A082044 -id:A082044 - cp's OEIS frontend

A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

1, 8, 75, 400, 1445, 4056, 9583, 20000, 38025, 67240, 112211, 178608, 273325, 404600, 582135, 817216, 1122833, 1513800, 2006875, 2620880, 3376821, 4298008, 5410175, 6741600, 8323225, 10188776, 12374883, 14921200, 17870525, 21268920, 25165831, 29614208

Offset: 0

Daniel Poveda Parrilla, Jun 11 2017

If we add a(n) and A288487(n) graphically we obtain a bigger cuboid which is a square of cubes (a cuboid with dimensions n^2 * n^2 * n).

a(10^n) is a palindrome in base 10.

Daniel Poveda Parrilla, Table of n, a(n) for n = 0..10000
Daniel Poveda Parrilla, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A000584, A002061, A002522, A014206, A059100, A082044, A288486.

Table[(1 + n)*(1 + n^2)^2, {n, 0, 28}] (* or *) CoefficientList[Series[(1 + 2 x + 42 x^2 + 50 x^3 + 25 x^4)/(1 - x)^6, {x, 0, 28}], x] (* or *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 75, 400, 1445, 4056}, 29]

Vec((1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6 + O(x^28))

G.f.: (1 + 2*x + 42*x^2 + 50*x^3 + 25*x^4)/(1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = (n + 1)*(n^2 + 1)^2 = (n + 1)*(A002522(n))^2 = (n + 1)*A082044(n).
a(n) = n^5 + A002061(A002061(n+1)).
a(n) = A000584(n+1) - A288486(n)
a(n) = (n + 1)*A059100(n-1) + 4*(n^2 -1)*A014206(n-1) for n > 1.

A234566 1/a(n) is the area of the smallest triangle delimited by 3 lines each passing through at least 2 points of an n X n unitary spaced grid.

Original entry on oeis.org

4, 30, 770, 5148, 30566, 89900, 219960, 614460, 1146596, 2624076, 4299916, 8432732, 11016390, 22391148, 28183214

Offset: 2

Giovanni Resta, Dec 28 2013

Only non-degenerate triangles are considered.
Conjecture: sequence is well-defined, i.e., a(n) is an integer for every n > 1.
For n > 1, n odd, a(n) >= 4(n-2)^2(n^2-3n+1)(n^2-3n+3), with equality for  n = 5,7,...,15. The bound is obtained considering the 3 lines passing by the 3 pairs of points {(0,1), (n-1, n-1)}, {(n-2, 0), (1, n-2)}, and {((n-1)/2, n-1), ((n-3)/2, 0)}.
We may consider the similar problem of finding the largest triangle. Here, the areas for n>=2 are 1/2, 9/2, 25, 100, 289, 676, 1369,... so it appears that for n >= 4 the maximal area is ((n-2)^2+1)^2, (cf. A082044) obtained via the lines passing through the points {(0,2), (1,n-1)}, {(n-2,0), (n-1,n-2)}, and {(0,0), (1,n-1)}.

			For n=2, consider the 2 X 2 grid formed by the points with coordinates (0,0), (0,1), (1,0) and (1,1). The two diagonals and the line passing through (0,0) and (1,0) form a triangle whose area is 1/4 and since no smaller triangle can be formed in this way, a(2) = 4.

Giovanni Resta, Illustration of the first 9 terms

Cf. A018808, A082044.

For n>1 odd, a(n) >= 4(n-2)^2 (n^2-3n+1)(n^2-3n+3).

A335717 a(n) is the smallest dividend m of the Euclidean division m = d*n + r such that m/d = r/n.

Original entry on oeis.org

25, 100, 162, 676, 400, 2500, 1156, 2352, 1458, 14884, 2601, 28900, 5202, 6084, 8712, 84100, 9408, 131044, 10816, 22500, 22275, 280900, 19602, 79380, 36517, 60516, 40000, 708964, 31827, 925444, 67600, 46128, 80937, 62500, 55112, 1876900, 112651, 88752, 83232

Offset: 2

Stéphane Rézel, Jun 18 2020

Such a Euclidean division exists, for all n>1, with r = n^2 + 1 because then m = r^2 and d = n*r. This solution leads to the largest possible dividend, given (for n>1) by A082044(n). This is also the only solution when n is prime, then a(n) = A082044(n).

Solutions with r = n^2 + k are only possible for most of the integers k < n such that any prime factor of k is also a prime factor of n. The largest such integers k are given by A079277(n). In the present sequence, k is not always this largest integer: a(18) is defined by k = 12 while it cannot be defined by k = 16 = A079277(18).

			a(2) = 25 because 25 = 10*2 + 5 and 25/10 = 5/2. Furthermore, there is no Euclidean division with a dividend less than 25 such that m/d = r/2.
a(4) = 162 because 162 = 36*4 + 18 and 162/36 = 18/4. The other Euclidean division with the same property has a larger dividend : 289 = 68*4 + 17 and 289/68 = 17/4.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A079277, A082044.

f:= proc(n) local m, r,k,d;
  for k from n-1 to 1 by -1 do
    r:= n^2+k;
    m:= r^2/k;
    if not m::integer then next fi;
    d:= n*r/k;
    if d::integer then return m fi;
  od
end proc:
map(f, [$2..50]); # Robert Israel, Sep 16 2020

f[n_] := Module[{m, r, k, d}, For[k = n-1, k >= 1, k--, r = n^2+k; m = r^2/k; If[!IntegerQ[m], Continue]; d = n*r/k; If[IntegerQ[d], Return[m]]]];
Table[f[n], {n, 2, 50}] (* Jean-François Alcover, Jul 31 2024, after Robert Israel *)

a(n) = k=n; until(k==1, k--; r=k + n^2; d=n*r/k; m=r^2/k; if(floor(d)==d && floor(m)==m, return(m); break));

A267707 a(n) = A000217(A000217(n)+1).

Original entry on oeis.org

1, 3, 10, 28, 66, 136, 253, 435, 703, 1081, 1596, 2278, 3160, 4278, 5671, 7381, 9453, 11935, 14878, 18336, 22366, 27028, 32385, 38503, 45451, 53301, 62128, 72010, 83028, 95266, 108811, 123753, 140185, 158203, 177906, 199396, 222778, 248160, 275653, 305371

Offset: 0

Waldemar Puszkarz, Jan 19 2016

It is the sequence of triangular numbers (A000217) with progressive gaps that grow as 0,1,2,3, ... (consecutive numbers), by which I mean that the 0,1,2,3, ... consecutive triangular numbers are removed from A000217 to form this sequence. For instance, (1), 6, a triangular number, is missing between 3 and 10, which is the gap with 1 triangular number removed, (2), 15 and 21 (two consecutive triangular numbers) are missing between 10 and 28, which is the gap with 2 triangular numbers removed, and so on.
The differences between the consecutive terms of this sequence can be expressed through the sum of cubes of two numbers separated by 2 as (n^3+(n+2)^3)/4, which is the same as A229183, except for the first term in there.
The same pattern when applied to squares, A000290(A000290(n)+1), gives A082044(n). Triangular numbers are also linked in a similar manner to A027927(n) = A000217(A000217(n)+2)/3.

			For n=0, a(0)=1*2/2=1. For n=2, a(2)=4*5/2=10.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000217 (triangular numbers), A229183 (consecutive terms differences), A082044 (related sequence for squares), A027927 (related sequence for triangular numbers).

I:=[1,3,10,28,66]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 22 2016

S[n_] :=n*(n+1)/2; Table[S[S[n]+1], {n, 0, 50}]
Table[(n*(n+1)/2+1)(n*(n+1)/2+2)/2, {n, 0, 50}]
Table[(n^4+2*n^3+7*n^2+6*n+8)/8, {n, 0, 50}]
CoefficientList[Series[(1 - 2 x + 5 x^2 - 2 x^3 + x^4) / (1 - x)^5, {x, 0, 33}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 3, 10, 28, 66}, 50] (* Vincenzo Librandi, Jan 22 2016 *)

for(n=0,50,print1((n^4+2*n^3+7*n^2+6*n+8)/8 ", "))

a(n) = A000217(A000217(n)+1) = (n*(n+1)/2+1)(n*(n+1)/2+2)/2.
a(n) = (n^4+2n^3+7n^2+6n+8)/8  = (n^2+n+2)(n^2+n+4)/8.
G.f.: (1-2*x+5*x^2-2*x^3+x^4)/(1-x)^5. - Vincenzo Librandi, Jan 22 2016
a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Vincenzo Librandi, Jan 22 2016

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A288487 Cuboids that fit in square rings from A288486 obtaining a fifth power.

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A234566 1/a(n) is the area of the smallest triangle delimited by 3 lines each passing through at least 2 points of an n X n unitary spaced grid.

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A335717 a(n) is the smallest dividend m of the Euclidean division m = d*n + r such that m/d = r/n.

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A267707 a(n) = A000217(A000217(n)+1).

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