A024215 -id:A024215 - cp's OEIS frontend

A213825 Rectangular array: (row n) = b**c, where b(h) = 3h-1, c(h) = 3n-5+3*h, n>=1, h>=1, and ** = convolution.

2, 13, 8, 42, 34, 14, 98, 87, 55, 20, 190, 176, 132, 76, 26, 327, 310, 254, 177, 97, 32, 518, 498, 430, 332, 222, 118, 38, 772, 749, 669, 550, 410, 267, 139, 44, 1098, 1072, 980, 840, 670, 488, 312, 160, 50, 1505, 1476, 1372

Offset: 1

Clark Kimberling, Jul 04 2012

Principal diagonal: A213826
Antidiagonal sums: A213827
Row 1, (2,5,8,13,...)**(1,4,7,10,13,...): (3*k^2 + k)/2
Row 2, (2,5,8,13,...)**(4,7,10,13,...): (3*k^3 + 9*k^2 - 2*k)/2
Row 3, (2,5,8,13,...)**(7,10,13,16,...): (3*k^3 + 18*k^2 - 5*k)/2
For a guide to related arrays, see A212500.

			Northwest corner (the array is read by falling antidiagonals):
2....13....42....98....190
8....34....87....176...310
14...55....132...254...430
20...76....177...332...550
26...97....222...410...670
32...118...267...488...790

Clark Kimberling, Antidiagonals n = 1..80, flattened

Cf. A212500

b[n_]:=3n-1;c[n_]:=3n-2;
t[n_,k_]:=Sum[b[k-i]c[n+i],{i,0,k-1}]
TableForm[Table[t[n,k],{n,1,10},{k,1,10}]]
Flatten[Table[t[n-k+1,k],{n,12},{k,n,1,-1}]]
r[n_]:=Table[t[n,k],{k,1,60}] (* A213825 *)
d=Table[t[n,n],{n,1,40}] (* A213826 *)
d/2 (* A024215 *)
s[n_]:=Sum[t[i,n+1-i],{i,1,n}]
s1=Table[s[n],{n,1,50}] (* A213827 *)

T(n,k) = 4*T(n,k-1)-6*T(n,k-2)+4*T(n,k-3)-T(n,k-4).

G.f. for row n: f(x)/g(x), where f(x) = x*((3*n-1) + (3*n+2)*x - (6*n-8)*x^2) and g(x) = (1-x)^4.

A213826 Principal diagonal of the convolution array A213825.

Original entry on oeis.org

2, 34, 132, 332, 670, 1182, 1904, 2872, 4122, 5690, 7612, 9924, 12662, 15862, 19560, 23792, 28594, 34002, 40052, 46780, 54222, 62414, 71392, 81192, 91850, 103402, 115884, 129332, 143782, 159270, 175832, 193504, 212322, 232322, 253540, 276012, 299774

Offset: 1

Clark Kimberling, Jul 04 2012

Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A213825.

[6*n^3-3*n^2-n: n in [1..40]]; // Vincenzo Librandi, Nov 23 2018

(See A213825.)
CoefficientList[Series[2 (1 + 13 x + 4 x^2) / (1 - x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Nov 23 2018 *)

a(n) = -n - 3*n^2 + 6*n*3.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4).
G.f.: f(x)/g(x), where f(x) = 2*x*(1 + 13*x + 4*x^2) and g(x) = (1-x)^4.
a(n) = 2*A024215(n).
E.g.f.: (2 + 32*x + 33*x^2 + 6*x^3)*exp(x). - Franck Maminirina Ramaharo, Nov 23 2018

A202676 Symmetric matrix based on (1,4,7,10,13,...), by antidiagonals.

Original entry on oeis.org

1, 4, 4, 7, 17, 7, 10, 32, 32, 10, 13, 47, 66, 47, 13, 16, 62, 102, 102, 62, 16, 19, 77, 138, 166, 138, 77, 19, 22, 92, 174, 232, 232, 174, 92, 22, 25, 107, 210, 298, 335, 298, 210, 107, 25, 28, 122, 246, 364, 440, 440, 364, 246, 122, 28, 31, 137, 282, 430

Offset: 1

Clark Kimberling, Dec 22 2011

Let s=(1,4,7,10,13,...) and let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s.  Let T' be the transpose of T.  Then A202676 represents the matrix product M=T'*T.  M is the self-fusion matrix of s, as defined at A193722.  See A202677 for characteristic polynomials of principal submatrices of M.
...
row 1 (1,4,7,10,...) A016777
diagonal (1,17,66,166,...) A024215
...
Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]: (1,25,144,484,..), the squares of the pentagonal numbers (A000326).

			Northwest corner:
1....4....7...10...13...16
4...17...32...47...62...77
7...32...66..102..138..174
10..47..102..166..232..298
13..62..138..232..335..440

Cf. A202677.

U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[3 k - 2, {k, 1, 15}]];
L = Transpose[U]; M = L.U; TableForm[M]
m[i_, j_] := M[[i]][[j]];
Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]

A350886 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the sorted values for X.

Original entry on oeis.org

54, 70, 1618, 2344, 2541, 27597, 48486, 73795, 184162, 320739, 648009, 766669, 990983, 1452962, 3816551, 4456264, 6287116, 23251921, 37396339, 43540374, 51136014, 53005618, 63668661, 147115419, 205943541, 236317895, 253970684, 275914803, 386480829, 629467300

Offset: 1

Thomas Scheuerle, Feb 25 2022

nonn

A generalization of the cannonball problem for pyramids with a slope of 1/A350888(n). In the cannonball problem, a square pyramid of stacked balls shall contain a square number of balls in total. Each layer of such a pyramid consists of a square number of balls, in the classic case the top layer has one ball, the layers below contain adjacent square numbers of balls. For this sequence we ignore the fact that if adjacent layers do not alternate between odd and even squares the stacking will become difficult at least for sphere-like objects. We start in the top layer always with one ball, but will skip a constant count of square numbers between each layer. This results in pyramids which slope <= 1.

a(n) may be interpreted as the length of a Pythagorean vector with gcd = 1 (over all coordinates) and no duplicate coordinate values. Such vectors may have applications in the theory of lattices.

			a(1) = 54 and A350887(1) = 4, A350888(1) = 14:
   54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 70 and  A350887(2) = 24, A350888(2) = 1:
   70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by A350887.
Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, An Equidistribution Result for Differences Associated with Square Pyramidal Numbers, arXiv:2412.10097 [math.NT], 2024. See p. 1.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares.

Cf. A350887 (number of layers), A350888 (denominator of slope).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.

sqtest(n,c)={q=1; for(t=2, c, t+=n; q+=(t*t); if(issquare(q), break)); q}
z=500000; a=[];for(n=0,z,r=sqtest(n,z); if(issquare(r), a=concat(a, sqrtint(r)))); a=vecsort(a) \\ Last valid value for z=500000 is 990983.

a(n)^2 = A350888(n)^2*binomial(2*A350887(n), 3)/4 + 2*A350888(n)*binomial(A350887(n), 2) + A350887(n).
a(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = A350887(n) and b = A350888(n). Expanded to see factors more clearly.
a(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.
(12*a(n)^2) mod A350887(n) = 0.
((12*a(n)^2/A350887(n)) - 12) mod A350888(n) = 0.
Choose n such that A350887(n) = 4 and a(n) = 54 and A350888(n) = 14, then we may find further solutions recursively for all A350887(m) = 4 with
  x = -A350888(n) = -14; y = -a(n) = -54 and also x = A350888(n) = 14; y = a(n) = 54. Recursive solutions:
  x_(n+1) = 15*x_n + 4*y_n + 6
  y_(n+1) = 56*x_n + 15*y_n + 24 and also:
  x_(n+1) = 15*x_n - 4*y_n + 6
  y_(n+1) = -56*x_n + 15*y_n - 24.
Choose n such that A350887(n) = 9 and a(n) = 27597 and A350888(n) = 1932, then we may find further solutions recursively for all A350887(m) = 9 with x = -A350888(n) = -1932; y = -a(n) = -27597 and also x = A350888(n) = 1932; y = a(n) = 27597. Recursive solutions:
  x_(n+1) = 4999*x_n + 350*y_n + 882
  y_(n+1) = 71400*x_n + 4999*y_n + 12600 and also:
  x_(n+1) = 4999*x_n - 350*y_n + 6
  y_(n+1) = -71400*x_n + 4999*y_n - 12600.
Further recursive solution formulas for other values of A350887(n) will be provided in a link as for some values the coefficients become very large sometimes with several hundred digits.
a(n) != a(m) if n != m.

A350887 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Y sorted by X.

Original entry on oeis.org

4, 24, 4, 64, 49, 9, 4, 50, 484, 3249, 81, 361, 49, 4, 289, 64, 16, 3938, 5041, 4, 36, 568, 441, 121, 4761, 33, 1936, 9, 49, 25872, 4, 64, 8257, 24, 361, 12024

Offset: 1

Thomas Scheuerle, Feb 25 2022

This sequence contains only numbers which appear in A186699, too. A number which is in A186699, does not appear in this sequence if it is of the form: 2^m*p where p is a prime of the form 12*k+1 or 12*k-1. There are also some perfect squares like 25 excluded, as these would only provide valid solutions in the case of A350888(n) = 0, which is not part of the sequence definition.

			a(1) = 4 and A350886(1) = 54, A350888(1) = 14: 54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 24 and A350886(2) = 70, A350888(2) = 1: 70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by a(n).
Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, An Equidistribution Result for Differences Associated with Square Pyramidal Numbers, arXiv:2412.10097 [math.NT], 2024. See p. 1.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares

Cf. A350886 (squareroot of pyramid), A350888 (denominator of slope).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.
Cf. A186699.

sqtest(n, c)={q=1; r=1; for(t=2, c, t+=n; r+=1; q+=(t*t); if(issquare(q), break)); [q, r]}
z=500000; b=[;]; for(n=0,z,r=sqtest(n,z); if(issquare(r[1]), b=concat(b,[sqrtint(r[1]); r[2]]))); b=vecsort(b,1); vector(#b, k, b[2, k]) \\ Last valid value for z=500000 is 49.

A350886(n)^2 = A350888(n)^2*binomial(2*a(n), 3)/4 + 2*A350888(n)*binomial(a(n), 2) + a(n).
A350886(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = a(n) and b = A350888(n). Expanded to see factors more clearly.
A350886(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.

A350888 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Z sorted by X.

Original entry on oeis.org

14, 1, 432, 8, 13, 1932, 12958, 367, 30, 3, 1554, 194, 5082, 388320, 1349, 15254, 178542, 163, 181, 11636654, 418782, 6791, 11928, 192638, 1086, 2209447, 5166, 19317900, 1981979, 262, 348711312, 4799102, 7379, 60240793

Offset: 1

Thomas Scheuerle, Feb 25 2022

This is a generalization of the cannonball problem for pyramids with a slope of 1/a(n). In the cannonball problem, a square pyramid of stacked balls shall contain a square number of balls in total. Each layer of such a pyramid consists of a square number of balls, in the classic case the top layer has one ball, the layers below contain adjacent square numbers of balls. For this sequence we ignore the fact that if adjacent layers do not alternate between odd and even squares the stacking will become difficult at least for sphere-like objects. We start in the top layer always with one ball, but will skip a constant count of square numbers between each layer. This results in pyramids which slope <= 1. A350887(n) is the number of layers needed to stack such a pyramid. As this sequence is only a list of solutions, n has no known relation to its definition.

For each slope 1/a(n) there exists exactly one or no such pyramid with a square number of balls.

			a(1) = 14 and A350886(1) = 54, A350887(1) = 4:
  54^2 = 1^2 + 15^2 + 29^2 + 43^2.
a(2) = 1 and A350886(2) = 70, A350887(2) = 24:
  70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.

Thomas Scheuerle, Some solutions to this problem sorted by A350887.
Anji Dong, Katerina Saettone, Kendra Song, and Alexandru Zaharescu, An Equidistribution Result for Differences Associated with Square Pyramidal Numbers, arXiv:2412.10097 [math.NT], 2024. See p. 1.
Thomas Scheuerle, Recursive solution formulas.
Index entries for sequences related to sums of squares.

Cf. A350886 (square root of pyramid), A350887 (number of layers).
Cf. A000330, A000447, A024215, A024381 (square pyramidal numbers for slope 1, 1/2, 1/3, 1/4).
Cf. A076215, A001032, A134419, A106521. Some related problems.
Cf. A186699.

sqtest(n, c)={q=1; for(t=2, c, t+=n; q+=(t*t); if(issquare(q), break)); q}
z=500000; b=[; ]; for(n=0, z, r=sqtest(n, z); if(issquare(r), b=concat(b, [sqrtint(r); n+1]))); b=vecsort(b, 1); vector(#b, k, b[2,k]) \\ Last valid value for z=500000 is 5082.

A350886(n)^2 = a(n)^2*binomial(2*A350887(n), 3)/4 + 2*a(n)*binomial(A350887(n), 2) + A350887(n).
A350886(n)^2 = c*((b*(4*b*c^2-(6*c-2)*b + 12*(c-1)) + 12)/12), with c = A350887(n) and b = a(n). Expanded to see factors more clearly.
A350886(n)^2 = c*b^2*( ((1/b) + (c-1)/2)^2 + (c^2-1)/12 ). Shorter form of above.
a(n) != a(m) if n != m.
Let s(n) be the sequence of numbers such that A350887(s(n)) = 4 and s(n) is sorted in ascending order, then a(s(n)) has the ordinary generating function (2*(-7 + z))/(-1 + 31*z - 31*z^2 + z^3).
Let s(n) be the sequence of numbers such that A350887(s(n)) = 49 and s(n) is sorted in ascending order, then a(s(n)) has the ordinary generating function (-13 + z)/(-1 + 391*z - 391*z^2 + z^3).

cp's OEIS Frontend

A213825 Rectangular array: (row n) = b**c, where b(h) = 3*h-1, c(h) = 3*n-5+3*h, n>=1, h>=1, and ** = convolution.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A213826 Principal diagonal of the convolution array A213825.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A202676 Symmetric matrix based on (1,4,7,10,13,...), by antidiagonals.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A350886 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the sorted values for X.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A350887 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Y sorted by X.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A350888 Let X,Y,Z be positive integer solutions to X^2 = Sum_{j=0..Y-1} (1+Z*j)^2, where solutions for Y or Z < 1 are excluded. This sequence lists the values for Z sorted by X.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

A213825 Rectangular array: (row n) = b**c, where b(h) = 3h-1, c(h) = 3n-5+3*h, n>=1, h>=1, and ** = convolution.