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.

Previous Showing 11-18 of 18 results.

A255844 a(n) = 2*n^2 + 6.

Original entry on oeis.org

6, 8, 14, 24, 38, 56, 78, 104, 134, 168, 206, 248, 294, 344, 398, 456, 518, 584, 654, 728, 806, 888, 974, 1064, 1158, 1256, 1358, 1464, 1574, 1688, 1806, 1928, 2054, 2184, 2318, 2456, 2598, 2744, 2894, 3048, 3206, 3368, 3534, 3704, 3878, 4056, 4238, 4424, 4614
Offset: 0

Views

Author

Avi Friedlich, Mar 08 2015

Keywords

Comments

This is the case k=3 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2. Also, it is noted that a(n)*n = (n + 1)^3 + (n - 1)^3.
Equivalently, numbers m such that 2*m-12 is a square.
For n = 0..16, 3*a(n)-1 is prime (see A087370); for n = 0..12, 3*a(n)-5 is prime (see A107303).

Crossrefs

Cf. A016825 (first differences), A087370, A107303, A114949, A117950.
Cf. A152811: nonnegative numbers of the form 2*m^2-6.
Subsequence of A000378.
Cf. similar sequences listed in A255843.

Programs

  • Magma
    [2*n^2+6: n in [0..50]];
  • Mathematica
    Table[2 n^2 + 6, {n, 0, 50}]
  • PARI
    vector(50, n, n--; 2*n^2+6)
    
  • Sage
    [2*n^2+6 for n in (0..50)]
    

Formula

G.f.: 2*(3-5*x+4*x^2)/(1 - x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*A117950(n).
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(3)*Pi*coth(sqrt(3)*Pi))/12.
Sum_{n>=0} (-1)^n/a(n) = (1 + (sqrt(3)*Pi)*cosech(sqrt(3)*Pi))/12. (End)
E.g.f.: 2*exp(x)*(3 + x + x^2). - Elmo R. Oliveira, Jan 25 2025

Extensions

Corrected and extended by Bruno Berselli, Mar 11 2015

A259054 a(n) = 4*n^2 - 4*n + 19, n >= 1.

Original entry on oeis.org

19, 27, 43, 67, 99, 139, 187, 243, 307, 379, 459, 547, 643, 747, 859, 979, 1107, 1243, 1387, 1539, 1699, 1867, 2043, 2227, 2419, 2619, 2827, 3043, 3267, 3499, 3739, 3987, 4243, 4507, 4779, 5059, 5347, 5643, 5947, 6259, 6579, 6907, 7243, 7587, 7939, 8299, 8667, 9043, 9427, 9819
Offset: 1

Views

Author

Keywords

Comments

a(n) gives twice the inverse radius of the circles touching the large Arbelos (2/3,1/3) circle (radius 1) and the n-th and (n-1)-th circles of the counterclockwise Pappus chain.
For twice the curvatures (inverse radii) of the counterclockwise Pappus chain of the (2/3,1/3) arbelos see A114949, also for the MathWorld link to Pappus chain.
For the small curvatures touching the left circle of the (2/3,1/3) arbelos and the n-th and (n-1)-st circles of the counterclockwise Pappus chain see A259555.
The curvatures of the circles can be computed with Descartes' three (actually 5) circle theorem. See A259555 for links to Descartes' theorem.

Crossrefs

Programs

Formula

a(n) = 4*n^2 - 4*n + 19, n >= 1.
O.g.f.: x*(19 - 30*x + 19*x^2)/(1 - x)^3.
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(4*x^2 + 19) - 19.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

Extensions

Terms a(37) and beyond from Andrew Howroyd, May 01 2020

A285766 Maximum spillway height for a zero or one bend minimal area lake in a number square.

Original entry on oeis.org

0, 0, 6, 10, 15, 22, 31, 42, 55, 70, 87, 106, 127, 150, 175, 202, 231, 262, 295, 330, 367, 406, 447, 490, 535, 582, 631, 682, 735, 790, 847, 906, 967, 1030, 1095, 1162, 1231, 1302, 1375, 1450, 1527, 1606, 1687, 1770, 1855, 1942, 2031, 2122, 2215, 2310, 2407
Offset: 0

Views

Author

Craig Knecht, May 04 2017

Keywords

Comments

The water retention model for mathematical surfaces led to definitions for a lake and a pond. These lakes and ponds divide the square up in interesting ways. This sequence looks at the spillway heights in zero or one bend minimal area lakes.
A lake has dimensions of (n-2) X (n-2) when the square is n X n. All other water retaining areas are ponds.
A number square contains the numbers 1 to n^2 without repeats.
The larger terms are a(n)= n^2+6 or A114949.

Examples

			For the 4 X 4 square a example of a smallest lake is shown. The values 1,2,3 form the lake. The pathway of least resistance off the square is the spillway value 10.
   ( 4  16  15   5)
   (10   1   2  14)
   ( 6  11   3  13)
   ( 7   8  12   9)
		

Crossrefs

Formula

Conjectures from Colin Barker, May 07 2017: (Start)
G.f.: x^2*(6 - 8*x + 3*x^2 + x^3) / (1 - x)^3.
a(n) = 7 - 2*n + n^2 for n>2.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>5.
(End)

A357995 Frobenius number for A = (n, n+1^2, n+2^2, n+3^2, ...) for n>=2.

Original entry on oeis.org

1, 5, 11, 13, 11, 20, 31, 24, 27, 29, 43, 37, 49, 52, 63, 58, 69, 53, 75, 61, 65, 84, 95, 98, 85, 96, 107, 115, 88, 121, 127, 122, 130, 136, 139, 134, 145, 148, 159, 151, 154, 157, 171, 174, 169, 180, 191, 194, 178, 181, 203, 198, 201, 212, 223, 210, 221, 232, 235, 214
Offset: 2

Views

Author

Michel Marcus, Oct 23 2022

Keywords

Crossrefs

Programs

  • Python
    def A357995(n):
        a, b = set([0]), set(range(1,n**2))
        for m in [n+k**2 for k in range(n+1)]:
            d=m
            while d < n**2:
                c2 = set([x for x in b if x-d in a])
                a |= c2 ; b -= c2 ; d*=2
        return max(b) # Bert Dobbelaere, Oct 30 2022

Extensions

More terms from Bert Dobbelaere, Oct 30 2022

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

Views

Author

Boris Putievskiy, Mar 11 2013

Keywords

Comments

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);
. . .

Examples

			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;
  ...
		

Crossrefs

Programs

  • Python
    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)

Formula

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).

A214871 Natural numbers placed in table T(n,k) layer by layer. The order of placement - T(n,n), T(1,n), T(n,1), T(2,n), T(n,2),...T(n-1,n), T(n,n-1). Table T(n,k) read by antidiagonals.

Original entry on oeis.org

1, 3, 4, 6, 2, 7, 11, 8, 9, 12, 18, 13, 5, 14, 19, 27, 20, 15, 16, 21, 28, 38, 29, 22, 10, 23, 30, 39, 51, 40, 31, 24, 25, 32, 41, 52, 66, 53, 42, 33, 17, 34, 43, 54, 67, 83, 68, 55, 44, 35, 36, 45, 56, 69, 84, 102, 85, 70, 57, 46, 26, 47, 58, 71, 86, 103, 123
Offset: 1

Views

Author

Boris Putievskiy, Mar 11 2013

Keywords

Comments

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(2,2), T(1,2), T(2,1);
. . .
T(n,n), T(1,n), T(n,1), T(2,n), T(n,2),...T(n-1,n), T(n,n-1);
. . .

Examples

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

Crossrefs

Cf. A060734, A060736, A185725, A213921, A213922; table T(n,k) contains: in rows A059100, A087475, A114949, A189833, A114948, A114962; in columns A117950, A117951, A117619, A189834, A189836; the main diagonal is A002522.

Programs

  • Python
    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-1)**2+1
    if i > j:
       result=(i-1)**2+2*j+1
    if i < j:
       result=(j-1)**2+2*i

Formula

As table
T(n,k) = (n-1)^2+1, if n=k;
T(n,k) = (n-1)^2+2*k+1, if n>k;
T(n,k) = (k-1)^2+2*n, if n
As linear sequence
a(n) = (i-1)^2+1, if i=j;
a(n) = (i-1)^2+2*j+1, if i>j;
a(n) = (j-1)^2+2*i, 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).

A255842 a(n) = 2*n^2 + 12.

Original entry on oeis.org

12, 14, 20, 30, 44, 62, 84, 110, 140, 174, 212, 254, 300, 350, 404, 462, 524, 590, 660, 734, 812, 894, 980, 1070, 1164, 1262, 1364, 1470, 1580, 1694, 1812, 1934, 2060, 2190, 2324, 2462, 2604, 2750, 2900, 3054, 3212, 3374, 3540, 3710, 3884, 4062, 4244, 4430
Offset: 0

Author

Avi Friedlich, Mar 08 2015

Keywords

Comments

This is the case k=6 of the form (n + sqrt(k))^2 + (n - sqrt(k))^2. Also, it is noted that a(n)*n = (n + sqrt(2))^3 + (n - sqrt(2))^3.
Equivalently, numbers m such that 2*m - 24 is a square.
For n = 0..10, a(n) - 1 is prime (see A092968).

Crossrefs

Cf. A016825 (first differences), A092968, A114949.
Subsequence of A047238 and A047406.
Cf. similar sequences listed in A255843.

Programs

  • Magma
    [2*n^2+12: n in [0..50]];
  • Mathematica
    Table[2 n^2 + 12, {n, 0, 50}]
  • PARI
    vector(50, n, n--; 2*n^2+12)
    
  • Sage
    [2*n^2+12 for n in (0..50)]
    

Formula

G.f.: 2*(6 - 11*x + 7*x^2)/(1 - x)^3.
a(n) = a(-n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
a(n) = 2*A114949(n).
From Amiram Eldar, Mar 28 2023: (Start)
Sum_{n>=0} 1/a(n) = (1 + sqrt(6)*Pi*coth(sqrt(6)*Pi))/24.
Sum_{n>=0} (-1)^n/a(n) = (1 + sqrt(6)*Pi*cosech(sqrt(6)*Pi))/24. (End)
E.g.f.: 2*exp(x)*(6 + x + x^2). - Elmo R. Oliveira, Jan 24 2025

Extensions

Edited by Bruno Berselli, Mar 11 2015

A259055 a(n) = 9*n^2 + 18*n + 7.

Original entry on oeis.org

7, 34, 79, 142, 223, 322, 439, 574, 727, 898, 1087, 1294, 1519, 1762, 2023, 2302, 2599, 2914, 3247, 3598, 3967, 4354, 4759, 5182, 5623, 6082, 6559, 7054, 7567, 8098, 8647, 9214, 9799, 10402, 11023, 11662, 12319, 12994, 13687, 14398, 15127
Offset: 0

Author

Keywords

Comments

a(n) gives twice the curvature of the n-th circle touching the two semicircles of the (2/3,1/3) arbelos and the (n-1)-th circle, with input circle of twice the curvature a(0) = A114949(1) = 7 (referring to the second circle of the counterclockwise Pappus chain).

Crossrefs

Programs

Formula

a(n) = 9*(n+1)^2 - 2, n >= 0.
O.g.f.: (-2*x^2+13*x+7)/(1-x)^3.
Recurrence: a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), n >= 3, with a(0)=7, a(1)=34, and a(2)=79.
Descartes' three (actually five) circle theorem (see links) leads to a nonlinear recurrence for twice the curvatures: a(n) = 2*(3 + 3/2) + a(n-1) + 4*sqrt((3 + 3/2)*a(n-1)/2 + 9/2) = 9 + a(n-1) + 6*sqrt(a(n-1) + 2), with input a(0) = 7 = 2*A114949(1). This leads to a quadratic equation with the relevant solution a(n) = 9*n^2 + 18*n + 7.
E.g.f.: exp(x)*(9*x*(x + 3) + 7). - Elmo R. Oliveira, Oct 20 2024
Previous Showing 11-18 of 18 results.