A184536 -id:A184536 - cp's OEIS frontend

A184537 a(n) = floor(1/{(2+n^4)^(1/4)}), where {} = fractional part.

5, 3, 16, 54, 128, 250, 432, 686, 1024, 1458, 2000, 2662, 3456, 4394, 5488, 6750, 8192, 9826, 11664, 13718, 16000, 18522, 21296, 24334, 27648, 31250, 35152, 39366, 43904, 48778, 54000, 59582, 65536, 71874, 78608, 85750, 93312, 101306, 109744, 118638, 128000, 137842, 148176, 159014, 170368, 182250, 194672, 207646, 221184, 235298, 250000, 265302, 281216, 297754, 314928, 332750, 351232, 370386, 390224, 410758, 432000, 453962, 476656

Offset: 0

Clark Kimberling, Jan 16 2011

Similar to A033431: replacing a(0) by 0 and a(1) by 2 gives A033431, see next comment.
From Bruno Berselli, Feb 04 2011: (Start)
For n >= 1, the value of (n^4+2)^(1/4) is just slightly above n, so the fractional part is (2+n^4)^(1/4)-n. For n > 1, then, 2*n^3 < 1/((2+n^4)^(1/4)-n) < 2*n^3+1.
The proof that 2*n^3*((2+n^4)^(1/4)-n) < 1 follows easily by isolating the quartic root and raising to the 4th power; similarly, 1 < (2*n^3+1)*((2+n^4)^(1/4)-n) needs a sign estimation of a 9th-order polynomial.
In conclusion, a(n)=A033431(n) for n > 1, with g.f. (34*x^2-12*x^3+x^4+x^5+5-17*x) / (x-1)^4. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A184536. Essentially the same as A033431.

I:=[5, 3, 16, 54, 128,250]; [n le 6 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..70]]; // Vincenzo Librandi, Jul 04 2012

f[n_] := Floor[1/FractionalPart[(n^4 + 2)^(1/4)]]; Array[f, 40, 0]
CoefficientList[Series[(34*x^2-12*x^3+x^4+x^5+5-17*x)/(x-1)^4,{x,0,50}],x] (* Vincenzo Librandi, Jul 04 2012 *)
LinearRecurrence[{4,-6,4,-1},{5,3,16,54,128,250},70] (* Harvey P. Dale, Apr 06 2018 *)

a(n) = floor( 1 / frac((2+n^4)^(1/4)) ).

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n > 5.

A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.

Original entry on oeis.org

2, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744, 3375, 4096, 4913, 5832, 6859, 8000, 9261, 10648, 12167, 13824, 15625, 17576, 19683, 21952, 24389, 27000, 29791, 32768, 35937, 39304, 42875, 46656, 50653, 54872, 59319, 64000, 68921, 74088, 79507, 85184, 91125, 97336, 103823, 110592, 117649, 125000, 132651, 140608, 148877, 157464, 166375

Offset: 1

Clark Kimberling, Jan 18 2011

Is a(n) = A066023(n) for n>=2? R. J. Mathar, Jan 28 2011

Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (4, -6, 4, -1).

Cf. A000578, A066023, A184536.

p[n_]:=FractionalPart[(n^4+4)^(1/4)];
  q[n_]:=Floor[1/p[n]]; Table[q[n],{n,1,80}]
  FindLinearRecurrence[Table[q[n],{n,1,1000}]]
Join[{2}, LinearRecurrence[{4, -6, 4, -1}, {8, 27, 64, 125}, 54]] (* Ray Chandler, Aug 01 2015 *)

a(n) = floor(1/{(4+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=4a(n-1)-6a(n-2)+4a(n-3)-a(n-4) for n>=6 and that a(n)=n^3 for n>=2.
Empirical g.f.: x*(x^4-4*x^3+7*x^2+2) / (x-1)^4. - Colin Barker, Sep 06 2014

A184629 Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

1, 7, 22, 51, 100, 173, 274, 409, 583, 800, 1064, 1382, 1757, 2195, 2700, 3276, 3930, 4665, 5487, 6400, 7408, 8518, 9733, 11059, 12500, 14060, 15746, 17561, 19511, 21600, 23832, 26214, 28749, 31443, 34300, 37324, 40522, 43897, 47455, 51200, 55136, 59270, 63605, 68147, 72900, 77868, 83058, 88473, 94119, 100000

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 1, -3, 3, -1).

Cf. A184536.

p[n_]:=FractionalPart[(n^4+5)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1, 7, 22, 51, 100, 173}, LinearRecurrence[{3, -3, 1, 0, 1, -3, 3, -1}, {274, 409, 583, 800, 1064, 1382, 1757, 2195}, 44]] (* Ray Chandler, Aug 01 2015 *)

a(n)=floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8) for n>=15.

A184630 Floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

1, 6, 18, 43, 83, 144, 228, 341, 486, 666, 887, 1152, 1464, 1829, 2250, 2730, 3275, 3888, 4572, 5333, 6174, 7098, 8111, 9216, 10416, 11717, 13122, 14634, 16259, 18000, 19860, 21845, 23958, 26202, 28583, 31104, 33768, 36581, 39546, 42666, 45947, 49392, 53004, 56789, 60750, 64890, 69215, 73728, 78432, 83333, 88434

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).

Cf. A184536.

p[n_]:=FractionalPart[(n^4+6)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,6,18,43},LinearRecurrence[{3,-3,2,-3,3,-1},{83,144,228,341,486,666},47]] (* Ray Chandler, Aug 02 2015 *)

a(n)=floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=11, which implies a(n) = (2*n^3-1+A049347(n))/3 for n>=5.

A184631 Floor(1/{(7+n^4)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

1, 5, 15, 36, 71, 123, 196, 292, 416, 571, 760, 987, 1255, 1568, 1928, 2340, 2807, 3332, 3919, 4571, 5292, 6084, 6952, 7899, 8928, 10043, 11247, 12544, 13936, 15428, 17023, 18724, 20535, 22459, 24500, 26660, 28944, 31355, 33896, 36571, 39383, 42336, 45432, 48676, 52071, 55620, 59327, 63195, 67228, 71428, 75800

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Ray Chandler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 0, 0, 1, -3, 3, -1).

Cf. A184536.

p[n_]:=FractionalPart[(n^4+7)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,5},LinearRecurrence[{3,-3,1,0,0,0,1,-3,3,-1},{15,36,71,123,196,292,416,571,760,987},49]] (* Ray Chandler, Aug 02 2015 *)

a(n)=floor(1/{(7+n^4)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-7)-3a(n-8)+3a(n-9)-a(n-10) for n>=13.

A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

Original entry on oeis.org

1, 4, 12, 28, 55, 96, 152, 227, 324, 444, 591, 768, 976, 1219, 1500, 1820, 2183, 2592, 3048, 3555, 4116, 4732, 5407, 6144, 6944, 7811, 8748, 9756, 10839, 12000, 13240, 14563, 15972, 17468, 19055, 20736, 22512, 24387, 26364, 28444, 30631, 32928, 35336, 37859, 40500, 43260, 46143, 49152, 52288, 55555, 58956, 62492, 66167, 69984, 73944, 78051, 82308, 86716, 91279, 96000, 100880, 105923, 111132, 116508, 122055, 127776, 133672, 139747, 146004, 152444, 159071, 165888, 172896, 180099, 187500

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Edward Jiang and Ray Chandler, Table of n, a(n) for n = 1..10000 (first 1000 terms from Edward Jiang)
Index entries for linear recurrences with constant coefficients, signature (3, -3, 2, -3, 3, -1).

Cf. A184536.

p[n_]:=FractionalPart[(n^4+9)^(1/4)]; q[n_]:=Floor[1/p[n]];
  Table[q[n], {n, 1, 80}]
  FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,4},LinearRecurrence[{3,-3,2,-3,3,-1},{12,28,55,96,152,227},73]] (* Ray Chandler, Aug 02 2015 *)

a(n)=my(t=(9+n^4)^(1/4)); 1\(t-t\1) \\ Charles R Greathouse IV, Sep 12 2014

a(n) = floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.
It appears that a(n) = 3a(n-1)-3a(n-2)+2a(n-3)-3a(n-4)+3a(n-5)-a(n-6) for n>=9.
Empirical g.f.: x*(x+1)*(x^6-3*x^5+3*x^4-x^3+3*x^2+1) / ((x-1)^4*(x^2+x+1)). - Colin Barker, Jun 13 2015

A184634 a(n) = floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

1, 3, 11, 25, 50, 86, 137, 204, 291, 400, 532, 691, 878, 1097, 1350, 1638, 1965, 2332, 2743, 3200, 3704, 4259, 4866, 5529, 6250, 7030, 7873, 8780, 9755, 10800, 11916, 13107, 14374, 15721, 17150, 18662, 20261, 21948, 23727, 25600, 27568, 29635, 31802, 34073, 36450, 38934, 41529, 44236, 47059, 50000, 53060, 56243, 59550, 62985, 66550, 70246, 74077, 78044, 82151, 86400, 90792, 95331, 100018, 104857, 109850, 114998, 120305, 125772, 131403, 137200

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Robert Israel, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (3, -3, 1, 0, 1, -3, 3, -1).

Cf. A184536.

f:= proc(n) local k, t;
  t:= n mod 5;
  k:= (n-t)/5;
  [50*k^3, 50*k^3 + 30*k^2 + 6*k,
50*k^3 + 60*k^2 + 24*k + 3,
50*k^3 + 90*k^2 + 54*k + 10,
50*k^3 + 120*k^2 + 96*k + 25][t+1]
end proc:
f(1):= 1: f(3):= 11:
map(f, [$1..100]); # Robert Israel, Feb 25 2019

p[n_]:=FractionalPart[(n^4+10)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{1,3,11},LinearRecurrence[{3,-3,1,0,1,-3,3,-1},{25,50,86,137,204,291,400,532},67]] (* Ray Chandler, Aug 02 2015 *)

a(n)=1\frac(sqrtn(n^4+10,4)) \\ Charles R Greathouse IV, Feb 07 2016

a(n)=floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.
It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3)+a(n-5)-3a(n-6)+3a(n-7)-a(n-8).
From Robert Israel, Feb 25 2019: (Start)
2*n^3/5 + 3/(2*n) > 1/{(10+n^4)^(1/4)} > 2*n^3/5 for all n.
From this we can show that a(5*k) = 50*k^3 for k >= 1,
a(5*k+1) = 50*k^3 + 30*k^2 + 6*k for k >= 1,
a(5*k+2) = 50*k^3 + 60*k^2 + 24*k + 3,
a(5*k+3) = 50*k^3 + 90*k^2 + 54*k + 10 for k >= 1,
a(5*k+4) = 50*k^3 + 120*k^2 + 96*k + 25.
This implies the conjectured recurrence for n >= 12.  (End)

A184635 a(n) = floor(1/{(n+n^4)^(1/4)}), where {} = fractional part.

Original entry on oeis.org

5, 16, 36, 64, 100, 144, 196, 256, 324, 400, 484, 576, 676, 784, 900, 1024, 1156, 1296, 1444, 1600, 1764, 1936, 2116, 2304, 2500, 2704, 2916, 3136, 3364, 3600, 3844, 4096, 4356, 4624, 4900, 5184, 5476, 5776, 6084, 6400, 6724, 7056, 7396, 7744, 8100, 8464

Offset: 1

Clark Kimberling, Jan 18 2011

nonn

Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A184536, A144916, A016742.

p[n_]:=FractionalPart[(n^4+n)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{5},LinearRecurrence[{3,-3,1},{16,36,64},45]] (* Ray Chandler, Aug 02 2015 *)

a(n) = floor(1/{(n+n^4)^(1/4)}), where {} = fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3) for n>=5, and that a(n)=4*n^2 for n>=2.

A184636 a(n) = floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.

Original entry on oeis.org

3, 8, 18, 32, 50, 72, 98, 128, 162, 200, 242, 288, 338, 392, 450, 512, 578, 648, 722, 800, 882, 968, 1058, 1152, 1250, 1352, 1458, 1568, 1682, 1800, 1922, 2048, 2178, 2312, 2450, 2592, 2738, 2888, 3042, 3200, 3362, 3528, 3698, 3872, 4050, 4232, 4418, 4608, 4802, 5000, 5202, 5408, 5618, 5832, 6050, 6272, 6498, 6728, 6962, 7200, 7442, 7688, 7938, 8192, 8450, 8712, 8978, 9248, 9522, 9800

Offset: 1

Clark Kimberling, Jan 18 2011

Is a(n) = A001105(n) for n>=2 ? R. J. Mathar, Jan 28 2011

Index entries for linear recurrences with constant coefficients, signature (3, -3, 1).

Cf. A184536, A184635, A184637.

p[n_]:=FractionalPart[(n^4+2*n)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n], {n, 1, 80}]
FindLinearRecurrence[Table[q[n], {n, 1, 1000}]]
Join[{3},LinearRecurrence[{3,-3,1},{8,18,32},69]] (* Ray Chandler, Aug 02 2015 *)

a(n)=floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.

It appears that a(n)=3a(n-1)-3a(n-2)+a(n-3) for n>=5, and that a(n)=2*n^2 for n>=2.

A184538 Floor[1/{(3+n^4)^(1/4)}], where {}=fractional part.

Original entry on oeis.org

2, 11, 36, 85, 166, 288, 457, 682, 972, 1333, 1774, 2304, 2929, 3658, 4500, 5461, 6550, 7776, 9145, 10666, 12348, 14197, 16222, 18432, 20833, 23434, 26244, 29269, 32518, 36000, 39721, 43690, 47916, 52405, 57166, 62208, 67537, 73162, 79092, 85333, 91894, 98784, 106009, 113578, 121500, 129781, 138430, 147456, 156865, 166666, 176868, 187477, 198502, 209952, 221833, 234154, 246924, 260149, 273838, 288000

Offset: 1

Clark Kimberling, Jan 16 2011

nonn

A184536.

p[n_]:=FractionalPart[(n^4+3)^(1/4)];
q[n_]:=Floor[1/p[n]];
Table[q[n],{n,1,80}]

a(n)=floor[1/{(3+n^4)^(1/4)}], where {}=fractional part.
Recurrence relation appears to be a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 3*a(n-4) + 3*a(n-5) - a(n-6).
Empirical G.f.: x*(x+1)*(x^6-3*x^5+3*x^4+6*x^2+3*x+2)/((x-1)^4*(x^2+x+1)). [Colin Barker, Sep 21 2012]

cp's OEIS Frontend

A184537 a(n) = floor(1/{(2+n^4)^(1/4)}), where {} = fractional part.

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A184628 Floor(1/frac((4+n^4)^(1/4))), where frac(x) is the fractional part of x.

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A184629 Floor(1/{(5+n^4)^(1/4)}), where {}=fractional part.

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A184630 Floor(1/{(6+n^4)^(1/4)}), where {}=fractional part.

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A184631 Floor(1/{(7+n^4)^(1/4)}), where {}=fractional part.

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A184633 Floor(1/{(9+n^4)^(1/4)}), where {} = fractional part.

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A184634 a(n) = floor(1/{(10+n^4)^(1/4)}), where {}=fractional part.

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A184635 a(n) = floor(1/{(n+n^4)^(1/4)}), where {} = fractional part.

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A184636 a(n) = floor(1/{(n^4+2*n)^(1/4)}), where {}=fractional part.