3*(7-3) = 5*51 - 4*19 + 3*4 = 191.

Evan Robinson, First 100 rows, flattened
Code Golf Stack Exchange, Related problem of finding the number of squares on a diagonal grid

Cf. A330805, A000447.

function T(n, k)
    (2*(n-k)*(4*k^3-k)+(4*k^4-k^2-3*k))÷6
end

T(n,k) = (n-k+1)*A330805(k-1) - (n-k)*T(k,k-1) + k*(n-k).
T(n,k) = (n-k+1)*(4*k^4-k^2-3*k)/6 - (n-k)*T(k,k-1) + k*(n-k).
T(n,k) = 1/3*(n-k)*(4*k^3-k) + (4*k^4-k^2-3*k)/6.
T(n,k) = (n-k)*A000447(k) + A330805(k-1).
T(n,1) = n-1.
T(n,n) = A330805(n-1).
T(n,n-1) = (4*n^4-8*n^3-n^2+5*n)/6.
T(n,k) = (n-1)*A000447(k) - T(k,k-1).

A372917 a(n) is the number of distinct rectangles with area n whose vertices lie on points of a unit square grid.

Original entry on oeis.org

1, 2, 1, 3, 2, 3, 1, 4, 2, 5, 1, 5, 2, 3, 3, 5, 2, 5, 1, 8, 2, 3, 1, 7, 3, 5, 2, 5, 2, 9, 1, 6, 2, 5, 3, 8, 2, 3, 3, 11, 2, 6, 1, 5, 5, 3, 1, 9, 2, 8, 3, 8, 2, 6, 3, 7, 2, 5, 1, 15, 2, 3, 3, 7, 5, 6, 1, 8, 2, 9, 1, 11, 2, 5, 5, 5, 2, 9, 1, 14, 3, 5, 1, 10, 5, 3

Offset: 1

Felix Huber, Jun 08 2024

nonn

A rectangle in the square unit grid has the sides W = w*sqrt(r) and H = h*sqrt(r). The area is therefore n = w*h*r. Let r be a squarefree divisor of n that can be written as the sum of two squares x^2 + y^2. The number of distinct rectangles is then the sum of the number of ways for each value of r to decompose n/r into two factors w and h (with w >= h).

			See also the linked illustrations of the terms a(4) = 3, a(8) = 4, a(15) = 3.
n = 4 has the three divisors 1, 2, 4. Since 4 is not squarefree, r can have the values 1 or 2. For r = 1 = 1^2 + 0^2 there are two rectangles (2,2), (4,1). For r = 2 = 1^2 + 1^2 and n/r = 4/2 = 2 = w*h there is the rectangle (2*sqrt(2), 1*sqrt(2)). That's a total of a(4) = 3 distinct rectangles.
n = 8 has the four divisors 1, 2, 4, 8. Since 4 and 8 are not squarefree, r can have the values 1 or 2. For r = 1 = 1^2 + 0^2 there are two rectangles (4,2), (8,1). For r = 2 = 1^2 + 1^2 and n/r = 8/2 = 4 = w*h there are the rectangles (4*sqrt(2), 1*sqrt(2)) and (2*sqrt(2), 2*sqrt(2)). That's a total of a(8) = 4 distinct rectangles.
n = 15 has the four divisors 1, 3, 5, 15. They are all squarefree, but 3 and 15 cannot be written as a sum of two squares, r can only have the values 1 or 5. For r = 1 = 1^2 + 0^2 there are two rectangles (5,3), (15,1). For r = 5 = 2^2 + 1^2 and n/r = 15/5 = 3 = w*h there is the rectangles (3*sqrt(5), 1*sqrt(5)). That's a total of a(15) = 3 distinct rectangles.

Felix Huber, Table of n, a(n) for n = 1..10000
Felix Huber, Illustrations of terms a(4), a(8), a(15)

Cf. A001481, A005117, A038548, A085582, A113751, A122225, A285956, A330805, A354704, A372915.

A372917:= proc(n)
    local f,i,prod;
    f:=ifactors(n)[2];
    prod:=1;
    for i from 1 to numelems(f) do
        if f[i][1] mod 4 = 3 then
            prod:=prod*(1*f[i][2]+1);
        else
            prod:=prod*(2*f[i][2]+1);
        end if;
    end do;
    return round(prod/2);
end proc;
seq(A372917(n),n=1..86);

a(n) = my(f=factor(n)); prod(i=1,#f[,1], if(f[i,1]%4==3,1,2)*f[i,2] + 1) \/ 2; \\ Kevin Ryde, Jun 09 2024

a(n) = ceiling(Product_{i=1..omega(n)}(k[i]*e[i] + 1)/2), with k[i] = 2 if p[i] mod 4 = 3 and k[i] = 1 else, where p[i]^e[i] is the prime factorization of n.

cp's OEIS Frontend

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A356356 Triangle of number of rectangles in the interior of the rectangle with vertices (k,0), (0,k), (n,n+k) and (n+k,n), read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Formula

A372917 a(n) is the number of distinct rectangles with area n whose vertices lie on points of a unit square grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

cp's OEIS Frontend

A338996 Numbers of squares and rectangles of all sizes in 3*n*(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A356356 Triangle of number of rectangles in the interior of the rectangle with vertices (k,0), (0,k), (n,n+k) and (n+k,n), read by rows.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Julia

Formula

A372917 a(n) is the number of distinct rectangles with area n whose vertices lie on points of a unit square grid.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Formula

A338996 Numbers of squares and rectangles of all sizes in 3n(n+1)/2-ominoes in form of three-quarters of Aztec diamonds.