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.

Showing 1-4 of 4 results.

A167268 Janet's sequence: Number of elements for each successively filled electronic subshell of an atom.

Original entry on oeis.org

2, 2, 6, 2, 6, 2, 10, 6, 2, 10, 6, 2, 14, 10, 6, 2, 14, 10, 6, 2, 18, 14, 10, 6, 2, 18, 14, 10, 6, 2, 22, 18, 14, 10, 6, 2, 22, 18, 14, 10, 6, 2, 26, 22, 18, 14, 10, 6, 2, 26, 22, 18, 14, 10, 6, 2, 30, 26, 22, 18, 14, 10, 6, 2, 30, 26, 22, 18, 14, 10, 6
Offset: 1

Views

Author

Paul Curtz, Oct 31 2009

Keywords

Comments

The s-block is placed at
* the left, except row 1 (for Mendeleev-Moseley-Seaborg periodic table)
* the right (for Janet periodic table)
The number of elements in each block of Janet's periodic table are
s-block (1s to 8s): 8*2 = 16 elements;
p-block (2p to 7p): 6*6 = 36 elements;
d-block (3d to 6d): 4*10 = 40 elements;
f-block (4f to 5f): 2*14 = 28 elements.
The atomic numbers of elements in each block of Janet's periodic table are
s-block: 1,2, 3,4, 11,12, 19,20, 37,38, 55,56, 87,88, 119,120 (cf. A160914)
p-block: 5..10, 13..18, 31..36, 49..54, 81..86, 113..118 (cf. A138469)
d-block: 21..30, 39..48, 71..80, 103..112 (cf. A199934)
f-block: 57..70, 89..102 (cf. A217923).
The number of elements for each successive filled subshell are
s-block (1s): a(0)=2 terms (for H and He, i.e., 1 and 2);
s-block (2s): a(1)=2 terms (for Li and Be, i.e., 3 and 4);
p-block (2p): a(2)=6 terms (for B,C,N,O,F,Ne, i.e., 5 to 10);
s-block (3s): a(3)=2 terms (for Na,Mg, i.e., 11 and 12);
p-block (3p): a(4)=6 terms (for Al,Si,P,S,Cl,Ar, i.e., 13 to 18);
...
Reference, 2 leaflet 2, with Janet form (5). Extended.
From Daniel Forgues, May 09 2011: (Start)
Janet also envisaged an 'element zero' ('neutronium'?) - whose 'atom' would consist of two neutrons (and thus zero electrons, which would give a(0) = 0) and he speculated that this would be the link to a mirror-image table of elements with negative atomic numbers - in effect anti-matter (which would give a(-n) = -a(n), since positrons are negated electrons).
Maximum number of electrons for successive subshells of each shell of an atom, in the building up order (per aufbau principle and Madelung's rule).
Every term is twice an odd number since each filled subshell in block l has m going from -l to + l (2l+1 values,) each with 2 electrons (spin +1/2 and spin -1/2).
Blocks:
l=0: s (2 electrons) (first subshell of a shell, new period of Mendeleev's table)
l=1: p (6 electrons) (except for first shell, last subshell of a shell)
l=2: d (10 electrons)
l=3: f (14 electrons)
l=4: g (18 electrons)
...
l=l: ... (2*(2l+1) electrons)
The first subshell of the k-th shell has l = 0, k >= 1.
The second subshell of the k-th shell has l = floor(k/2), k >= 2.
The following subshells of the k-th shell have l decrementing down to 1, k >= 2.
(End)
Concatenation of finite arithmetic sequences, each followed by 2: { }, 2, { }, 2, {6}, 2, {6}, 2, {10, 6}, 2, {10, 6}, 2, {14, 10, 6}, 2, {14, 10, 6}, 2, {18, 14, 10, 6}, 2, {18, 14, 10, 6}, 2, ... - Daniel Forgues, May 15 2011
Named after the French mathematician Charles Janet (1849-1932). - Amiram Eldar, Jun 22 2021

References

  • Charles Janet, Considérations sur la structure du noyau de l'atome, N 5, Décembre 1929, Beauvais, 2+45 pages, 4 leaflets.

Crossrefs

Programs

  • Mathematica
    row[n_] := Range[2, 4*Floor[n/2] + 2, 4]; Flatten[ Table[ row[n] // Reverse, {n, 0, 15}]] (* Jean-François Alcover, Oct 02 2012 *)
  • Python
    from mpmath.libmp import sqrtrem
    def a(n):
        s, r = sqrtrem(n)
        return 4 * (-n % (s + (r>s))) + 2
    # Christoph B. Kassir, Apr 07 2022

Formula

a(n) = 4*((-n) mod round(sqrt(n))) + 2. - Jon E. Schoenfield, Sep 08 2013
a(n) = 4*A216607(n) + 2. - Szymon Lukaszyk, Oct 27 2023

Extensions

Edited by Daniel Forgues, May 09 2011

A384688 Runs of t in the range 0 <= t <= k and the same parity as k, for successive k >= 0.

Original entry on oeis.org

0, 1, 0, 2, 1, 3, 0, 2, 4, 1, 3, 5, 0, 2, 4, 6, 1, 3, 5, 7, 0, 2, 4, 6, 8, 1, 3, 5, 7, 9, 0, 2, 4, 6, 8, 10, 1, 3, 5, 7, 9, 11, 0, 2, 4, 6, 8, 10, 12, 1, 3, 5, 7, 9, 11, 13, 0, 2, 4, 6, 8, 10, 12, 14, 1, 3, 5, 7, 9, 11, 13, 15, 0, 2, 4, 6, 8, 10, 12, 14, 16
Offset: 0

Views

Author

Kevin Ryde, Jun 07 2025

Keywords

Comments

The corresponding k is A055086(n), or k+1 = A000267(n).
A run is 0, 2, 4, ..., k when k even, or 1, 3, 5, ..., k when k odd, and has length floor(k/2) + 1.
Runs start at quarter squares n = A002620(k+1), with those beginning 0 at oblong numbers n = A002378(i) and those starting 1 at the squares n = (i+1)^2 (for i >= 0 in both cases).
Starts to differ from A025643 at n=109.

Examples

			Runs and their corresponding k = A055086(n) begin,
  n          = 0  1  2    4    6      9
  a(n)       = 0, 1, 0,2, 1,3, 0,2,4, 1,3,5, ...
  A055086(n) = 0, 1, 2,2, 3,3, 4,4,4, 5,5,5, ...
		

Crossrefs

Cf. A002620, A002378 (indices of 0's), A000290 (indices of 1's).

Programs

  • Mathematica
    ClearAll[a] a[n_Integer]:=Module[{s,r},s=Floor[Sqrt[n]]; r=n-s^2; If[rVincenzo Librandi, Jul 06 2025 *)
  • PARI
    a(n) = my(r,s=sqrtint(n,&r)); if(r
    				

Formula

a(n) = 2*r+1 if r < s or a(n) = 2*(r-s) otherwise, where square root and remainder n = s^2 + r being s=A000196(n), r=A053186(n).
a(n) = ceiling(A053186(4*n+1) / 2).
a(n) = A055086(n) - 2*A216607(n+1).
a(n) = 2*A055087(n) + A079813(n+1).

A232091 Smallest square or promic (oblong) number greater than or equal to n.

Original entry on oeis.org

0, 1, 2, 4, 4, 6, 6, 9, 9, 9, 12, 12, 12, 16, 16, 16, 16, 20, 20, 20, 20, 25, 25, 25, 25, 25, 30, 30, 30, 30, 30, 36, 36, 36, 36, 36, 36, 42, 42, 42, 42, 42, 42, 49, 49, 49, 49, 49, 49, 49, 56, 56, 56, 56, 56, 56, 56, 64, 64, 64, 64, 64, 64, 64, 64, 72, 72, 72, 72, 72, 72, 72, 72, 81
Offset: 0

Views

Author

L. Edson Jeffery, Nov 18 2013

Keywords

Comments

Result attributed to the students Daring, et al., in the links section.
a(n) appears in floor(sqrt(a(n))) = A000194(n) successive terms.
Counting successive equal terms give sequence: 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ... (see A008619). - Michel Marcus, Jan 10 2014

Crossrefs

Cf. A000290 (squares), A002378 (promic or oblong numbers), A002620 (A000290 union A002378).

Programs

  • Magma
    [(Ceiling(n /Ceiling(Sqrt(n)))*Ceiling(Sqrt(n))): n in [1..80]]; // Vincenzo Librandi, Jun 22 2015
  • Mathematica
    Join[{0}, Table[Ceiling[n/Ceiling[Sqrt[n]]] Ceiling[Sqrt[n]], {n, 100}]] (* Alonso del Arte, Nov 18 2013 *)
  • PARI
    a(n)=my(t=sqrtint(n-1)+1);t*((n-1)\t+1) \\ Charles R Greathouse IV, Nov 18 2013
    

Formula

a(n) = ceiling(n/ceiling(sqrt(n)))*ceiling(sqrt(n)).
a(n) = min(k : k >= n, k in A002620).
a(k^2) = k^2; a(k*(k+1)) = k*(k+1).
It appears that a(n) = A216607(n) + n. (Verified for all n<10^9 by Lars Blomberg, Jan 09 2014.) This conjecture now follows from a proof given by David Applegate, Jan 10 2014 (see [Applegate]).
a(n) = min(A048761(n), A259225(n)). - Michel Marcus, Jun 22 2015
Sum_{n>=1} 1/a(n)^2 = 2 - Pi^2/6 + zeta(3). - Amiram Eldar, Aug 16 2022

Extensions

Extended by Charles R Greathouse IV, Nov 18 2013
a(0)=0 prepended by Michel Marcus, Jun 22 2015

A366421 a(n) is the floor of the n-th improper fraction (including the ones of the type n/n) sorted by increasing numerator+denominator, then by numerator.

Original entry on oeis.org

1, 2, 1, 3, 1, 4, 1, 2, 5, 1, 2, 6, 1, 1, 3, 7, 1, 2, 3, 8, 1, 1, 2, 4, 9, 1, 1, 2, 4, 10, 1, 1, 2, 3, 5, 11, 1, 1, 2, 3, 5, 12, 1, 1, 1, 2, 3, 6, 13, 1, 1, 2, 2, 4, 6, 14, 1, 1, 1, 2, 3, 4, 7, 15, 1, 1, 1, 2, 3, 4, 7, 16, 1, 1, 1, 2, 2, 3, 5, 8, 17, 1, 1, 1, 2, 2, 3, 5, 8, 18
Offset: 1

Views

Author

Mariano Masiello, Nov 16 2023

Keywords

Examples

			The first such fraction is 1/1, followed by 2/1, 2/2, 3/1, 3/2 and so on, then a(1)=[1/1]=1, a(2)=[2/1]=2, a(3)=[2/2]=1, a(4)=[3/1]=3, a(5)=[3/2]=1 and so on, where [.] is the floor function.
Terms begin:
  [1/1]=1,
  [2/1]=2,
  [2/2]=1, [3/1]=3,
  [3/2]=1, [4/1]=4,
  [3/3]=1, [4/2]=2, [5/1]=5,
  [4/3]=1, [5/2]=2, [6/1]=6,
  ...
		

Formula

a(n) = floor[A138099(n)/(A216607(n)+1)]
Showing 1-4 of 4 results.