cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A322699 Square array A(n,k), n >= 0, k >= 0, read by antidiagonals, where A(n,k) is 1/2 * (-1 + Sum_{j=0..k} binomial(2*k,2*j)*(n+1)^(k-j)*n^j).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 8, 2, 0, 0, 49, 24, 3, 0, 0, 288, 242, 48, 4, 0, 0, 1681, 2400, 675, 80, 5, 0, 0, 9800, 23762, 9408, 1444, 120, 6, 0, 0, 57121, 235224, 131043, 25920, 2645, 168, 7, 0, 0, 332928, 2328482, 1825200, 465124, 58080, 4374, 224, 8, 0
Offset: 0

Views

Author

Seiichi Manyama, Dec 23 2018

Keywords

Examples

			Square array begins:
   0, 0,   0,    0,      0,       0,        0, ...
   0, 1,   8,   49,    288,    1681,     9800, ...
   0, 2,  24,  242,   2400,   23762,   235224, ...
   0, 3,  48,  675,   9408,  131043,  1825200, ...
   0, 4,  80, 1444,  25920,  465124,  8346320, ...
   0, 5, 120, 2645,  58080, 1275125, 27994680, ...
   0, 6, 168, 4374, 113568, 2948406, 76545000, ...

Links

Crossrefs

Columns 0-5 give A000004, A001477, A033996, A322675, A322677, A322745.
Rows 0-9 give A000004, A001108, A132596, A007654(n+1), A132584, A322707, A322708, A322709, A132592, A132593.
Main diagonal gives A322746.
Cf. A173175 (A(n,2*n)), A322790.

Programs

  • Mathematica
    Unprotect[Power]; 0^0 := 1; Protect[Power]; Table[(-1 + Sum[Binomial[2 k, 2 j] (# + 1)^(k - j)*#^j, {j, 0, k}])/2 &[n - k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Jan 01 2019 *)
nmax = 9; row[n_] := LinearRecurrence[{4n+3, -4n-3, 1}, {0, n, 4n(n+1)}, nmax+1]; T = Array[row, nmax+1, 0]; A[n_, k_] := T[[n+1, k+1]];
Table[A[n-k, k], {n, 0, nmax}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Jan 06 2019 *)
  • Ruby
    def ncr(n, r)
  return 1 if r == 0
  (n - r + 1..n).inject(:*) / (1..r).inject(:*)
end
def A(k, n)
  (0..n).map{|i| (0..k).inject(-1){|s, j| s + ncr(2 * k, 2 * j) * (i + 1) ** (k - j) * i ** j} / 2}
end
def A322699(n)
  a = []
  (0..n).each{|i| a << A(i, n - i)}
  ary = []
  (0..n).each{|i|
    (0..i).each{|j|
      ary << a[i - j][j]
    }
  }
  ary
end
p A322699(10)

Formula

sqrt(A(n,k)+1) + sqrt(A(n,k)) = (sqrt(n+1) + sqrt(n))^k.
sqrt(A(n,k)+1) - sqrt(A(n,k)) = (sqrt(n+1) - sqrt(n))^k.
A(n,0) = 0, A(n,1) = n and A(n,k) = (4*n+2) * A(n,k-1) - A(n,k-2) + 2*n for k > 1.
A(n,k) = (T_{k}(2*n+1) - 1)/2 where T_{k}(x) is a Chebyshev polynomial of the first kind.
T_1(x) = x. So A(n,1) = (2*n+1-1)/2 = n.

A221076 Continued fraction expansion of product_{n>=0} (1-sqrt(5)*[sqrt(5)-2]^{4n+3})/(1-sqrt(5)*[sqrt(5)-2]^{4n+1}).

Original entry on oeis.org

2, 16, 1, 32, 1, 320, 1, 608, 1, 5776, 1, 10944, 1, 103680, 1, 196416, 1, 1860496, 1, 3524576, 1, 33385280, 1, 63245984, 1, 599074576, 1, 1134903168, 1, 10749957120, 1, 20365011072, 1, 192900153616, 1, 365435296160, 1
Offset: 0

Views

Author

Peter Bala, Jan 06 2013

Keywords

Comments

Simple continued fraction expansion of product {n >= 0} {1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+3)}/{1 - sqrt(m)*[sqrt(m) - sqrt(m-1)]^(4*n+1)} at m = 5. For other cases see A221073 (m = 2), A221074 (m = 3) and A221075 (m = 4).
If we denote the present sequence by [2; 16, 1, c(3), 1, c(4), 1, ...] then for k >= 1 the sequence [1; c(2*k+1), 1, c(2*(2*k+1)), 1, c(3*(2*k+1)), 1, ...] gives the simple continued fraction expansion of product {n >= 0} [1-sqrt(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+3)]/[1 - sqrt(5)*{(sqrt(5)-2)^(2*k+1)}^(4*n+1)]. An example is given below.

Examples

			Product {n >= 0} {1 - sqrt(5)*(sqrt(5) - 2)^(4*n+3)}/{1 - sqrt(5)*(sqrt(5) - 2)^(4*n+1)} = 2.05892 54859 32105 82744 ...
= 2 + 1/(16 + 1/(1 + 1/(32 + 1/(1 + 1/(320 + 1/(1 + 1/(608 + ...))))))).
Since (sqrt(5) - 2)^3 = 17*sqrt(5) - 38 we have the following simple continued fraction expansion:
product {n >= 0} {1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+3)}/{1 - sqrt(5)*(17*sqrt(5) - 38)^(4*n+1)} = 1.03030 31892 29728 52318 ... = 1 + 1/(32 + 1/(1 + 1/(5776 + 1/(1 + 1/(196416 + 1/(1 + 1/(33385280 + ...))))))).

Links

Crossrefs

Cf. A049664, A049863, A053606, A132584, A174500, A221073 (m = 2), A221074 (m = 3), A221075 (m = 4).

Programs

  • Mathematica
    LinearRecurrence[{0,1,0,18,0,-18,0,-1,0,1},{2,16,1,32,1,320,1,608,1,5776,1},40] (* or *) Join[{2},Riffle[LinearRecurrence[{1,18,-18,-1,1},{16,32,320,608,5776},20],1]] (* Harvey P. Dale, Jun 05 2023 *)

Formula

a(2*n) = 1 for n >= 1. For n >= 1 we have:
a(4*n - 3) = (sqrt(5) + 2)^(2*n) + (sqrt(5) - 2)^(2*n) - 2;
a(4*n - 1) = 1/sqrt(5)*{(sqrt(5) + 2)^(2*n + 1) + (sqrt(5) - 2)^(2*n + 1)} - 2.
a(4*n - 3) = 16*A049863(n) = 4*A132584(n);
a(4*n - 1) = 32*A049664(n) = 4*A053606(n).
O.g.f.: 2 + x^2/(1 - x^2) + 16*x*(1 + x^2)^2/(1 - 19*x^4 + 19*x^8 - x^12) = 2 + 16*x + x^2 + 32*x^3 + x^4 + 320*x^5 + ....
O.g.f.: (x^10-2*x^8-18*x^6+36*x^4-16*x^3+x^2-16*x-2) / ((x-1)*(x+1)*(x^4-4*x^2-1)*(x^4+4*x^2-1)). - Colin Barker, Jan 10 2014

A292443 a(n) = (5/32)*A000045(6*n)^2.

Original entry on oeis.org

0, 10, 3240, 1043290, 335936160, 108170400250, 34830532944360, 11215323437683690, 3611299316401203840, 1162827164557749952810, 374426735688279083601000, 120564246064461307169569210, 38821312806020852629517684640, 12500342159292650085397524884890
Offset: 0

Views

Author

Felix Fröhlich, Sep 16 2017

Keywords

Comments

Every term is a triangular number. [Problem B-967 in Euler and Sadek, 2003; solution in Euler and Sadek, 2004]

Links

Crossrefs

Cf. A000045, A132584, A298101.
Subsequence of A000217.

Programs

  • GAP
    List([0..20], n-> 5*Fibonacci(6*n)^2/32); # G. C. Greubel, Feb 03 2019
  • Magma
    [5*Fibonacci(6*n)^2/32: n in [0..20]]; // G. C. Greubel, Feb 03 2019
  • Mathematica
    Table[(5/32) Fibonacci[6 n]^2, {n, 0, 13}] (* Michael De Vlieger, Sep 16 2017 *)
LinearRecurrence[{323,-323,1},{0,10,3240},20] (* Harvey P. Dale, Aug 31 2024 *)
  • PARI
    a(n) = (5/32)*fibonacci(6*n)^2
  • Sage
    [5*fibonacci(6*n)^2/32 for n in (0..20)] # G. C. Greubel, Feb 03 2019

Formula

From Colin Barker, Sep 16 2017: (Start)
G.f.: 10*x*(1 + x) / ((1 - x)*(1 - 322*x + x^2)).
a(n) = ((161+72*sqrt(5))^(-n)*(-1+(161+72*sqrt(5))^n)^2) / 32.
a(n) = 323*a(n-1) - 323*a(n-2) + a(n-3) for n > 2.
(End)
a(n) = A000217(A132584(n)). - Amiram Eldar, Jan 11 2022
a(n) = 10*A298101(n). - Pontus von Brömssen, Jan 06 2025

A174906 a(n) is the index of the first triangular number T_i exceeding T_n such that the product of T_i*T_n is a perfect square.

Original entry on oeis.org

24, 48, 80, 120, 168, 224, 49, 360, 440, 528, 624, 728, 840, 960, 1088, 1224, 1368, 1520, 1680, 1848, 2024, 2208, 242, 2600, 2808, 3024, 3248, 3480, 3720, 3968, 4224, 4488, 4760, 5040, 5328, 5624, 5928, 6240, 6560, 6888, 7224, 7568, 7920, 8280, 8648
Offset: 2

Views

Author

Robert G. Wilson v, Apr 01 2010

Keywords

Comments

"You can find an infinite number of [different] triangular numbers such that when multipled together form a square number. For example, for every triangular number, T_n, there are an infinite number of other triangular numbers, T_m, such that T_n*T_m is a square. For example, T_2 * T_24 = 30^2."

References

  • Clifford A. Pickover, The Loom of God, Tapestries of Mathematics and Mysticism, Sterling, NY, 2009, page 33.

Crossrefs

Cf. A132596, A007654, A132584.

Programs

  • Mathematica
    tri[n_] := n (n + 1)/2; f[n_] := Block[{k = n + 1, t = tri@n}, While[ !IntegerQ@ Sqrt[ t*tri@k], k++ ]; k]; Table[ f@n, {n, 2, 46}]
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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