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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.
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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
COMMENTS
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.
FORMULA
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.
EXAMPLE
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
MATHEMATICA
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 *) s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
s1=Table[s[n], {n, 1, 50}] (* A213827 *)
Principal diagonal of the convolution array A213825.
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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
FORMULA
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.
MATHEMATICA
CoefficientList[Series[2 (1 + 13 x + 4 x^2) / (1 - x)^4, {x, 0, 30}], x] (* Vincenzo Librandi, Nov 23 2018 *)
Symmetric matrix based on (1,4,7,10,13,...), by antidiagonals.
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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
COMMENTS
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. ...
...
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).
EXAMPLE
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
MATHEMATICA
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}]]
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.
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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
COMMENTS
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.
FORMULA
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.
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.
EXAMPLE
54^2 = 1^2 + 15^2 + 29^2 + 43^2.
70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.
PROG
(PARI)
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.
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.
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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
COMMENTS
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) 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. 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.
FORMULA
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.
EXAMPLE
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.
PROG
(PARI)
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.
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.
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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
COMMENTS
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.
FORMULA
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).
EXAMPLE
54^2 = 1^2 + 15^2 + 29^2 + 43^2.
70^2 = 1^2 + 2^2 + 3^2 + ... + 23^2 + 24^2. This is the classic solution for the cannonball problem.
PROG
(PARI)
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.
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