A028884 -id:A028884 - cp's OEIS frontend

A164604 a(n) = ((1+4sqrt(2))(3+2sqrt(2))^n + (1-4sqrt(2))(3-2sqrt(2))^n)/2.

1, 19, 113, 659, 3841, 22387, 130481, 760499, 4432513, 25834579, 150574961, 877615187, 5115116161, 29813081779, 173763374513, 1012767165299, 5902839617281, 34404270538387, 200522783613041, 1168732431139859

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com), Aug 17 2009

Binomial transform of A164603. Third binomial transform of A164702. Inverse binomial transform of A164605.
From Klaus Purath, Mar 14 2024: (Start)
For any two consecutive terms (a(n), a(n+1)) = (x,y): x^2 - 6xy + y^2 = 248 = A028884(13). In general, the following applies to all recursive sequences (t) with constant coefficients (6,-1) and t(0) = 1 and two consecutive terms (x,y): x^2 - 6xy + y^2 = A028884(t(1)-6). This includes and interprets the Feb 04 2014 comment on A001541 by Colin Barker as well as the Mar 17 2021 comment on A054489 by John O. Oladokun.
By analogy to this, for three consecutive terms (x,y,z) of any recursive sequence (t) of form (6,-1) with t(0) = 1: y^2 - xz = A028884(t(1)-6). (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..155 from Vincenzo Librandi)
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A164603, A164702, A164605.

Z:=PolynomialRing(Integers()); N:=NumberField(x^2-2); S:=[ ((1+4*r)*(3+2*r)^n+(1-4*r)*(3-2*r)^n)/2: n in [0..19] ]; [ Integers()!S[j]: j in [1..#S] ]; // Klaus Brockhaus, Aug 23 2009

LinearRecurrence[{6,-1}, {1,19}, 50] (* G. C. Greubel, Aug 11 2017 *)

Vec((1+13*x)/(1-6*x+x^2)+O(x^99)) \\ Charles R Greathouse IV, Jun 12 2011

a(n) = 6*a(n-1) - a(n-2) for n > 1; a(0) = 1, a(1) = 19.
G.f.: (1+13*x)/(1-6*x+x^2).
E.g.f.: exp(3*x)*( cosh(2*sqrt(2)*x) + 4*sqrt(2)*sinh(2*sqrt(2)*x) ). - G. C. Greubel, Aug 11 2017

Edited and extended beyond a(5) by Klaus Brockhaus, Aug 23 2009

A190576 a(n) = n^2 + 5*n - 5.

Original entry on oeis.org

1, 9, 19, 31, 45, 61, 79, 99, 121, 145, 171, 199, 229, 261, 295, 331, 369, 409, 451, 495, 541, 589, 639, 691, 745, 801, 859, 919, 981, 1045, 1111, 1179, 1249, 1321, 1395, 1471, 1549, 1629, 1711, 1795, 1881, 1969, 2059, 2151, 2245, 2341

Offset: 1

Vladimir Joseph Stephan Orlovsky, May 12 2011

Also a(n) = n^2 + 9*n + 9 if the offset is changed to -1. - R. J. Mathar, May 18 2011

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. sequences of the form n^2 + k*n - k : A000290 (k=0), A028387 (k=1), A028872 (k=2), A082111 (k=3), A028884 (k=4).

[n^2+5*n-5: n in [1..50]]; // Vincenzo Librandi, Sep 30 2011

k = 5; Table[n^2 + k*n - k, {n, 100}]
LinearRecurrence[{3,-3,1},{1,9,19},50] (* Harvey P. Dale, May 28 2015 *)

a(n)=n^2+5*n-5 \\ Charles R Greathouse IV, Jun 17 2017

G.f.: x*(-1 - 6*x + 5*x^2) / (x-1)^3. - R. J. Mathar, May 18 2011
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(1)=1, a(2)=9, a(3)=19. - Harvey P. Dale, May 28 2015
Sum_{n>=1} 1/a(n) = 199/495 + Pi*tan(3*sqrt(5)*Pi/2)/(3*sqrt(5)). - Amiram Eldar, Jan 18 2021

A213922 Natural numbers placed in table T(n,k) layer by layer. The order of placement: T(n,n), T(n-1,n), T(n,n-1), ... T(1,n), T(n,1). Table T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 3, 4, 8, 2, 9, 15, 6, 7, 16, 24, 13, 5, 14, 25, 35, 22, 11, 12, 23, 36, 48, 33, 20, 10, 21, 34, 49, 63, 46, 31, 18, 19, 32, 47, 64, 80, 61, 44, 29, 17, 30, 45, 62, 81, 99, 78, 59, 42, 27, 28, 43, 60, 79, 100, 120, 97, 76, 57, 40, 26, 41, 58, 77, 98, 121

Offset: 1

Boris Putievskiy, Mar 05 2013

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1). Enumeration table T(n,k) is layer by layer. The order of the list:
T(1,1)=1;
T(2,2), T(1,2), T(2,1);
...
T(n,n), T(n-1,n), T(n,n-1), ... T(1,n), T(n,1);
...

			The start of the sequence as a table:
   1,  3,  8, 15, 24, 35, ...
   4,  2,  6, 13, 22, 33, ...
   9,  7,  5, 11, 20, 31, ...
  16, 14, 12, 10, 18, 29, ...
  25, 23, 21, 19, 17, 27, ...
  36, 34, 32, 30, 28, 26, ...
...
The start of the sequence as triangular array read by rows:
   1;
   3,  4;
   8,  2,  9;
  15,  6,  7, 16;
  24, 13,  5, 14, 25;
  35, 22, 11, 12, 23, 36;
  ...

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Eric W. Weisstein, MathWorld: Pairing functions
Index entries for sequences that are permutations of the natural numbers

Cf. A060734, A060736; table T(n,k) contains: in rows A005563, A028872, A028875, A028881, A028560, A116711; in columns A000290, A008865, A028347, A028878, A028884.

f[n_, k_] := n^2 - 2*k + 2 /; n >= k; f[n_, k_] := k^2 - 2*n + 1 /; n < k; TableForm[Table[f[n, k], {n, 1, 5}, {k, 1, 10}]]; Table[f[n - k + 1, k], {n, 5}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Aug 19 2017 *)

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
if i >= j:
   result=i*i-2*j+2
else:
   result=j*j-2*i+1

As a table,
  T(n,k) = n*n - 2*k + 2, if n >= k;
  T(n,k) = k*k - 2*n + 1, if n < k.
As a linear sequence,
  a(n) = i*i - 2*j + 2, if i >= j;
  a(n) = j*j - 2*i + 1, if i < j
  where
    i = n - t*(t+1)/2,
    j = (t*t + 3*t + 4)/2 - n,
    t = floor((-1 + sqrt(8*n-7))/2).

A175019 a(0)=a(1)=1; a(n) = a(n-a(n-2)) + 2.

Original entry on oeis.org

1, 1, 3, 5, 3, 3, 7, 5, 3, 5, 7, 9, 5, 5, 7, 9, 7, 5, 11, 9, 7, 7, 11, 9, 7, 9, 11, 13, 7, 9, 11, 13, 9, 9, 11, 13, 11, 9, 15, 13, 11, 9, 15, 13, 11, 11, 15, 13, 11, 13, 15, 17, 11, 13, 15, 17, 11, 13, 15, 17, 13, 13, 15, 17, 15, 13, 19, 17, 15, 13, 19, 17, 15, 13, 19, 17, 15, 15, 19, 17

Offset: 0

José María Grau Ribas, Apr 04 2010

nonn

Conjectures from Chai Wah Wu, Jun 12 2016: (Start)
All odd numbers appear in sequence.
2n+1 appears 2n+2 times.
a((n+3)^2-8) = 2n+1 is the last time 2n+1 appears.
For n > 0, a((n-1)^2+2) = 2n+1 is the first time 2n+1 appears. (End)

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A028884.

a[0] = a[1] = 1; a[n_] := a[n] = a[n - a[n - 2]] + 2; Table[a[n], {n, 0, 200}]

A193517 T(n,k) = number of ways to place any number of 5X1 tiles of k distinguishable colors into an nX1 grid.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 3, 1, 1, 1, 1, 4, 5, 4, 1, 1, 1, 1, 5, 7, 7, 5, 1, 1, 1, 1, 6, 9, 10, 9, 6, 1, 1, 1, 1, 7, 11, 13, 13, 11, 8, 1, 1, 1, 1, 8, 13, 16, 17, 16, 17, 11, 1, 1, 1, 1, 9, 15, 19, 21, 21, 28, 27, 15, 1, 1, 1, 1, 10, 17, 22, 25, 26, 41, 49, 41, 20, 1, 1, 1

Offset: 1

R. H. Hardin, with proof and formula from Robert Israel in the Sequence Fans Mailing List, Jul 29 2011

Table starts:
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..1..1...1...1...1...1...1....1....1....1....1....1....1....1....1....1....1
..2..3...4...5...6...7...8....9...10...11...12...13...14...15...16...17...18
..3..5...7...9..11..13..15...17...19...21...23...25...27...29...31...33...35
..4..7..10..13..16..19..22...25...28...31...34...37...40...43...46...49...52
..5..9..13..17..21..25..29...33...37...41...45...49...53...57...61...65...69
..6.11..16..21..26..31..36...41...46...51...56...61...66...71...76...81...86
..8.17..28..41..56..73..92..113..136..161..188..217..248..281..316..353..392
.11.27..49..77.111.151.197..249..307..371..441..517..599..687..781..881..987
.15.41..79.129.191.265.351..449..559..681..815..961.1119.1289.1471.1665.1871
.20.59.118.197.296.415.554..713..892.1091.1310.1549.1808.2087.2386.2705.3044
.26.81.166.281.426.601.806.1041.1306.1601.1926.2281.2666.3081.3526.4001.4506

			Some solutions for n=11 k=3; colors=1, 2, 3; empty=0
..0....2....2....0....0....1....0....3....3....0....0....0....0....3....1....0
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..0....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....2....2....0....1....1....3....3....3....3....2....3....1....3....1....1
..3....1....0....2....1....0....3....3....0....3....2....3....1....0....0....1
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..3....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....1....3....2....1....1....2....3....2....0....0....3....0....3....0....2
..0....0....3....0....1....1....2....0....2....0....0....3....0....3....0....2

R. H. Hardin, Table of n, a(n) for n = 1..9999

Column 1 is A003520,
Column 2 is A143447(n-4),
Column 3 is A143455(n-4),
Row 10 is A028884.

T:= proc(n, k) option remember;
      `if`(n<0, 0,
      `if`(n<5 or k=0, 1, k*T(n-5, k) +T(n-1, k)))
    end:
seq(seq(T(n, d+1-n), n=1..d), d=1..13); # Alois P. Heinz, Jul 29 2011

T[n_, k_] := T[n, k] = If[n<0, 0, If[n < 5 || k == 0, 1, k*T[n-5, k]+T[n-1, k]]]; Table[Table[T[n, d+1-n], {n, 1, d}], {d, 1, 14}] // Flatten (* Jean-François Alcover, Mar 04 2014, after Alois P. Heinz *)

With z X 1 tiles of k colors on an n X 1 grid (with n >= z), either there is a tile (of any of the k colors) on the first spot, followed by any configuration on the remaining (n-z) X 1 grid, or the first spot is vacant, followed by any configuration on the remaining (n-1) X 1. So T(n,k) = T(n-1,k) + k*T(n-z,k), with T(n,k) = 1 for n=0,1,...,z-1. The solution is T(n,k) = sum_r r^(-n-1)/(1 + z k r^(z-1)) where the sum is over the roots of the polynomial k x^z + x - 1.
T(n,k) = sum {s=0..[n/5]} (binomial(n-4*s,s)*k^s).
For z X 1 tiles, T(n,k,z) = sum{s=0..[n/z]} (binomial(n-(z-1)*s,s)*k^s). - R. H. Hardin, Jul 31 2011

A214870 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places counterclockwise. T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 2, 3, 5, 4, 7, 10, 9, 8, 13, 17, 16, 6, 14, 21, 26, 25, 11, 12, 22, 31, 37, 36, 18, 15, 20, 32, 43, 50, 49, 27, 24, 23, 30, 44, 57, 65, 64, 38, 35, 19, 33, 42, 58, 73, 82, 81, 51, 48, 28, 29, 45, 56, 74, 91, 101, 100, 66, 63, 39, 34, 41, 59, 72, 92, 111

Offset: 1

Boris Putievskiy, Mar 11 2013

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Layer is pair of sides of square from T(1,n) to T(n,n) and from T(n,n) to T(n,1).
Enumeration table T(n,k) layer by layer. The order of the list:
  T(1,1)=1;
  T(1,2), T(2,1), T(2,2);
  . . .
  T(1,n), T(3,n), ... T(n,3), T(n,1); T(n,2), T(n,4), ... T(4,n), T(2,n);
  . . .

			The start of the sequence as table:
   1   2   5  10  17  26 ...
   3   4   9  16  25  36 ...
   7   8   6  11  18  27 ...
  13  14  12  15  24  35 ...
  21  22  20  23  19  28 ...
  31  32  30  33  29  34 ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   2,  3;
   5,  4,  7;
  10,  9,  8, 13;
  17, 16,  6, 14, 21;
  26, 25, 11, 12, 22, 31;
  ...

Boris Putievskiy, Rows n = 1..140 of triangle, flattened
Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.
Eric Weisstein's World of Mathematics, Pairing functions
Index entries for sequences that are permutations of the natural numbers

Cf. A060734, A060736, A185725, A213921, A213922; table T(n,k) contains: in rows A002522, A000290, A059100, A005563, A117950, A008865, A087475, A028872, A117951, A028347, A114949, A028875, A117619, A028878, A189833, A028881, A189834, A028884, A114948, A028560, A189836; in columns A002061, A014206, A002378, A027688, A028387, A027689, A028552, A027690, A014209, A027691, A027692, A082111, A027693, A028557, A027694, A108195, A187710, A048058, A048840.

t=int((math.sqrt(8*n-7) - 1)/ 2)
i=n-t*(t+1)/2
j=(t*t+3*t+4)/2-n
if i > j:
   result=i*i-i+(j%2)*(2-(j+1)/2)+((j+1)%2)*(j/2+1)
else:
   result=j*j-2*(i%2)*j + (i%2)*((i+1)/2+1) + ((i+1)%2)*(-i/2+1)

As table
T(n,k) = k*k-2*(n mod 2)*k+(n mod 2)*((n+1)/2+1)+((n+1) mod 2)*(-n/2+1), if n<=k;
T(n,k) = n*n-n+(k mod 2)*(2-(k+1)/2)+((k+1) mod 2)*(k/2+1), if n>k.
As linear sequence
a(n) = j*j-2*(i mod 2)*j+(i mod 2)*((i+1)/2+1)+((i+1) mod 2)*(-i/2+1), if i<=j;
a(n) = i*i-i+(j mod 2)*(2-(j+1)/2)+((j+1) mod 2)*(j/2+1), if i>j; where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2).

A221131 Table, T, read by antidiagonals where T(-j,k) = ((1+sqrt(j))^k + (1-sqrt(j))^k)/2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, -1, -2, 1, 1, 1, -2, -5, -4, 1, 1, 1, -3, -8, -7, -4, 1, 1, 1, -4, -11, -8, 1, 0, 1, 1, 1, -5, -14, -7, 16, 23, 8, 1, 1, 1, -6, -17, -4, 41, 64, 43, 16, 1, 1, 1, -7, -20, 1, 76, 117, 64, 17, 16, 1, 1, 1, -8, -23, 8, 121, 176, 29, -128, -95, 0, 1

Offset: 0

Al Hakanson (hawkuu(AT)gmail.com) and Robert G. Wilson v, Jan 02 2013

.j\k.........0..1...2....3...4....5....6......7.......8......9......10
.0: A000012..1..1...1....1...1....1....1......1.......1......1.......1
-1: A146559..1..1...0...-2..-4...-4....0......8......16.....16.......0
-2: A087455..1..1..-1...-5..-7....1...23.....43......17....-95....-241
-3: A138230..1..1..-2...-8..-8...16...64.....64....-128...-512....-512
-4: A006495..1..1..-3..-11..-7...41..117.....29....-527..-1199.....237
-5: A138229..1..1..-4..-14..-4...76..176...-104...-1264..-1904....3776
-6: A090592..1..1..-5..-17...1..121..235...-377...-2399..-2159...12475
-7: A090590..1..1..-6..-20...8..176..288...-832...-3968..-1280...29184
-8: A025172..1..1..-7..-23..17..241..329..-1511...-5983...1633...57113
-9: A120743..1..1..-8..-26..28..316..352..-2456...-8432...7696...99712
-10: ........1..1..-9..-29..41..401..351..-3709..-11279..18241..160551

Cf. A084097,
Rows: A000012, A146559, A087455, A138230, A006495, A138229, A090592, A090590, A025172, A120743.
Columns: A000012, A000012, A024000, 1-3n, A028884, A134593.

T[j_, k_] := Expand[((1 + Sqrt[j])^k + (1 - Sqrt[j])^k)/2]; Table[ T[ -j + k, k], {j, 0, 11}, {k, 0, j}] // Flatten

A382310 Array read by ascending antidiagonals: A(n,m) is the squared distance between the roots of the 2nd degree equations z^2 +- n*z + m = 0 on the complex plane.

Original entry on oeis.org

0, 1, 4, 4, 3, 8, 9, 0, 7, 12, 16, 5, 4, 11, 16, 25, 12, 1, 8, 15, 20, 36, 21, 8, 3, 12, 19, 24, 49, 32, 17, 4, 7, 16, 23, 28, 64, 45, 28, 13, 0, 11, 20, 27, 32, 81, 60, 41, 24, 9, 4, 15, 24, 31, 36, 100, 77, 56, 37, 20, 5, 8, 19, 28, 35, 40, 121, 96, 73, 52, 33, 16, 1, 12, 23, 32, 39, 44

Offset: 0

Stefano Spezia, Mar 21 2025

			The array begins as:
   0,  4,  8, 12, 16, 20, 24, 28, 32, 36, 40, 44, ...
   1,  3,  7, 11, 15, 19, 23, 27, 31, 35, 39, 43, ...
   4,  0,  4,  8, 12, 16, 20, 24, 28, 32, 36, 40, ...
   9,  5,  1,  3,  7, 11, 15, 19, 23, 27, 31, 35, ...
  16, 12,  8,  4,  0,  4,  8, 12, 16, 20, 24, 28, ...
  25, 21, 17, 13,  9,  5,  1,  3,  7, 11, 15, 19, ...
  ...
A(2,0) = 4 since z^2 - 2*z = 0 and z^2 + 2*z = 0 have respectively roots 0, 2, and -2, 0 with squared distance equal to 4;
A(1,2) = 7 since z^2 - z + 2 = 0 and z^2 + z + 2 = 0 have respectively roots (1 +- i*sqrt(7))/2 and (-1 +- i*sqrt(7))/2 with squared distance equal to 7, where i denotes the imaginary unit.

Cf. A000290 (m=0), A008586 (n=0), A028347, A028566, A028884, A131098, A134594, A145917, A382311 (antidiagonal sums).

A[n_,m_]:=Abs[n^2-4m]; Table[A[n-m,m],{n,0,11},{m,0,n}]//Flatten

A(n,m) = abs(n^2 - 4*m).
A(n,n) = A028347(n-2) for n > 3.
A(n,1) = A028347(n) for n > 1.
A(n,2) = A028884(n-3) for n > 2.
A(n,4) = A028566(n-4) for n > 3.
A(n,5) = A134594(n-5) for n > 4.
A(1,n) = A131098(n+1).

A374584 Numbers k such that 7*k + 2 is a square.

Original entry on oeis.org

1, 2, 14, 17, 41, 46, 82, 89, 137, 146, 206, 217, 289, 302, 386, 401, 497, 514, 622, 641, 761, 782, 914, 937, 1081, 1106, 1262, 1289, 1457, 1486, 1666, 1697, 1889, 1922, 2126, 2161, 2377, 2414, 2642, 2681, 2921, 2962, 3214, 3257, 3521, 3566, 3842, 3889

Offset: 1

Juri-Stepan Gerasimov, Aug 12 2024

Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

The numbers k such that (m + (9-m)*k) is a square: A000217 (m = 1), this sequence (m = 2), A003154 (m = 3), A195162 (m = 4), A028387 (m = 5), A100536 (m = 6), A059993 (m = 7), A028884 (m = 8).

Cf. A047341.

[k: k in [0..4000] | IsSquare(7*k + 2)];

((Table[7*n + {3, 4}, {n, 0, 23}] // Flatten)^2 - 2)/7 (* Amiram Eldar, Aug 12 2024 *)

a(n) = (A047341(n)^2 - 2)/7. - Amiram Eldar, Aug 12 2024

cp's OEIS Frontend

A164604 a(n) = ((1+4*sqrt(2))*(3+2*sqrt(2))^n + (1-4*sqrt(2))*(3-2*sqrt(2))^n)/2.

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A190576 a(n) = n^2 + 5*n - 5.

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A213922 Natural numbers placed in table T(n,k) layer by layer. The order of placement: T(n,n), T(n-1,n), T(n,n-1), ... T(1,n), T(n,1). Table T(n,k) read by antidiagonals.

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Mathematica

Python

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A175019 a(0)=a(1)=1; a(n) = a(n-a(n-2)) + 2.

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Mathematica

A193517 T(n,k) = number of ways to place any number of 5X1 tiles of k distinguishable colors into an nX1 grid.

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Maple

Mathematica

Formula

A214870 Natural numbers placed in table T(n,k) layer by layer. The order of placement: at the beginning filled odd places of layer clockwise, next - even places counterclockwise. T(n,k) read by antidiagonals.

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Python

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A221131 Table, T, read by antidiagonals where T(-j,k) = ((1+sqrt(j))^k + (1-sqrt(j))^k)/2.

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Mathematica

A382310 Array read by ascending antidiagonals: A(n,m) is the squared distance between the roots of the 2nd degree equations z^2 +- n*z + m = 0 on the complex plane.

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A164604 a(n) = ((1+4sqrt(2))(3+2sqrt(2))^n + (1-4sqrt(2))(3-2sqrt(2))^n)/2.