A061037 -id:A061037 - cp's OEIS frontend

A136017 a(n) = 36n^2 - 1.

35, 143, 323, 575, 899, 1295, 1763, 2303, 2915, 3599, 4355, 5183, 6083, 7055, 8099, 9215, 10403, 11663, 12995, 14399, 15875, 17423, 19043, 20735, 22499, 24335, 26243, 28223, 30275, 32399, 34595, 36863, 39203, 41615, 44099, 46655, 49283, 51983

Offset: 1

Artur Jasinski, Dec 10 2007

The least common multiple of 6*n+1 and 6*n-1. - Colin Barker, Feb 11 2017

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
X. Gourdon and P. Sebah, Collection of series for Pi
Eric Weisstein's World of Mathematics, Pell Equation
Edward Everett Withford, Pell Equation
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A088878, A023208, A136016, A061037, A016910.

[36*n^2 - 1: n in [1..50]]; // Vincenzo Librandi, Jul 09 2012

Mathematica
```
Table[36n^2 - 1, {n, 1, 100}]
```

a(n)=36*n^2-1 \\ Jaume Oliver Lafont, Oct 20 2009

O.g.f.: x*(-35-38*x+x^2)/(-1+x)^3 = 1-35/(-1+x)-108/(-1+x)^2-72/(-1+x)^3. - R. J. Mathar, Dec 12 2007
a(n) = A061037(12n+10)=(6n-1)*(6n+1). - Paul Curtz, Sep 25 2008
Sum_{k>=1} (-1)^(k+1)/a(k) = (Pi-3)/6. - Jaume Oliver Lafont, Oct 20 2009
E.g.f.: 1 + (36 x^2 + 26 x - 1) exp(x). - Robert Israel, Jun 09 2016
Product_{n >= 1} A016910(n) / a(n) = Pi / 3. - Fred Daniel Kline, Jun 09 2016
Sum_{n>=1} 1/a(n) = 1/2 - sqrt(3)*Pi/12. - Amiram Eldar, Jun 27 2020

A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

Original entry on oeis.org

1, 2, 2, 1, 2, 4, 4, 4, 8, 8, 1, 4, 8, 4, 16, 2, 2, 1, 8, 32, 32, 1, 2, 4, 4, 16, 32, 64, 8, 8, 16, 16, 64, 64, 128, 128, 1, 8, 16, 8, 32, 64, 128, 32, 256, 2, 2, 8, 16, 64, 64, 128, 64, 512, 512, 1, 2, 4, 8, 32, 64, 128, 16, 128, 512, 1024

Offset: 0

Paul Curtz, Jan 18 2013

1/2^n and successive rows are
1,       1/2,   1/4,   1/8,  1/16,  1/32,   1/64, 1/128, 1/256,...
1/2,     1/2,   3/8,   1/4,  5/32,  3/32,  7/128,  1/32,...       = A000265/A075101, the Oresme numbers n/2^n. Paul Curtz, Jan 18 2013 and May 11 2016
0,       1/4,   3/8,   3/8,  5/16, 15/64, 21/128,...              = (0 before A069834)/new,
-1/4,   -1/4,     0,   1/4, 25/64, 27/64,...
0,      -1/2,  -3/4, -9/16, -5/32,...
1/2,     1/2, -9/16, -13/8,...
0,      17/8, 51/16,...
-17/8, -17/8,...
0
The first column is A198631/(A006519?), essentially the fractional Euler numbers 1, -1/2, 0, 1/4, 0,...  in A060096.
Numerators b(n): 1, 1, 1, 0, 1, 1, -1, 1, 3, 1, ... .
Coll(n+1) - 2*Coll(n) = -1/2, -5/8, -1/2, -11/32, -7/32, -17/128, -5/64, -23/512, ... = -A075677/new, from Collatz problem.
There are three different Bernoulli numbers:
The first Bernoulli numbers are  1, -1/2, 1/6, 0,... = A027641(n)/A027642(n).
The second Bernoulli numbers are 1,  1/2, 1/6, 0,... = A164555(n)/A027642(n). These are the binomial transform of the first one.
The third Bernoulli numbers are  1,   0,  1/6, 0,... = A176327(n)/A027642(n), the half sum. Via A177427(n) and A191567(n), they yield the Balmer series A061037/A061038.
There are three different fractional Euler numbers:
1) The first are  1, -1/2, 0, 1/4, 0, -1/2,... in A060096(n).
Also Akiyama-Tanigawa algorithm for ( 1, 3/2, 7/4, 15/8, 31/16, 63/32,... = A000225(n+1)/A000079(n) ).
2) The second are 1, 1/2, 0, -1/4, 0,  1/2,... , mentioned by Wolfdieter Lang in A198631(n).
3) The third are  0, 1/2, 0, -1/4, 0,  1/2,... , half difference of 2) and 1).
Also Akiyama-Tanigawa algorithm for ( 0, -1/2, -3/4, -7/8, -15/16, -31/32,... =  A000225(n)/A000079(n) ). See A097110(n).

			Triangle begins:
  1,
  2, 2,
  1, 2,  4,
  4, 4,  8,  8,
  1, 4,  8,  4, 16,
  2, 2,  1,  8, 32, 32,
  1, 2,  4,  4, 16, 32,  64,
  8, 8, 16, 16, 64, 64, 128, 128,
  ...

G. C. Greubel, Table of n, a(n) for n = 0..5049
A. F. Horadam, Oresme Numbers, Fibonacci Quarterly, 12, #3, 1974, pp. 267-271.

Cf. Second Bernoulli numbers A164555(n)/A027642(n) via Akiyama-Tanigawa algorithm for 1/(n+1), A272263.

max = 10; t[0, k_] := 1/2^k; t[n_, k_] := t[n, k] = (k + 1)*(t[n - 1, k] - t[n - 1, k + 1]); denoms = Table[t[n, k] // Denominator, {n, 0, max}, {k, 0, max - n}]; Table[denoms[[n - k + 1, k]], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Feb 05 2013 *)

A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

Original entry on oeis.org

1, 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 845, 226, 965, 257, 1093, 290, 1229, 325, 1373, 362, 1525, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677

Offset: 0

Paul Curtz, Aug 15 2015

Using (n+sqrt(4+n^2))/2, after the integer 1 for n=0, the reduced metallic means are b(1) = (1+sqrt(5))/2, b(2) = 1+sqrt(2), b(3) = (3+sqrt(13))/2, b(4) = 2+sqrt(5), b(5) = (5+sqrt(29))/2, b(6) = 3+sqrt(10), b(7) = (7+sqrt(53))/2, b(8) = 4+sqrt(17), b(9) = (9+sqrt(85))/2, b(10) = 5+sqrt(26), b(11) = (11+sqrt(125))/2 = (11+5*sqrt(5))/2, ... . The last value yields the radicals in a(n) or A013946.
b(2) = 2.41, b(3) = 3.30, b(4) = 4.24, b(5) = 5.19 are "good" approximations of fractal dimensions corresponding to dimensions 3, 4, 5, 6: 2.48, 3.38, 4.33 and 5.45 based on models. See "Arbres DLA dans les espaces de dimension supérieure: la théorie des peaux entropiques" in Queiros-Condé et al. link. DLA: beginning of the title of the Witten et al. link.
Consider the symmetric array of the half extended Rydberg-Ritz spectrum of the hydrogen atom:
   0,    1/0,     1/0,     1/0,     1/0,      1/0,      1/0,     1/0, ...
-1/0,      0,     3/4,     8/9,   15/16,    24/25,    35/36,   48/49, ...
-1/0,   -3/4,       0,    5/36,    3/16,   21/100,      2/9,  45/196, ...
-1/0,   -8/9,   -5/36,       0,   7/144,   16/225,     1/12,  40/441, ...
-1/0, -15/16,   -3/16,  -7/144,       0,    9/400,    5/144,  33/784, ...
-1/0, -24/25, -21/100, -16/225,  -9/400,        0,   11/900, 24/1225, ...
-1/0, -35/36,    -2/9,   -1/12,  -5/144,  -11/900,        0, 13/1764, ...
-1/0, -48/49, -45/196, -40/441, -33/784, -24/1225, -13/1764,       0, ... .
The numerators are almost A165795(n).
Successive rows: A000007(n)/A057427(n), A005563(n-1)/A000290(n), A061037(n)/A061038(n), A061039(n)/A061040(n), A061041(n)/A061042(n), A061043(n)/A061044(n), A061045(n)/A061046(n), A061047(n)/A061048(n), A061049(n)/A061050(n).
A144433(n) or A195161(n+1) are the numerators of the second upper diagonal (denominators: A171522(n)).
c(n+1) = a(n) + a(n+1) = 6, 7, 15, 18, 34, 39, 63, 70, 102, 111, ... .
c(n+3) - c(n+1) = 9, 11, 19, 21, 29, 31, ... = A090771(n+2).
The final digit of a(n) is neither 4 nor 8. - Paul Curtz, Jan 30 2019

Colin Barker, Table of n, a(n) for n = 0..1000
Diogo Queiros-Condé, Jean Chaline, and Jacques Dubois, Le monde des fractales La Nature trans-échelles, 478p., ellipses, Paris, 2015, page 220.
T. A. Witten, Jr. and L. M. Sander, Diffusion-Limited Aggregation, a Kinetic Critical Phenomenom, Phys. Rev. Lett., Vol. 47 (Nov 09 1981), pp. 1400-1403.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A000007, A000290, A001622, A002522, A005563, A010685, A010698, A013946, A014176, A057427, A061035-A061050, A078370, A087475, A090771, A098316, A098317, A098318, A144433, A168077, A171621, A176398, A176439, A176458, A176522, A176537, A195161.

Cf. A069894.

[Numerator(1+n^2/4): n in [0..60]]; // Vincenzo Librandi, Aug 15 2015

A261327:=n->numer((4 + n^2)/4); seq(A261327(n), n=0..60); # Wesley Ivan Hurt, Aug 15 2015

LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 60] (* Vincenzo Librandi, Aug 15 2015 *)
a[n_] := (n^2 + 4) / 4^Mod[n + 1, 2]; Table[a[n], {n, 0, 52}] (* Peter Luschny, Mar 18 2022 *)

vector(60, n, n--; numerator(1+n^2/4)) \\ Michel Marcus, Aug 15 2015

Vec((1+5*x-x^2-2*x^3+2*x^4+5*x^5)/(1-x^2)^3 + O(x^60)) \\ Colin Barker, Aug 15 2015

a(n)=if(n%2,n^2+4,(n/2)^2+1) \\ Charles R Greathouse IV, Oct 16 2015

[(n*n+4)//4**((n+1)%2) for n in range(60)] # Gennady Eremin, Mar 18 2022

[numerator(1+n^2/4) for n in (0..60)] # G. C. Greubel, Feb 09 2019

a(n) = numerator(1 + n^2/4). (Previous name.) See A010685 (denominators).
a(2*k) = 1 + k^2.
a(2*k+1) = 5 + 4*k*(k+1).
a(2*k+1) = 4*a(2*k) + 4*k + 1.
a(4*k+2) = A069894(k). - Paul Curtz, Jan 30 2019
a(-n) = a(n).
a(n+2) = a(n) + A144433(n) (or A195161(n+1)).
a(n) = A168077(n) + period 2: repeat 1, 4.
a(n) = A171621(n) + period 2: repeat 2, 8.
From Colin Barker, Aug 15 2015: (Start)
a(n) = (5 - 3*(-1)^n)*(4 + n^2)/8.
a(n) = n^2/4 + 1 for n even;
a(n) = n^2 + 4 for n odd.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.
G.f.: (1 + 5*x - x^2 - 2*x^3 + 2*x^4 + 5*x^5)/ (1 - x^2)^3. (End)
E.g.f.: (5/8)*(x^2 + x + 4)*exp(x) - (3/8)*(x^2 - x + 4)*exp(-x). - Robert Israel, Aug 18 2015
Sum_{n>=0} 1/a(n) = (4*coth(Pi)+tanh(Pi))*Pi/8 + 1/2. - Amiram Eldar, Mar 22 2022

New name by Peter Luschny, Mar 18 2022

A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

Original entry on oeis.org

-1, 0, -3, 2, 3, 6, 5, 12, 15, 20, 21, 30, 35, 42, 45, 56, 63, 72, 77, 90, 99, 110, 117, 132, 143, 156, 165, 182, 195, 210, 221, 240, 255, 272, 285, 306, 323, 342, 357, 380, 399, 420, 437, 462, 483, 506, 525, 552, 575, 600, 621, 650, 675, 702, 725, 756, 783, 812, 837, 870, 899, 930, 957, 992, 1023, 1056, 1085, 1122, 1155, 1190

Offset: 0

Paul Curtz, Jul 11 2012

Let a(n)/A000290(n) = [-1/0, 0/1, -3/4, 2/9, 3/16, 6/25, 5/36, 12/49, 15/64, 20/81, 21/100, 30/121, ...] = a(n)/b(n) (say).
Then b(n)-4*a(n)=4, 1, 16, 1 (period of length 4).
Permutation from a(n) to A061037(n): 1, 3, 2, 7, 5, 11, 4, 15, 9, 19, 6, ... = shifted A145979 + 1.
A061037(n) - a(n) = 0, 3, -3, -3, 0, -15, 3, -33, 0 -57, 15, -87, 0, -123, ...
First 3 rows:
-1 0 -3 2  3  6  5 12 15 20 21 30 35
1 -3  5 1  3 -1  7  3  5  1  9  5  7
-4 8 -4 2 -4  8 -4  2 -4  8 -4  2 -4.
Note that the terms of a(n) increase from 12. Compare to increasing terms permutation of A061037(n): -3,-1,0,2,3,5,6,12,15, .... and A129647.
c(n) =  0, -1, 0, -1, 2, 1, 2, 1, 4, 3, 4, 3, 6, 5, 6, 5, ... (cf. A134967)
d(n) = -1,  1, 1,  3, 1, 3, 3, 5, 3, 5, 5, 7, 5, 7, 7, 9, ..., hence:
a(n) = c(n+1) * d(n+1).

G. C. Greubel, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,1,-2,1).

[(2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8: n in [0..100]]; // G. C. Greubel, Sep 19 2018

A214297 := proc(n)
    option remember;
    if n <=5 then
        op(n+1,[-1,0,-3,2,3,6]) ;
    else
        2*procname(n-1)-procname(n-2)+procname(n-4)-2*procname(n-5)+procname(n-6) ;
    end if;
end proc: # R. J. Mathar, Jun 28 2013

Table[(2 n^2 - 11 - 9 (-1)^n + 6 ((-1)^((2 n + 1 - (-1)^n)/4) + (-1)^((2 n - 1 + (-1)^n)/4)))/8, {n, 0, 69}] (* or *)
CoefficientList[Series[-(1 - 2 x + 4 x^2 - 8 x^3 + 3 x^4)/((1 - x)^2*(1 - x^4)), {x, 0, 69}], x] (* Michael De Vlieger, Mar 24 2017 *)

vector(100, n, n--; (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8) \\ G. C. Greubel, Sep 19 2018

a(k+4)  - a(k)    =  2*k + 4.
a(k+2)  - a(k-2)  =  2*k.
a(k+6)  - a(k-6)  =  6*k.
a(k+10) - a(k-10) = 10*k.
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(2*k)   = -1, -3, followed by 3, 5, 15, 21, 35, 45, ... (A142717);
a(2*k+1) = k*(k+1) (see A002378).
A198442(n) = -1,0,0,2,3,6,8,12,  minus 3 at A198442(4*n+2).
G.f. -( 1-2*x+4*x^2-8*x^3+3*x^4 )/( (1-x)^2*(1-x^4) ). - R. J. Mathar, Jul 17 2012; edited by N. J. A. Sloane, Jul 22 2012
From R. J. Mathar, Jun 28 2013: (Start)
a(4*k)   = A000466(k);
a(4*k+1) = A002943(k);
a(4*k+2) = A078371(k-1) for k>0;
a(4*k+3) = A002939(k+1). (End)
a(n) = (2*n^2-11-9*(-1)^n+6*((-1)^((2*n+1-(-1)^n)/4)+(-1)^((2*n-1+(-1)^n)/4)))/8. - Luce ETIENNE, Oct 27 2016

Edited by N. J. A. Sloane, Jul 22 2012

A142705 Numerator of 1/4 - 1/(2n)^2.

Original entry on oeis.org

0, 3, 2, 15, 6, 35, 12, 63, 20, 99, 30, 143, 42, 195, 56, 255, 72, 323, 90, 399, 110, 483, 132, 575, 156, 675, 182, 783, 210, 899, 240, 1023, 272, 1155, 306, 1295, 342, 1443, 380, 1599, 420, 1763, 462, 1935, 506, 2115, 552, 2303, 600, 2499, 650, 2703, 702

Offset: 1

Paul Curtz, Sep 24 2008

Read modulo 10 (the last digits), a sequence with period length 10 results: 0, 3, 2, 5, 6, 5, 2, 3, 0, 9. Read modulo 9, a sequence with period length 18 results.
Denominators are in A154615.
a(n) is the numerator of (n-1)*(n+1)/4. - Altug Alkan, Apr 19 2018

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

Essentially the same as A070260. Cf. A078371 (second bisection of A061037), A142888 (first differences), A154615 (denominators), A225948.

[-(3/4)*(-1)^n*n-(3/8)*(-1)^n*n^2+(5/8)*n^2+(5/4)*n: n in [0..60]]; // Vincenzo Librandi, Jul 02 2011

Numerator[Table[(1/4)*(1 - 1/n^2), {n,1,50}]] (* G. C. Greubel, Jul 20 2017 *)

for(n=1, 50, print1(numerator((1/4)*(1 - 1/n^2)), ", ")) \\ G. C. Greubel, Jul 20 2017

a(n) = if(n%2,(n^2-1)/4,n^2-1); \\ Altug Alkan, Apr 19 2018

a(n) = A061037(2*n).
a(n) = A070260(n-1), n>1.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(2^(n-1)) = a(1+A000225(n-1)) = 4^(n-1)-1 = A024036(n-1).
G.f.: x^2*(3+2x+6x^2-x^4)/(1-x^2)^3. - R. J. Mathar, Oct 24 2008
E.g.f.: 1 + (1/4)*((4*x^2 + x - 4)*cosh(x) + (x^2 + 4*x -1)*sinh(x)). - G. C. Greubel, Jul 20 2017
Sum_{n>=2} 1/a(n) = 3/2. - Amiram Eldar, Aug 11 2022

Edited by R. J. Mathar, Oct 24 2008

A144433 Multiples of 8 interleaved with the sequence of odd numbers >= 3.

Original entry on oeis.org

8, 3, 16, 5, 24, 7, 32, 9, 40, 11, 48, 13, 56, 15, 64, 17, 72, 19, 80, 21, 88, 23, 96, 25, 104, 27, 112, 29, 120, 31, 128, 33, 136, 35, 144, 37, 152, 39, 160, 41, 168, 43, 176, 45, 184, 47, 192, 49, 200, 51, 208, 53, 216, 55, 224, 57, 232, 59, 240, 61, 248, 63, 256, 65, 264

Offset: 1

Paul Curtz, Oct 04 2008

For n >= 2, these are the numerators of 1/n^2 - 1/(n+1)^2: A061037(4), A061039(5), A061041(6), A061043(7), A061045(8), A061047(9), A061049(10), etc.

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

Cf. A120070.

[5+3/2*(-1)^(n-1)*(n-1)+3*(-1)^(n-1)+5/2*(n-1): n in [1..70]]; // Vincenzo Librandi, Jul 30 2011

A144433:=n->(n+1)*4^(n mod 2); seq(A144433(n), n=1..100); # Wesley Ivan Hurt, Nov 27 2013

Table[(n + 1)* 4^Mod[n, 2], {n, 100}] (* Wesley Ivan Hurt, Nov 27 2013 *)

x='x+O('x^50); Vec( x*(8+3*x-x^3)/((1-x)^2*(1+x)^2)) \\ G. C. Greubel, Sep 19 2018

a(2*n+1) = A008590(n+1), a(2*n) = A005408(n).
a(2*n+1) + a(2*n+2) = A017281(n+1).
From R. J. Mathar, Apr 01 2009: (Start)
a(n) = 2*a(n-2) - a(n-4).
G.f.: x*(8+3*x-x^3)/((1-x)^2*(1+x)^2). (End)
a(n) = (n + 1) * 4^(n mod 2). - Wesley Ivan Hurt, Nov 27 2013

Edited by R. J. Mathar, Apr 01 2009

A172157 Triangle T(n,m) = numerator of 1/n^2 - 1/m^2, read by rows, with T(n,0) = -1.

Original entry on oeis.org

-1, -1, -3, -1, -8, -5, -1, -15, -3, -7, -1, -24, -21, -16, -9, -1, -35, -2, -1, -5, -11, -1, -48, -45, -40, -33, -24, -13, -1, -63, -15, -55, -3, -39, -7, -15, -1, -80, -77, -8, -65, -56, -5, -32, -17, -1, -99, -6, -91, -21, -3, -4, -51, -9, -19, -1, -120, -117, -112

Offset: 1

Paul Curtz, Jan 27 2010

The triangle obtained by negating the values of the triangle A120072 and adding a row T(n,0) = -1.

			The full array of numerators starts in row n=1 with columns m>=0 as:
-1...0...3...8..15..24..35..48..63..80..99. A005563
-1..-3...0...5...3..21...2..45..15..77...6. A061037, A070262
-1..-8..-5...0...7..16...1..40..55...8..91. A061039
-1.-15..-3..-7...0...9...5..33...3..65..21. A061041
-1.-24.-21.-16..-9...0..11..24..39..56...3. A061043
-1.-35..-2..-1..-5.-11...0..13...7...5...4. A061045
-1.-48.-45.-40.-33.-24.-13...0..15..32..51. A061047
-1.-63.-15.-55..-3.-39..-7.-15...0..17...9. A061049
The triangle is the portion below the main diagonal, left from the zeros, 0<=m<n.

G. C. Greubel, Rows n=1..100 of triangle, flattened

Cf. A172370, A174233, A165795.

T[n_, 0] := -1; T[n_, k_] := 1/n^2 - 1/k^2; Table[Numerator[T[n, k]], {n, 1, 100}, {k, 0, n - 1}] // Flatten (* G. C. Greubel, Sep 19 2018 *)

A172370 Mirrored triangle A120072 read by rows.

Original entry on oeis.org

3, 5, 8, 7, 3, 15, 9, 16, 21, 24, 11, 5, 1, 2, 35, 13, 24, 33, 40, 45, 48, 15, 7, 39, 3, 55, 15, 63, 17, 32, 5, 56, 65, 8, 77, 80, 19, 9, 51, 4, 3, 21, 91, 6, 99, 21, 40, 57, 72, 85, 96, 105, 112, 117, 120, 23, 11, 7, 5, 95, 1, 119, 1, 5, 35, 143, 25, 48, 69, 88, 105, 120, 133, 144

Offset: 2

Paul Curtz, Feb 01 2010

A table of numerators of 1/n^2 - 1/m^2 extended to negative m looks as follows, stacked such that values of common m are aligned
and the central column of -1 is defined for m=0:
.............................0..-1...0...3...8..15..24..35..48..63..80..99. A005563
.........................0..-3..-1..-3...0...5...3..21...2..45..15..77...6. A061037
.....................0..-5..-8..-1..-8..-5...0...7..16...1..40..55...8..91. A061039
.................0..-7..-3.-15..-1.-15..-3..-7...0...9...5..33...3..65..21. A061041
.............0..-9.-16.-21.-24..-1.-24.-21.-16..-9...0..11..24..39..56...3. A061043
.........0.-11..-5..-1..-2.-35..-1.-35..-2..-1..-5.-11...0..13...7...5...4. A061045
.....0.-13.-24.-33.-40.-45.-48..-1.-48.-45.-40.-33.-24.-13...0..15..32..51. A061047
.0.-15..-7.-39..-3.-55.-15.-63..-1.-63.-15.-55..-3.-39..-7.-15...0..17...9. A061049
The row-reversed variant of A120072 appears (negated) after the leftmost 0.
Equals A061035 with the first column removed. - Georg Fischer, Jul 26 2023

			The table starts
   3
   5   8
   7   3  15
   9  16  21  24
  11   5   1   2  35
  13  24  33  40  45  48
  15   7  39   3  55  15  63
  17  32   5  56  65   8  77  80
  19   9  51   4   3  21  91   6  99

G. C. Greubel, Rows n = 2..100 of triangle, flattened

Lower diagonal gives: A070262, A061037(n+2).

Cf. A061035, A172157, A165795.

[[Numerator(1/(n-k)^2 -1/n^2): k in [1..n-1]]: n in [2..20]]; // G. C. Greubel, Sep 20 2018

Table[Numerator[1/(n-k)^2 -1/n^2], {n, 2, 20}, {k, 1, n-1}]//Flatten (* G. C. Greubel, Sep 20 2018 *)

for(n=2,20, for(k=1,n-1, print1(numerator(1/(n-k)^2 -1/n^2), ", "))) \\ G. C. Greubel, Sep 20 2018

T(n,m) = numerator of 1/(n-m)^2 - 1/n^2, n >= 2, 1 <= m < n. - R. J. Mathar, Nov 23 2010

Comment rewritten and offset set to 2 by R. J. Mathar, Nov 23 2010

A225948 a(0) = -1; for n>0, a(n) = numerator(1/4 - 4/n^2).

Original entry on oeis.org

-1, -15, -3, -7, 0, 9, 5, 33, 3, 65, 21, 105, 2, 153, 45, 209, 15, 273, 77, 345, 6, 425, 117, 513, 35, 609, 165, 713, 12, 825, 221, 945, 63, 1073, 285, 1209, 20, 1353, 357, 1505, 99, 1665, 437, 1833, 30, 2009, 525, 2193, 143

Offset: 0

Paul Curtz, May 21 2013

Denominators are in A226008.
Fractions in lowest terms for n>0: -15/4, -3/4, -7/36, 0/1, 9/100, 5/36, 33/196, 3/16, 65/324, 21/100, 105/484, 2/9, 153/676, 45/196, 209/900, 15/64,...
If t(n) is the sequence with period 8: 4, 64, 16, 64, 1, 64, 16, 64, 4, 64, 16, ... (see A226044), then A226008(n) = 4*a(n) + t(n).

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

Cf. A000466, A028566, A061037, A061041, A078371, A106609, A142705, A145923, A226008.

[-1] cat [Numerator(1/4-4/n^2): n in [1..50]]; // Bruno Berselli, May 22 2013

Join[{-1}, Table[Numerator[1/4 - 4/n^2], {n, 50}]] (* Bruno Berselli, May 24 2013 *)

concat([-1], vector(100, n, numerator(1/4 - 4/n^2))) \\ G. C. Greubel, Sep 19 2018

a(n) = 3*a(n-8) -3*a(n-16) +a(n-24).
a(2n) = A061037(n), a(2n+1) = A145923(n-2) for A145923(-2)=-15, A145923(-1)=-7.
a(4n)   = A142705(n) for A142705(0)=-1, a(8n) = A000466(n);
a(4n+1) = A028566(4n-3) for A028566(-3)=-15;
a(4n+2) = A078371(n-1) for A078371(-1)=-3;
a(4n+3) = A028566(4n-1) for A028566(-1)=-7.
a(n+4)  = A106609(n) * A106609(n+8).
G.f.: -(1 +15*x +3*x^2 +7*x^3 -9*x^5 -5*x^6 -33*x^7 -6*x^8 -110*x^9 -30*x^10 -126*x^11 -2*x^12 -126*x^13 -30*x^14 -110*x^15 -3*x^16 -33*x^17 -5*x^18 -9*x^19 +7*x^21 +3*x^22 +15*x^23)/(1-x^8)^3. - Bruno Berselli, May 22 2013
a(n) = (n^2-16)*(6*cos(Pi*n/4)-54*cos(Pi*n/2)+6*cos(3*Pi*n/4)-219*(-1)^n+293)/512. - Bruno Berselli, May 22 2013
a(n+10) = a(n+2)*(n+14)/(n-2) for n=0,1 and n>2. - Bruno Berselli, May 22 2013

Edited by Bruno Berselli, May 22 2013

A126252 Wavenumbers of red, turquoise, blue, indigo and violet in the spectrum of hydrogen, as first measured by Robert Bunsen and Gustav Kirchhoff in 1859.

Original entry on oeis.org

1523310, 2056410, 2303240, 2437290, 2518130

Offset: 1

Reinhard Zumkeller, Dec 22 2006

How Johann Jakob Ballmer found his formula in 1885 by analyzing and manipulating the ratios of these data:
r(1) = a(1)/a(1) = 1,
a(2)/a(1) = 1.349961..., rounded: r(2) = 135/100 = 27/20,
a(3)/a(1) = 1.511996..., rounded: r(3) = 1512/1000 = 189/125,
a(4)/a(1) = 1.599996..., rounded: r(4) = 16/10 = 8/5,
a(5)/a(1) = 1.6530647..., r(5) = 81/49 = 2-1/(3-1/(9-1/2)), derived from a(5)/a(1) = 2-1/(3-1/(9-3095/6216)) when replacing 3095/6216 by 1/2;
the multiplication of these fractions by 5/36 is the key trick to get more handy figures to see eventually increasing squares in the denominators by an appropriate expansion:
b(1) = r(1)*5/36 = 5 / 36,
b(2) = r(2)*5/36 = 3 / 16,
b(3) = r(3)*5/36 = 21 / 100,
b(4) = r(4)*5/36 = 2 / 9,
b(5) = r(5)*5/36 = 45 / 196;
... b(1) .|.... b(2) ..|.... b(3) ..|.... b(4) ..|.... b(5),
... 5/36 .|.... 3/16 ..|... 21/100 .|.... 2/9 ...|... 45/196,
... 5/36 .|... 12/64 ..|... 21/100 .|... 32/144 .|... 45/196,
(9-4)/9*4 |(16-4)/16*4 |(25-4)/25*4 |(36-4)/36*4 |(49-4)/49*4,
this last step was the crowning achievement: the discovery of the pattern (x-y)/x*y,
b(n) = ((n+2)^2 - 4)/(4*(n+2)^2) = 1/4 - 1/(n+2)^2;
1<=n<=5: b(n) = A061037(n+2)/A061038(n+2) = A120072(n+2,2)/A120073(n+2,2).

R. Taschner, Der Zahlen gigantischer Schatten, Vieweg 2005, 137-143.

Science Trek, Balmer Formula
Eric Weisstein's World of Physics, Balmer Formula
Wikipedia, Empirical formula
Wikipedia, Johann Jakob Balmer
Wikipedia, Robert Wilhelm Bunsen
Wikipedia, Gustav Robert Kirchhoff

cp's OEIS Frontend

A136017 a(n) = 36n^2 - 1.

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A209308 Denominators of the Akiyama-Tanigawa algorithm applied to 2^(-n), written by antidiagonals.

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A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

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A214297 a(0)=-1, a(1)=0, a(2)=-3; thereafter a(n+2) - 2*a(n+1) + a(n) has period 4: repeat -4, 8, -4, 2.

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A142705 Numerator of 1/4 - 1/(2n)^2.

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A144433 Multiples of 8 interleaved with the sequence of odd numbers >= 3.

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