A008590 -id:A008590 - cp's OEIS frontend

A243883 Numerator of circle radius r(n) at constant unit length sagitta and chord length = n.

5, 1, 13, 5, 29, 5, 53, 17, 85, 13, 125, 37, 173, 25, 229, 65, 293, 41, 365, 101, 445, 61, 533, 145, 629, 85, 733, 197, 845, 113, 965, 257, 1093, 145, 1229, 325, 1373, 181, 1525, 401, 1685, 221, 1853, 485, 2029, 265, 2213, 577, 2405, 313, 2605, 677, 2813, 365, 3029

Offset: 1

Kival Ngaokrajang, Jun 13 2014

Denominator of circle radius r(n) is A143025(n+2). The integral radius appearing at n = 2, 6, 10, 14, ..., = 1, 5, 13, 25, ..., respectively which is A001844(n/4 - 1/2). Floor (r(n)) = A001971(n). For the case of sagitta = n and chord length = 1, the numerator and the denominator will be A053755(n) and A008590(n) respectively. See illustration in links.

Colin Barker, Table of n, a(n) for n = 1..1000
Kival Ngaokrajang, Illustration for n = 1..5
Wikipedia, Sagitta

Cf. A143025, A001844, A001971.

PARI
```
a(n) = numerator(n^2/8+1/2);
```

a(n) = numerator(n^2/8 + 1/2).

Empirical g.f.: -x*(x^11 +5*x^10 +x^9 +13*x^8 +2*x^7 +14*x^6 +2*x^5 +14*x^4 +5*x^3 +13*x^2 +x +5) / ((x -1)^3*(x +1)^3*(x^2 +1)^3). - Colin Barker, Jan 17 2015

A267958 4 times A042965.

Original entry on oeis.org

0, 4, 12, 16, 20, 28, 32, 36, 44, 48, 52, 60, 64, 68, 76, 80, 84, 92, 96, 100, 108, 112, 116, 124, 128, 132, 140, 144, 148, 156, 160, 164, 172, 176, 180, 188, 192, 196, 204, 208, 212, 220, 224, 228, 236, 240, 244, 252, 256, 260, 268, 272, 276, 284, 288, 292, 300, 304

Offset: 1

Matthew Burch, Jan 22 2016

Ordered list of differences between even squares.

			(2*0)^2 - (2*0)^2 = 0,
(2*1)^2 - (2*0)^2 = 4,
(2*2)^2 - (2*1)^2 = 12,
(2*2)^2 - (2*0)^2 = 16,
(2*3)^2 - (2*2)^2 = 20,
...

Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A008590, A042965.

a := proc(n) option remember; if n = 1 then 0 elif n = 2 then 4 elif n = 3 then 12 else a(floor((1/2)*n)) + a(1+ceil((1/2)*n)) end if; end proc:
seq(a(n), n = 1..50); # Peter Bala, Aug 03 2022

def DifferenceOfEvenSquares(maximumBound):
    sequence = set([0])
    for x in range(0, maximumBound+1, 4):
        if x % 16 != 8:
            sequence.add(x)
    print(sorted(sequence))

Numbers of the form (2m)^2 - (2n)^2, sorted.
From Chai Wah Wu, Sep 01 2024: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n > 4.
G.f.: 4*x^2*(x + 1)^2/(x^4 - x^3 - x + 1). (End)

A277780 a(n) is the least k > n such that n*k^2 is a cube.

Original entry on oeis.org

8, 16, 24, 32, 40, 48, 56, 27, 72, 80, 88, 96, 104, 112, 120, 54, 136, 144, 152, 160, 168, 176, 184, 81, 200, 208, 64, 224, 232, 240, 248, 108, 264, 272, 280, 288, 296, 304, 312, 135, 328, 336, 344, 352, 360, 368, 376, 162, 392, 400, 408, 416, 424, 128, 440

Offset: 1

Peter Kagey, Oct 30 2016

a(n) is bounded above by 8*n (A008590) because n*(8*n)^2 = (4*n)^3.
If and only if n is cubefree, a(n) = 8n. - David A. Corneth, Nov 01 2016
Theorem: If n = q*m^3 with q cubefree then k = q*(m+1)^3. - Hartmut F. W. Hoft, Nov 02 2016
Proof: let q have u distinct prime divisors p_i. Then q = Product_{i=1..u}(p_i^e_i) where e_i > 0 since p_i|q and e_i < 3 since q is cubefree. Therefore, e_i = 1 or e_i = 2. This yields q|k, i.e., q*t = k. Now for n*k^2 = q*m^3*q^2*t^2 = (q*m)^3 * t^2 to be a cube, t must be a cube. Now, k > n, so q*t/(q*m^3) = t/m^3. The least cube > m^3 is (m+1)^3 so k = q*(m+1)^3 which completes the proof. - David A. Corneth, Nov 03 2016

			a(24) = 81  because 24 *  81^2 =  54^3;
a(25) = 200 because 25 * 200^2 = 100^3;
a(26) = 208 because 26 * 208^2 = 104^3;
a(27) = 64  because 27 *  64^2 =  48^3.
The cubefree part of 144 is 18. The cubefull part of 144 is 8 = 2^3. Therefore, a(144) = 18 * 3^3 = 486. - _David A. Corneth_, Nov 01 2016

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000578, A008590, A008834, A048766, A050985, A072905, A254767, A277781.

Cf. A002117, A013662, A013663, A013664.

Table[k = n + 1; While[! IntegerQ[(n k^2)^(1/3)], k++]; k, {n, 55}] (* Michael De Vlieger, Nov 04 2016 *)

a(n) = {my(k = n+1); while (!ispower(n*k^2, 3), k++); k;} \\ Michel Marcus, Oct 31 2016

a(n) = {my(f = factor(n)); f[, 2] = f[, 2]%3; f=factorback(f); n = sqrtnint(n/f,3); (n+1)^3 * f} \\ David A. Corneth, Nov 01 2016

a(n) = A050985(n) * A000578(1+A048766(A008834(n))). [Formula given in comments expressed with A-numbers] - Antti Karttunen, Nov 02 2016.

Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + (3*zeta(4) + 3*zeta(5) + zeta(6))/zeta(3) = 7.13539675963975495073... . - Amiram Eldar, Feb 17 2024

A317095 a(n) = 40*n.

Original entry on oeis.org

0, 40, 80, 120, 160, 200, 240, 280, 320, 360, 400, 440, 480, 520, 560, 600, 640, 680, 720, 760, 800, 840, 880, 920, 960, 1000, 1040, 1080, 1120, 1160, 1200, 1240, 1280, 1320, 1360, 1400, 1440, 1480, 1520, 1560, 1600, 1640, 1680, 1720, 1760, 1800, 1840, 1880

Offset: 0

Felix Fröhlich, Sep 07 2018

a(n) is equal to the freshwater zone below sea level for a water table of elevation n above sea level in a simplified freshwater-saltwater interface in a coastal water-table aquifer (cf. Barlow, 2003, p. 14, eq. (2) and p. 15, Fig. B-1 and B-2).
From Bruno Berselli, Sep 10 2018: (Start)
After 0, subsequence of A065607: 1/a(n)^2 + 1/(30*n)^2 = 1/(24*n)^2, with n > 0 and a(n) > 30*n.
Also, all positive terms belong to A049094: 2^(40*n)-1 = 1024^(4*n)-1 and (25*41-1)^(4*n)-1 is divisible by 25. (End)

P. M. Barlow, Ground Water in Freshwater-Saltwater Environments of the Atlantic Coast, Circular 1262 (2003).
Wikipedia, Aquifer
Wikipedia, Saltwater intrusion
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A005843, A007395, A008586, A008592, A008602, A010692, A049094, A065607.
Row n = 40 of A004247. Intersection of A008587 and A008590.
After 0, subsequence of A005101.

Table[40 n, {n, 0, 50}] (* or *)
LinearRecurrence[{2, -1}, {0, 40}, 50] (* or *)
CoefficientList[Series[40*x/(1 - x)^2, {x, 0, 50}], x] (* Stefano Spezia, Sep 07 2018 *)

PARI
```
a(n) = 40*n
```

a(n) = if(n==0, 0, if(n==1, 40, 2*a(n-1)-a(n-2)))

PARI
```
concat(0, Vec(40*x/(1-x)^2 + O(x^60)))
```

O.g.f.: 40*x/(1 - x)^2.
E.g.f.: 40*x*exp(x). - Bruno Berselli, Sep 10 2018
a(n) = 2*a(n - 1) - a(n - 2) for n > 1. - Stefano Spezia, Sep 07 2018
a(n) = A008586(A008592(n)) = 4*A008592(n).
a(n) = A010692(n)*A008586(n) = 10*A008586(n).
a(n) = A008602(A005843(n)) = 20*A005843(n).
a(n) = A007395(n)*A008602(n) = 2*A008602(n).

A361692 a(n) = 17*n - 1.

Original entry on oeis.org

16, 33, 50, 67, 84, 101, 118, 135, 152, 169, 186, 203, 220, 237, 254, 271, 288, 305, 322, 339, 356, 373, 390, 407, 424, 441, 458, 475, 492, 509, 526, 543, 560, 577, 594, 611, 628, 645, 662, 679, 696, 713, 730, 747, 764, 781, 798, 815, 832, 849, 866, 883, 900, 917, 934, 951, 968, 985, 1002, 1019

Offset: 1

Leo Tavares, Mar 20 2023

Leo Tavares, Illustration: Double Diamonds.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008590, A008599, A017257.

17*Range[100] - 1 (* Paolo Xausa, Aug 30 2024 *)
LinearRecurrence[{2,-1},{16,33},90] (* Harvey P. Dale, Jun 03 2025 *)

a(n) = 17*n - 1 = A008599(n) - 1.
a(n) = 2*A008590(n) + n - 1.
a(n) = A008590(n) + A017257(n-1).
From Elmo R. Oliveira, Apr 03 2025: (Start)
G.f.: x*(16 + x)/(x - 1)^2.
E.g.f.: exp(x)*(17*x - 1) + 1.
a(n) = 2*a(n-1) - a(n-2) for n > 2. (End)

A181390 Absolute difference between (sum of previous terms) and (n-th-odd square) with a(1) = 1.

Original entry on oeis.org

1, 0, 8, 16, 24, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 136, 144, 152, 160, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 272, 280, 288, 296, 304, 312, 320, 328, 336, 344, 352, 360, 368, 376, 384, 392, 400, 408, 416, 424

Offset: 1

Giovanni Teofilatto, Oct 17 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A008590, A022144.

Module[{lst={1}},Do[AppendTo[lst,Abs[Total[lst]-n^2]],{n,1,111,2}];lst] (* or *) Join[{1},LinearRecurrence[{2,-1},{0,8},60]] (* Harvey P. Dale, Aug 23 2012 *)
ad[{t_,n_,a_}]:=Module[{c=Abs[t-(2n-1)^2]},{t+c,n+1,c}]; NestList[ad,{1,1,1},60][[All,3]] (* or  *) Join[{1}, NestList[8 + # &, 0, 60]] (* Harvey P. Dale, May 14 2019 *)

a(n)=if(n>1,8*n-8,1) \\ Charles R Greathouse IV, Jul 31 2013

a(n) = 8*(n-2) = A008590(n-2), n>1. - R. J. Mathar, Oct 18 2010
G.f.: x*(1 - 2*x + 9*x^2)/(-1 + x)^2. -Alexander R. Povolotsky, Oct 18 2010
a(1)=1, a(2)=0, a(3)=8, a(n)=2*a(n-1)-a(n-2). -Harvey P. Dale, Aug 23 2012
E.g.f.: 16 + 9*x + 8*exp(x)*(x - 2). - Stefano Spezia, Apr 03 2023

Adapted g.f. to the offset from Bruno Berselli, Aug 01 2013

A236203 Interleave A005563(n), A028347(n).

Original entry on oeis.org

0, 0, 3, 5, 8, 12, 15, 21, 24, 32, 35, 45, 48, 60, 63, 77, 80, 96, 99, 117, 120, 140, 143, 165, 168, 192, 195, 221, 224, 252, 255, 285, 288, 320, 323, 357, 360, 396, 399, 437, 440, 480, 483, 525, 528, 572, 575, 621, 624, 672, 675, 725, 728, 780, 783, 837, 840, 896

Offset: 2

Paul Curtz, Jan 20 2014

A175628 gives the numerators of interleaved Lyman and Balmer series, i.e., A005563(n)/A000290(n+1) and A061037(n+2)/A061038(n+2).
Difference table of a(n):
   -1,  -3,  0,  0,  3,   5,   8,  12,  15,  21,  24, ...
   -2,   3,  0,  3,  2,   3,   4,   3,   6,   3,   8, ...
    5,  -3,  3, -1,  1,   1,  -1,   3,  -3,   5,  -5, ...
   -8,   6, -4,  2,  0,  -2,   4,  -6,   8, -10,  12, ...
   14, -10,  6, -2, -2,   6, -10,  14, -18,  22, -26, ...
  -24,  16, -8,  0,  8, -16,  24, -32,  40, -48,  56, ... .
a(n+2) gives the numerators of 0/1, 0/16, 3/4, 5/36, 8/9, 12/64, 15/16, 21/100, 24/25, 32/144, ... . The denominators are A097362(n+1)^2. (Compare A097362 to A029578.)
Note the particular distribution of a(-n). Example:
a(n-9) = 12,15, 5,8, 0,3, -3,0, -4,-1, -3,0, 0,3, 5,8, 12,15, ... .
a(2n) + a(2n+1) = a(-2n-1) + a(-2n-2) = -4,0,8,20,36,56,80,... = 4*A000096(n-1).
a(2n) + a(2n-1) = a(-2n) + a(-2n-1) = -5,-3,3,13,... = A001105(n) - A010716(n).

Vincenzo Librandi, Table of n, a(n) for n = 2..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A005843, A008590, A016825, A109613, A147658.

List([2..60], n-> (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 ); # G. C. Greubel, Dec 04 2019

[(2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8: n in [2..60]]; // Vincenzo Librandi, Jul 27 2014

seq( (2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8, n=2..60); # G. C. Greubel, Dec 04 2019

CoefficientList[Series[x^2(3x^2-2x-3)/((x-1)^3(x+1)^2), {x, 0, 60}], x] (* Vincenzo Librandi, Jul 27 2014 *)
LinearRecurrence[{1,2,-2,-1,1},{0,0,3,5,8},60] (* Harvey P. Dale, Aug 30 2018 *)

concat([0,0], Vec(x^4*(3*x^2-2*x-3)/((x-1)^3*(x+1)^2) + O(x^60))) \\ Colin Barker, Jan 26 2014

[(2*n^2 +2*n -19 -(2*n-11)*(-1)^n)/8 for n in (2..60)] # G. C. Greubel, Dec 04 2019

a(n+2) = (period 8: repeat 1, 16, 1, 1, 1, 4, 1, 1)*A175628(n+1).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n+4) - a(n-4) = 0, 8, 8, ... = A168397.
From Colin Barker, Jan 26 2014: (Start)
a(n) = (n^2 -4)/4 for n even, a(n) = (n^2 +2*n -15)/4 for n odd.
G.f.: x^4*(3 + 2*x - 3*x^2)/ ((1-x)^3*(1+x)^2). (End)
a(n) = (2*n^2 + 2*n - 19 - (2*n - 11)*(-1)^n)/8. - Luce ETIENNE, Jul 26 2014
Sum_{n>=4} (-1)^n/a(n) = 11/48. - Amiram Eldar, Aug 21 2022

More terms from Colin Barker, Jan 26 2014

A239794 5n^2 + 4n - 15.

Original entry on oeis.org

-6, 13, 42, 81, 130, 189, 258, 337, 426, 525, 634, 753, 882, 1021, 1170, 1329, 1498, 1677, 1866, 2065, 2274, 2493, 2722, 2961, 3210, 3469, 3738, 4017, 4306, 4605, 4914, 5233, 5562, 5901, 6250, 6609, 6978, 7357, 7746, 8145, 8554, 8973, 9402, 9841, 10290

Offset: 1

Katherine Guo, Mar 26 2014

Follows the integer values from 1 on the quadratic equation 5*x^2 + 4*n - 15, this is the case x=n.

			For n=3, a(3) = 5*3^2 + 4*3 - 15 = 42; for n=6, a(6) = 5*6^2 + 4*6 - 15 = 189.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
WolframAlpha, Table of 5n^2+4n-15
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A008590, A017377.

Magma
```
[5*n^2+4*n-15: n in [1..50]];
```

Table[5 n^2 + 4 n - 15, {n, 50}]
CoefficientList[Series[(6 - 31 x + 15 x^2)/(x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, Mar 29 2014 *)
LinearRecurrence[{3,-3,1},{-6,13,42},50] (* Harvey P. Dale, May 30 2025 *)

a(n)=5*n^2+4*n-15 \\ Charles R Greathouse IV, Jun 17 2017

From Bruno Berselli, Mar 27 2014: (Start)
G.f.: -x*(6 - 31*x + 15*x^2)/(1 - x)^3.
a(n+1) - a(n) = A017377(n).
a(n) - a(-n) = A008590(n). (End)

A247555 A permutation of the nonnegative numbers: a(4n) = 8n, a(4n+1) = 2n + 1, a(4n+2) = 4n + 2, a(4n+3) = 8n + 4.

Original entry on oeis.org

0, 1, 2, 4, 8, 3, 6, 12, 16, 5, 10, 20, 24, 7, 14, 28, 32, 9, 18, 36, 40, 11, 22, 44, 48, 13, 26, 52, 56, 15, 30, 60, 64, 17, 34, 68, 72, 19, 38, 76, 80, 21, 42, 84, 88, 23, 46, 92, 96, 25, 50, 100, 104, 27, 54, 108, 112, 29, 58, 116, 120

Offset: 0

Paul Curtz, Sep 19 2014

A permutation of the nonnegative integers.

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A005408, A008590, A016825, A017113, A225055, A001477, A111284.

&cat[[4*(i-1),i,2*i,4*i]: i in [1..50 by 2]]; // Bruno Berselli, Sep 19 2014

a[n_]:=Switch[Mod[n,4],0,2 n,1,(n+1)/2,2,n,3,2 n-2]; Table[a[n],{n,0,60}] (* Jean-François Alcover, Oct 09 2014 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1}, {0,1,2,4,8,3,6,12}, 50] (* G. C. Greubel, May 01 2018 *)

Vec(x*(4*x^6+2*x^5+x^4+8*x^3+4*x^2+2*x+1)/((x-1)^2*(x+1)^2*(x^2+1)^2) + O(x^100)) \\ Colin Barker, Sep 19 2014

a(n) = a(n-4) + a(n-8) - a(n-12).
a(n) = 2*a(n-4) - a(n-8). - Colin Barker, Sep 19 2014
G.f.: x*(4*x^6 + 2*x^5 + x^4 + 8*x^3 + 4*x^2 + 2*x + 1) / ((x-1)^2*(x+1)^2*(x^2+1)^2). - Colin Barker, Sep 19 2014
a(n) = (11*n-3+(n+3)*(-1)^n+(4*n-1+(-1)^n)*cos(n*Pi/2)+2*(9-3*n+4(-1)^n)* sin(n*Pi/2))/8. - Wesley Ivan Hurt, May 07 2021

A329549 Numbers 4k such that 1 is the last integer obtained when 4k is successively divided by its divisors in increasing order.

Original entry on oeis.org

8, 24, 40, 56, 64, 120, 144, 280, 320, 448, 704, 720, 832, 1008, 1024, 1152, 2240, 3200, 4928, 5040, 5760, 5824, 6272, 8064, 9152, 10368, 11264, 13312, 17408, 19456, 22400, 23552, 29696, 31744, 32768, 35200, 40320, 41600, 51200, 51840, 64064, 68992, 72576, 81536, 100352, 114048

Offset: 1

David A. Corneth, Nov 16 2019

nonn

At sequence A076933, the question is asked: "What is the longest string of ones in this sequence?" As A076933(4*n) is rarely 1, such a string is not very long. The longest starting below 4*10^8 has length 6 and starts at 141. Checking multiples of 4 may help in finding longer such strings.

Terms are also a multiple of 8. Proof: If m = 8*k + 4 then its divisors are 1, 2, 4 (and maybe 3). After dividing by 4 we have a fraction with denominator 2. Before that we did not see 1.

			The divisors of 8 are 1, 2, 4 and 8. Dividing from left to right gives 8/1 = 8, 8/2 = 4, 4/4 = 1, and then 1/8 isn't an integer so as the last integer we see is 1, 8 is in the sequence.

Cf. A076933, A240694 (partial products of divisors of n).

Subsequence of A008586 (multiples of 4) and of A008590 (multiples of 8).

cp's OEIS Frontend

A243883 Numerator of circle radius r(n) at constant unit length sagitta and chord length = n.

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A267958 4 times A042965.

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A277780 a(n) is the least k > n such that n*k^2 is a cube.

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A317095 a(n) = 40*n.

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A361692 a(n) = 17*n - 1.

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A181390 Absolute difference between (sum of previous terms) and (n-th-odd square) with a(1) = 1.

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A236203 Interleave A005563(n), A028347(n).

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A239794 5n^2 + 4n - 15.

A329549 Numbers 4k such that 1 is the last integer obtained when 4k is successively divided by its divisors in increasing order.