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.

User: David Williams

David Williams's wiki page.

David Williams has authored 10 sequences.

A354445 Number of polynomials per row where the minimum number of rows and polynomials per row necessary to transform A335105 into a triangular array are present.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 2, 3, 4, 5, 4, 5, 4, 5, 6, 7, 8, 9, 8, 9, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 16, 17, 18, 19, 20, 21, 20, 21, 20, 21, 22, 23, 24, 25, 26, 29, 28, 29, 28, 29, 30, 31, 32, 33, 32, 33, 32, 31, 34, 35, 36, 37, 38, 37, 40, 41, 42, 43, 44, 45, 44, 45, 46, 47
Offset: 1

Author

David Williams, May 29 2022

Keywords

Comments

This array treats A335105, an irregular triangle, as a subset of a symmetrical one. It is only necessary to add one row in order to transform A335105 into a triangular array. Rows two, four and six, which correspond to Hydrogen, Lithium and Boron in A335105, are the only rows composed entirely of numerical terms; for these rows the terminal number divided by two and then squared equals the sum of terms left of the right edge. Polynomials within a row may change places with numerical terms within the same row without changing the number of polynomials per row. Given that the summands of A335105 (shell and number of shell's electrons) are necessarily added in multiples of two, the parity of this sequence is alternating.
All the above statements apply to A350597.

Examples

			                  X                1
  1 2             1 2              0      Thus, 1, 0, 1, 0, 1, 0, 1, 2, ...
  1 3             1 3 X            1
  1 3 5 6         1 3 5 6          0
  1 3 5 7         1 3 5 7 X        1
  1 3 5 7 9 10    1 3 5 7 9 10     0
		

Crossrefs

A350597 Irregular triangle read by rows in which row n lists the partial sums of shell numbers and respective number of electrons for all occupied shells of the n-th element of the periodic table of the elements where electron configurations are ordered according to Madelung's rule.

Original entry on oeis.org

1, 2, 1, 3, 1, 3, 5, 6, 1, 3, 5, 7, 1, 3, 5, 7, 9, 10, 1, 3, 5, 7, 9, 11, 1, 3, 5, 7, 9, 12, 1, 3, 5, 7, 9, 13, 1, 3, 5, 7, 9, 14, 1, 3, 5, 7, 9, 15, 1, 3, 5, 7, 9, 15, 18, 19, 1, 3, 5, 7, 9, 15, 18, 20, 1, 3, 5, 7, 9, 15, 18, 20, 23, 24, 1, 3, 5, 7, 9, 15, 18, 20, 23, 25
Offset: 1

Author

David Williams, Jan 08 2022

Keywords

Comments

This sequence is a variant of A335105.
This sequence agrees with A335105 until the row for scandium; specifically, divergence occurs at t(20,10).
Electron configurations are either written in spectroscopic order as in A335105 where configurations are listed in order of increasing values of the main quantum number, or by order of Madelung's rule, where they are written in order of increasing values of the sums of the main quantum number and respective angular momentum numbers (s,p,d or f) with s=0, p=1, d=2, f=3. Thus in Madelung order 4s (sum 4+0) is written ahead of 3d (sum 3+2).
The final term of each row of this sequence agrees with final term of each row of A335105.

Examples

			The configuration for lithium is 1s2 2s1 whether written in spectroscopic order or by order of the Madelung rule. Thus the row for lithium in A335105 and the row for lithium in this sequence agree being derived as follows:
  1s2 2s1
  1+2+2+1
  1, 3, 5, 6
  However: in spectroscopic order scandium is written 1s2, 2s2, 2p6, 3s2, 3p6, 3d1, 4s2 yielding a row in A335105 of 1, 3, 5, 7, 9, 15, 18, 20, 23, 29, 32, 33, 37, 39, whereas in Madelung order the same algorithm applied to the configuration for scandium yields a row of 1, 3, 5, 7, 9, 15, 18, 20, 23, 29, 33, 35, 38, 39.
		

References

  • Gary L. Miessler and Donald A. Tarr, Inorganic Chemistry, 4th Edition. Prentice Hall. Upper Saddle River, New Jersey, 2011.
  • John R. Rumble (ed.), CRC Handbook of Chemistry and Physics, 100th edition. CRC Press. Boca Raton, Florida, 2019; Section 1, Electron Configurations and Ionization Energy of Neutral Atoms in the Ground State.
  • Eric Scerri, The Periodic Table, 2nd edition. Oxford University Press. New York, New York, 2019.

Crossrefs

Cf. A335105.

A335105 Irregular triangle read by rows in which row n lists the partial sums of shell numbers and respective number of electrons for all occupied shells of the n-th element of the periodic table of the elements.

Original entry on oeis.org

1, 2, 1, 3, 1, 3, 5, 6, 1, 3, 5, 7, 1, 3, 5, 7, 9, 10, 1, 3, 5, 7, 9, 11, 1, 3, 5, 7, 9, 12, 1, 3, 5, 7, 9, 13, 1, 3, 5, 7, 9, 14, 1, 3, 5, 7, 9, 15, 1, 3, 5, 7, 9, 15, 18, 19, 1, 3, 5, 7, 9, 15, 18, 20, 1, 3, 5, 7, 9, 15, 18, 20, 23, 24, 1, 3, 5, 7, 9, 15, 18, 20, 23, 25
Offset: 1

Author

David Williams, May 23 2020

Keywords

Comments

The rows provide a way to distinguish the elements.
Electron Configurations for elements after Nobelium (atomic number 102), whose configuration of [RN] 5f14 7s2 comprises the last row of sequence, are at present unknown (see CRC Handbook in References).

Examples

			Lithium 1s2 2s1
        1+2+2+1
        1,3,5,6
Thus:
Hydrogen  1s1           1,2
Helium    1s2           1,3
Lithium   1s2 2s1       1,3,5,6
Beryllium 1s2 2s2       1,3,5,7
Boron     1s2 2s2 2p1   1,3,5,7,9,10
		

References

  • Darleane C. Hoffman, Diana M. Lee, and Valeria Pershina, Transactinides and Future Elements, in Morss; Edelstein, Norman M.; Fuger, Jean (eds.). The Chemistry of Actinide and Transactinide Elements (3rd ed.). Dordrecht, The Netherlands (2006).
  • Catherine E. Housecroft and Alan G. Sharpe, Inorganic Chemistry (5th ed.). Pearson Education Limited. Harlow, England. (2018).
  • John R. Rumble (ed.), CRC Handbook of Chemistry and Physics, 100th edition. CRC Press. Boca Raton, Florida, 2019; Section 1, Electron Configuration and Ionization Energy of Neutral Atoms in the Ground State.

Crossrefs

A330384 Single-digit numbers in order of appearance in decimal expansions of composite numbers, followed by two-digit numbers in order of appearance, then three-digit numbers and so forth.

Original entry on oeis.org

4, 6, 8, 9, 1, 0, 2, 5, 7, 3, 10, 12, 14, 15, 16, 18, 20, 21, 22, 24, 25, 26, 27, 28, 30, 32, 33, 34, 35, 36, 38, 39, 40, 42, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 62, 63, 64, 65, 66, 68, 69, 70, 72, 74, 75, 76, 77, 78, 80, 81, 82, 84, 85, 86
Offset: 1

Author

David Williams, Dec 12 2019

Keywords

Examples

			From _Rémy Sigrist_, Dec 24 2019: (Start)
The first terms, alongside the corresponding composite numbers, are:
  n   a(n)  Composite
  --  ----  ---------
   1     4          4
   2     6          6
   3     8          8
   4     9          9
   5     1         10
   6     0         10
   7     2         12
   8     5         15
   9     7         27
  10     3         30
  11    10         10
  12    12         12
  13    14         14
  ...
  79    99         99
  80    11        110
  81    17        117
  82    19        119
  83    23        123
(End)
		

Crossrefs

Programs

  • PARI
    \\ See Links section.

Extensions

More terms from Rémy Sigrist, Dec 24 2019

A321043 Single-digit numbers in the order in which they first appear in the decimal expansions of powers of 2, followed by the two-digit numbers in the order in which they appear, then the three-digit numbers, and so on.

Original entry on oeis.org

1, 2, 4, 8, 6, 3, 5, 0, 9, 7, 16, 32, 64, 12, 28, 25, 56, 51, 10, 24, 20, 48, 40, 96, 81, 19, 92, 63, 38, 84, 27, 76, 68, 65, 55, 53, 36, 13, 31, 72, 26, 62, 21, 14, 44, 52, 42, 88, 85, 57, 97, 71, 15, 41, 94, 43, 30, 83, 86, 60, 67, 77, 33, 35, 54, 34, 17, 45
Offset: 1

Author

David Williams, Oct 26 2018

Keywords

Comments

Apparently this algorithm applied to most sequences will produce a fractal scatterplot graph. - David Williams, Jan 20 2019

Examples

			1,2,4,8,16,32,64,128,256,512,1024, ..., 4096, ..., 32768, ... gives 1,2,4,8,6,3,5,0,9,7.
Then we get 16,32,64,12,28,25,56,51,10,24,20,48,40,96,81,19,92,...
11 does not appear until 2^40 = 1099511627776.
		

Crossrefs

See A030000 for an inverse.

Programs

  • PARI
    \\ See Links section.

Extensions

Edited by N. J. A. Sloane, Oct 27 2018
More terms from Rémy Sigrist, Oct 27 2018

A218450 Number of digits of n plus number of digits of n equal to 1, 2, 4, or 8.

Original entry on oeis.org

2, 2, 1, 2, 1, 1, 1, 2, 1, 3, 4, 4, 3, 4, 3, 3, 3, 4, 3, 3, 4, 4, 3, 4, 3, 3, 3, 4, 3, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 3, 4, 4, 3, 4, 3, 3, 3, 4, 3, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 2, 3, 3, 2, 3, 2, 2, 2, 3, 2, 3, 4, 4, 3, 4, 3, 3, 3
Offset: 1

Author

David Williams, Oct 28 2012

Keywords

Programs

  • Maple
    f:= proc(n) local L;
      L:= convert(n,base,10);
      nops(L) + numboccur(L,{1,2,4,8})
    end proc:
    map(f, [$1..100]); # Robert Israel, Apr 29 2021
  • Mathematica
    Table[d = IntegerDigits[n]; Length[d] + Count[d, 1] + Count[d, 2] + Count[d, 4] + Count[d, 8], {n, 100}] (* T. D. Noe, Nov 16 2012 *)
    Table[IntegerLength[n]+Total[DigitCount[n,10,{1,2,4,8}]],{n,100}] (* Harvey P. Dale, Aug 03 2021 *)
  • PARI
    a(n) = my(d = digits(n)); #d + #select(x->((x==1) || (x==2) || (x==4) || (x==8)), d); \\ Michel Marcus, Jan 31 2016

Formula

lim inf a(n)/log_10 n = 1; lim sup a(n)/log_10 n = 2.

A135635 Triangle read by rows, constructed by the Pascal rule, with top entry 2, left edge = odd numbers, right edge = squares plus 1.

Original entry on oeis.org

2, 3, 5, 5, 8, 10, 7, 13, 18, 17, 9, 20, 31, 35, 26, 11, 29, 51, 66, 61, 37, 13, 40, 80, 117, 127, 98, 50, 15, 53, 120, 197, 244, 225, 148, 65, 17, 68, 173, 317, 441, 469, 373, 213, 82, 19, 85, 241, 490, 758, 910, 842, 586, 295, 101
Offset: 1

Author

David Williams (davidwilliams(AT)paxway.com), Mar 01 2008

Keywords

Examples

			............2
...........3.5
..........5.8.10
........7.13.18.17
.......9.20.31.35.26
.....11.29.51.66.61.37
		

Crossrefs

Cf. A002522.

Programs

  • Maple
    T:=proc(n,k)if n=1 and k=1 then 2 elif k=1 then 2*n-1 elif n < k then 0 elif k =n then n^2+1 else T(n-1,k)+T(n-1,k-1) end if end proc: for n to 10 do seq(T(n,k),k=1..n) end do; # yields sequence in triangular form - Emeric Deutsch, Mar 03 2008
  • Mathematica
    T[1, 1]:= 2; T[n_, 1]:= 2*n - 1; T[n_, n_]:= n^2 + 1; T[n_, k_] := T[n - 1, k] + T[n - 1, k - 1]; Table[T[n, k], {n, 1, 10}, {k, 1, n}]//Flatten (* G. C. Greubel, Oct 25 2016 *)

Extensions

More terms from Emeric Deutsch, Mar 03 2008

A133493 Base 8 version of A102363.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 10, 11, 14, 17, 20, 21, 25, 33, 37, 40, 41, 46, 60, 72, 77, 100, 101, 107, 126, 152, 171, 177, 200, 201, 210, 235, 300, 343, 370, 377, 400, 401, 411, 445, 535, 643, 733, 767, 777, 1000, 1001, 1012, 1056, 1202, 1400, 1576, 1722, 1766, 1777
Offset: 1

Author

David Williams (davidwilliams(AT)paxway.com), Dec 01 2007

Keywords

Crossrefs

Cf. A102363.

A102363 Triangle read by rows, constructed by a Pascal-like rule with left edge = 2^k, right edge = 2^(k+1)-1 (k >= 0).

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 8, 9, 12, 15, 16, 17, 21, 27, 31, 32, 33, 38, 48, 58, 63, 64, 65, 71, 86, 106, 121, 127, 128, 129, 136, 157, 192, 227, 248, 255, 256, 257, 265, 293, 349, 419, 475, 503, 511, 512, 513, 522, 558, 642, 768, 894, 978, 1014, 1023, 1024, 1025, 1035, 1080, 1200, 1410, 1662, 1872, 1992, 2037, 2047
Offset: 0

Author

David Williams, Mar 15 2005, Oct 05 2007

Keywords

Comments

First column right of center divided by 3 equals powers of 4.
Right of left edge, sums of rows are divisible by 3.
Apparently the number of terms per row plus the number of numbers in natural order skipped per row equals a power of 2. - David Williams, Jun 27 2009

Examples

			This triangle begins:
                            1
                         2     3
                      4     5     7
                   8     9    12    15
               16    17    21    27    31
            32    33    38    48    58    63
         64    65    71    86   106   121   127
     128   129   136   157   192   227   248   255
  256   257   265   293   349   419   475   503   511
G.f. of this sequence in flattened form:
A(x) = 1 + 2*x + 3*x^2 + 4*x^3 + 5*x^4 + 7*x^5 + 8*x^6 + 9*x^7 + 12*x^8 + 15*x^9 + 16*x^10 + 17*x^11 + 21*x^12 + 27*x^13 + 31*x^14 + 32*x^15 + ...
such that
A(x) = (1+x) + x*(1+x)^2 + x^2*(1+x)^2 + x^3*(1+x)^3 + x^4*(1+x)^3 + x^5*(1+x)^3 + x^6*(1+x)^4 + x^7*(1+x)^4 + x^8*(1+x)^4 + x^9*(1+x)^4 + x^10*(1+x)^5 + x^11*(1+x)^5 + x^12*(1+x)^5 + x^13*(1+x)^5 + x^14*(1+x)^5 + x^15*(1+x)^6 + ...
		

Crossrefs

Cf. A000079, A053220 (row sums), A265939 (central terms).

Programs

  • Maple
    T:=proc(n,k) if k=0 then 2^n elif k=n then 2^(n+1)-1 else T(n-1,k)+T(n-1,k-1) fi end: for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form - Emeric Deutsch, Mar 26 2005
  • Mathematica
    t[n_, 0] := 2^n; t[n_, n_] := 2^(n+1)-1; t[n_, k_] := t[n, k] = t[n-1, k] + t[n-1, k-1]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 15 2013 *)
  • PARI
    /* Print in flattened form: Sum_{n>=0} x^n*(1+x)^tr(n) */
    {tr(n) = ceil( (sqrt(8*n+9)-1)/2 )}
    {a(n) = polcoeff( sum(m=0,n, x^m * (1+x +x*O(x^n))^tr(m) ),n)}
    for(n=0,78, print1(a(n),", ")) \\ Paul D. Hanna, Feb 19 2016

Formula

G.f.: Sum_{n>=0} x^n * (1+x)^tr(n) = Sum_{n>=0} a(n)*x^n, where tr(n) = A002024(n+1) = floor(sqrt(2*n+1) + 1/2). - Paul D. Hanna, Feb 19 2016
G.f.: Sum_{n>=1} x^(n*(n-1)/2) * (1-x^n)/(1-x) * (1+x)^n = Sum_{n>=0} a(n)*x^n. - Paul D. Hanna, Feb 20 2016
a(n) = A007318(n-1) + a(n-1). - Jon Maiga, Dec 22 2018

Extensions

More terms from Emeric Deutsch, Mar 26 2005

A067076 Numbers k such that 2*k + 3 is a prime.

Original entry on oeis.org

0, 1, 2, 4, 5, 7, 8, 10, 13, 14, 17, 19, 20, 22, 25, 28, 29, 32, 34, 35, 38, 40, 43, 47, 49, 50, 52, 53, 55, 62, 64, 67, 68, 73, 74, 77, 80, 82, 85, 88, 89, 94, 95, 97, 98, 104, 110, 112, 113, 115, 118, 119, 124, 127, 130, 133, 134, 137, 139, 140, 145, 152, 154, 155
Offset: 1

Author

David Williams, Aug 17 2002

Keywords

Comments

The following sequences (allowing offset of first term) all appear to have the same parity: A034953, triangular numbers with prime indices; A054269, length of period of continued fraction for sqrt(p), p prime; A082749, difference between the sum of next prime(n) natural numbers and the sum of next n primes; A006254, numbers n such that 2n-1 is prime; A067076, 2n+3 is a prime. - Jeremy Gardiner, Sep 10 2004
n is in the sequence iff none of the numbers (n-3k)/(2k+1), 1 <= k <= (n-1)/5, is positive integer. - Vladimir Shevelev, May 31 2009
Zeta(s) = Sum_{n>=1} 1/n^s = 1/1 - 2^(-s) * Product_{p=prime=(2*A067076)+3} 1/(1 - (2*A067076+3)^(-s)). - Eric Desbiaux, Dec 15 2009
This sequence is a subsequence of A047949. - Jason Kimberley, Aug 30 2012

Crossrefs

Numbers n such that 2n+k is prime: A005097 (k=1), this seq(k=3), A089038 (k=5), A105760 (k=7), A155722 (k=9), A101448 (k=11), A153081 (k=13), A089559 (k=15), A173059 (k=17), A153143 (k=19). - Jason Kimberley, Sep 07 2012
Numbers n such that 2n-k is prime: A006254 (k=1), A098090 (k=3), A089253 (k=5), A089192 (k=7), A097069 (k=9), A097338 (k=11), A097363 (k=13), A097480 (k=15), A098605 (k=17), A097932 (k=19).

Programs

Formula

a(n) = A006254(n) - 2 = A086801(n+1)/2. [Corrected by M. F. Hasler, Feb 14 2024]
a(n) = A089253(n) - 4. - Giovanni Teofilatto, Dec 14 2003
Conjecture: a(n) = A008507(n) + n - 1 = A005097(n) - 1 = A102781(n+1) - 1. - R. J. Mathar, Jul 07 2009
a(n) = A179893(n) - A000040(n). - Odimar Fabeny, Aug 24 2010

Extensions

Offset changed from 0 to 1 in 2008: some formulas here and elsewhere may need to be corrected.