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.

A168342 Even atomic numbers in the Janet table of the PSE, read right to left along rows.

Original entry on oeis.org

2, 4, 12, 10, 8, 6, 20, 18, 16, 14, 38, 36, 34, 32, 30, 28, 26, 24, 22, 56, 54, 52, 50, 48, 46, 44, 42, 40, 88, 86, 84, 82, 80, 78, 76, 74, 72, 70, 68, 66, 64, 62, 60, 58, 120, 118, 116, 114, 112, 110, 108, 106, 104, 102, 100, 98, 96, 94, 92, 90
Offset: 1

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Author

Paul Curtz, Nov 23 2009

Keywords

Comments

In the Janet arrangement, the elements appear in groups of twice 2, twice 8,... twice 2*k^2, and are here right-aligned:
...............................1,.2;
...............................3,.4;
.............5,.6,.7,.8,.9,10,11,12;
............13,14,15,16,17,18,19,20;
...28,39.30,31,32,33,34,35,36,37,38;
The even numbers in the table are read top-down, right-to-left and entered into the sequence (which, in consequence, is a permutation of the even numbers.)

Examples

			Skipping each second (i.e., each odd) element in the table, the result is
2;
4;
12,10,8,6;
20,18,16,14;
38,36,34,32,30,28,..
counting down the even numbers restarting at indices provided by A168380.
		

Crossrefs

A338429 Maximum number of copies of a 1234 permutation pattern in an alternating (or zig-zag) permutation of length n + 5.

Original entry on oeis.org

4, 8, 28, 48, 104, 160, 280, 400, 620, 840, 1204, 1568, 2128, 2688, 3504, 4320, 5460, 6600, 8140, 9680, 11704, 13728, 16328, 18928, 22204, 25480, 29540, 33600, 38560, 43520, 49504, 55488, 62628, 69768, 78204, 86640, 96520, 106400, 117880, 129360
Offset: 1

Views

Author

Lara Pudwell, Dec 01 2020

Keywords

Comments

The maximum number of copies of 123 in an alternating permutation is motivated in the Notices reference, and the argument here is analogous.

Examples

			a(1) = 4. The alternating permutation of length 1+5=6 with the maximum number of copies of 1234 is 132546.  The four copies are 1246, 1256, 1346, and 1356.
a(2) = 8. The alternating permutation of length 2+5=7 with the maximum number of copies of 1234 is 1325476. The eight copies are 1246, 1256, 1247, 1257, 1346, 1356, 1347, and 1357.
		

Crossrefs

Formula

a(2n) = A072819(n+1) = (2*n*(n + 2)*(n + 1)^2)/3.
a(2n-1) = 4*A006325(n+1) = (2*n*(n + 1)*(n^2 + n + 1))/3.
G.f.: 4*x*(1 + x^2)/((1 - x)^5*(1 + x)^3). - Stefano Spezia, Dec 12 2021
E.g.f.: (x*(75 + 51*x + 14*x^2 + x^3)*cosh(x) + (21 + 45*x + 57*x^2 + 14*x^3 + x^4)*sinh(x))/24. - Stefano Spezia, Aug 29 2025

A217927 Elements of the horizontal ADOMAH periodic table written from right to left, from bottom to top.

Original entry on oeis.org

2, 1, 4, 3, 10, 9, 8, 7, 6, 5, 12, 11, 18, 17, 16, 15, 14, 13, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 36, 35, 34, 33, 32, 31, 48, 47, 46, 45, 44, 43, 42, 41, 40, 39, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 38, 37, 54, 53, 52, 51, 50, 49, 80, 79, 78, 77, 76, 75, 74, 73, 72, 71, 102, 101, 100, 99, 98, 97, 96, 95, 94, 93, 92, 91, 90, 89
Offset: 1

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Author

Paul Curtz, Oct 15 2012

Keywords

Comments

We first write a variant of the ADOMAH periodic table:
1 2
5 6 7 8 9 10 3 4
21 22 23 24 25 26 27 28 29 30 13 14 15 16 17 18 11 12
57 to 70 39 40 41 42 43 44 45 46 47 48 31 32 33 34 35 36 19 20 (B)
89 to 102 71 72 73 74 75 76 77 78 79 80 49 50 51 52 53 54 37 38
103104105106107108109110111112 81 82 83 84 85 86 55 56
113114115116117118 87 88
119120
(See A219388).
It could be written vertically.
Differences of (2,4,12,20,38,56,88,120,... = A168380) = 2,8,8,18,18,... = A093907 from a 118 terms table.
The number of elements in the n-th period (2,8,18,32,32,18,8,2) is in A168281. Compare to A093907 = 2,8,8,18,18,32,32,50,...(extension of the Mendeleyev-Moseley-Seaborg table) and A137583 = 2,2,8,8,18,18,32,32. See the possible element 120 in A168208 (which must be clarified).
The horizontal ADOMAH periodic table (2006) is
119120
113114115116117118 87 88
103104105106107108109110111112 81 82 83 84 85 86 55 56
89 to 102 71 72 73 74 75 76 77 78 79 80 49 50 51 52 53 54 37 38 (A)
57 to 70 39 40 41 42 43 44 45 46 47 48 31 32 33 34 35 36 19 20
21 22 23 24 25 26 27 28 29 30 13 14 15 16 17 18 11 12
5 6 7 8 9 10 3 4
1 2
Generally it is written vertically.

References

  • Philip J. Stewart, "Charles Janet, Unrecognized genius of the periodic system", Foundations of Chemistry, January, 2009. ISSN 1386-4238.

Crossrefs

Cf. A137325.

Extensions

Reference given by Jean-François Alcover, Oct 22 2012
Typos corrected in comments by Jean-François Alcover, Nov 16 2012

A219527 a(n) = (6*n^2 + 7*n - 9 + 2*n^3)/12 - (-1)^n*(n+1)/4.

Original entry on oeis.org

1, 3, 11, 19, 37, 55, 87, 119, 169, 219, 291, 363, 461, 559, 687, 815, 977, 1139, 1339, 1539, 1781, 2023, 2311, 2599, 2937, 3275, 3667, 4059, 4509, 4959, 5471, 5983, 6561, 7139, 7787, 8435, 9157, 9879, 10679, 11479, 12361, 13243
Offset: 1

Views

Author

Paul Curtz, Nov 21 2012

Keywords

Comments

First column of the Mendeleyev-Moseley-Seaborg table (with alkali metals) or 31st column of the Janet table. See A138726.
(a(n+10) - a(n))/10 = 29, 36, 45, 54, ... = A061925(n+7) + 3.
b(n) = a(n+1) - 2*a(n) = 1, 5, -3, -1, -19, -23, -55, -69, -119, -147, -219, -265, -363, -431, ... contains -a(2*n).
b(2*n-1) - b(2*n-2) = 4, 2, -4, -14, -28, -46, -68, ... = A147973(n+3).

Crossrefs

Cf. A147973.

Programs

  • Mathematica
    a[n_] := (6*n^2 + 7*n - 9 + 2*n^3)/12 - (-1)^n*(n + 1)/4; Table[ a[n], {n, 1, 42}] (* Jean-François Alcover, Apr 05 2013 *)
    LinearRecurrence[{2,1,-4,1,2,-1},{1,3,11,19,37,55},50] (* Harvey P. Dale, Apr 01 2018 *)

Formula

a(n) = A168380(n+1) - 1.
a(n+2) - a(n+1) = A093907(n) = A137583(n+1).
a(n+3) - a(n+1) = 10,16,26,36,... = A137928(n+3).
G.f. x*(1 + x + 4*x^2 - 2*x^3 + x^5 - x^4) / ( (1+x)^2*(x-1)^4 ). - R. J. Mathar, Mar 27 2013

A272000 Coinage sequence: a(n) = A018227(n)-7.

Original entry on oeis.org

3, 11, 29, 47, 79, 111, 161, 211, 283, 355, 453, 551, 679, 807, 969, 1131, 1331, 1531, 1773, 2015, 2303, 2591, 2929, 3267, 3659, 4051, 4501, 4951, 5463, 5975, 6553, 7131, 7779, 8427, 9149, 9871, 10671, 11471, 12353, 13235, 14203, 15171, 16229, 17287, 18439
Offset: 1

Views

Author

Natan Arie Consigli, Jul 02 2016

Keywords

Comments

Terms from 29 to 111 are the atomic numbers of the elements of group 11 in the periodic table. The group is also known as the coinage metals since copper (element 29), silver (element 47) and gold (element 79) are in group 11.

Crossrefs

Other groups: 1(A219527), 2(A168380), 3(A168388), 12(A271998), 13(A271997), 14(A271996), 15(A271995), 16(A271994), 17(A271999), 18(A018227).

Programs

  • Mathematica
    LinearRecurrence[{2,1,-4,1,2,-1},{3,11,29,47,79,111},50] (* Harvey P. Dale, Nov 26 2018 *)
  • PARI
    Vec(x*(3+5*x+4*x^2-10*x^3-3*x^4+5*x^5)/((1-x)^4*(1+x)^2) + O(x^60)) \\ Colin Barker, Oct 25 2016

Formula

From Colin Barker, Oct 25 2016: (Start)
G.f.: x*(3 + 5*x + 4*x^2 - 10*x^3 - 3*x^4 + 5*x^5)/((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
a(n) = (n^3 + 9*n^2 + 26*n - 30)/6 for n even.
a(n) = (n^3 + 9*n^2 + 29*n - 21)/6 for n odd. (End)

A339355 Maximum number of copies of a 12345 permutation pattern in an alternating (or zig-zag) permutation of length n + 7.

Original entry on oeis.org

8, 16, 64, 112, 272, 432, 832, 1232, 2072, 2912, 4480, 6048, 8736, 11424, 15744, 20064, 26664, 33264, 42944, 52624, 66352, 80080, 99008, 117936, 143416, 168896, 202496, 236096, 279616, 323136, 378624, 434112, 503880, 573648, 660288, 746928, 853328, 959728, 1089088, 1218448
Offset: 1

Views

Author

Lara Pudwell, Dec 01 2020

Keywords

Comments

The maximum number of copies of 123 in an alternating permutation is motivated in the Notices reference, and the argument here is analogous.

Examples

			a(1) = 8. The alternating permutation of length 1 + 7 = 8 with the maximum number of copies of 12345 is 13254768. The eight copies are 12468, 12478, 12568, 12578, 13468, 13478, 13568, and 13578.
		

Crossrefs

Programs

  • Maple
    a := proc(n2) local n; n:= floor(n2/2): if n2 = 2*n then 32*binomial(n+4,5) - 16*binomial(n+3,4) else n:=n+1; (4*n*(n^4+5*n^3+10*n^2+10*n+4))/15 fi end; seq(a(n), n=1..20); # Georg Fischer, Nov 25 2022

Formula

a(2*n) = 16*A005585(n) = 32*binomial(n+4, 5) - 16*binomial(n+3, 4).
a(2*n-1) = 8*A033455(n) = (4*n*(n^4 + 5*n^3 + 10*n^2 + 10*n + 4))/15.
D-finite with recurrence: (n-1)*((n-3)^2+9*n-6)*a(n) - (2*(n-3)^2+20*n-16)*a(n-1) - (n+5)*((n-3)^2+11*n-2)*a(n-2) = 0. - Georg Fischer, Nov 25 2022

A339358 Maximum number of copies of a 1234567 permutation pattern in an alternating (or zig-zag) permutation of length n + 11.

Original entry on oeis.org

32, 64, 320, 576, 1696, 2816, 6400, 9984, 19392, 28800, 50304, 71808, 116160, 160512, 244992, 329472, 480480, 631488, 887744, 1144000, 1560416, 1976832, 2629120, 3281408, 4271488, 5261568, 6723840, 8186112, 10294656, 12403200, 15379968, 18356736, 22480800, 26604864
Offset: 1

Views

Author

Lara Pudwell, Dec 01 2020

Keywords

Comments

The maximum number of copies of 123 in an alternating permutation is motivated in the Notices reference, and the argument here is analogous.

Examples

			a(1) = 32. The alternating permutation of length 1+11=12 with the maximum number of copies of 1234567 is 132547698(11)(10)(12).  The 32 copies are 12468(10)(12), 12469(10)(12), 12478(10)(12), 12479(10)(12), 12568(10)(12), 12569(10)(12), 12578(10)(12), 12579(10)(12), 13468(10)(12), 13469(10)(12), 13478(10)(12), 13479(10)(12), 13568(10)(12), 13569(10)(12), 13578(10)(12), 13579(10)(12), 12468(11)(12), 12469(11)(12), 12478(11)(12), 12479(11)(12), 12568(11)(12), 12569(11)(12), 12578(11)(12), 12579(11)(12), 13468(11)(12), 13469(11)(12), 13478(11)(12), 13479(11)(12), 13568(11)(12), 13569(11)(12), 13578(11)(12), and 13579(11)(12).
		

Crossrefs

Cf. A168380.

Programs

  • Maple
    A339358 := proc(n)
        nhalf := ceil(n/2) ;
        if type(n,'even') then
            128*binomial(nhalf+6,7)-64*binomial(nhalf+5,6) ;
        else
            128*binomial(nhalf+4,7)+128*binomial(nhalf+4,6)+32*binomial(nhalf+4,5) ;
        end if;
    end proc:
    seq(A339358(n),n=1..40) ; # R. J. Mathar, Jan 11 2024

Formula

a(2n) = 64*A050486(n-1) = 128*C(n+6,7) - 64*C(n+5,6).
a(2n-1) = 128*C(n+4,7) + 128*C(n+4,6) + 32*C(n+4,5).
D-finite with recurrence (-n+1)*a(n) +2*a(n-1) +16*a(n-2) +2*a(n-3) +(n+7)*a(n-4)=0. - R. J. Mathar, Jan 11 2024

A339356 Maximum number of copies of a 123456 permutation pattern in an alternating (or zig-zag) permutation of length n + 9.

Original entry on oeis.org

16, 32, 144, 256, 688, 1120, 2352, 3584, 6496, 9408, 15456, 21504, 32928, 44352, 64416, 84480, 117744, 151008, 203632, 256256, 336336, 416416, 534352, 652288, 821184, 990080, 1226176, 1462272, 1785408, 2108544, 2542656, 2976768, 3550416, 4124064, 4870992, 5617920, 6577648
Offset: 1

Views

Author

Lara Pudwell, Dec 01 2020

Keywords

Comments

The maximum number of copies of 123 in an alternating permutation is motivated in the Notices reference, and the argument here is analogous.

Examples

			a(1) = 16. The alternating permutation of length 1+9=10 with the maximum number of copies of 123456 is 132547698(10). The sixteen copies are 12468(10), 12469(10), 12478(10), 12479(10), 12568(10), 12569(10), 12578(10), 12579(10), 13468(10), 13469(10), 13478(10), 13479(10), 13568(10), 13569(10), 13578(10), and 13579(10).
		

Crossrefs

Cf. A168380.

Formula

a(2n) = 32*A040977(n-1) = 64*C(n+5,6) - 32*C(n+4,5).
a(2n-1) = 16*A259181(n) = (2*n*(n + 1)*(n + 2)*(n + 3)*(2*n^2 + 6*n + 7))/45.
From Chai Wah Wu, Jul 06 2025: (Start)
a(n) = 2*a(n-1) + 4*a(n-2) - 10*a(n-3) - 5*a(n-4) + 20*a(n-5) - 20*a(n-7) + 5*a(n-8) + 10*a(n-9) - 4*a(n-10) - 2*a(n-11) + a(n-12) for n > 12.
G.f.: x*(-16*x^2 - 16)/((x - 1)^7*(x + 1)^5). (End)
Previous Showing 11-18 of 18 results.