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.

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A376033 Number A(n,k) of binary words of length n avoiding distance (i+1) between "1" digits if the i-th bit is set in the binary representation of k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 4, 5, 16, 1, 2, 3, 6, 8, 32, 1, 2, 4, 4, 9, 13, 64, 1, 2, 3, 8, 6, 15, 21, 128, 1, 2, 4, 5, 12, 9, 25, 34, 256, 1, 2, 3, 6, 7, 18, 13, 40, 55, 512, 1, 2, 4, 4, 8, 11, 27, 19, 64, 89, 1024, 1, 2, 3, 8, 5, 11, 16, 45, 28, 104, 144, 2048
Offset: 0

Views

Author

Alois P. Heinz, Sep 09 2024

Keywords

Comments

Also the number of subsets of [n] avoiding distance (i+1) between elements if the i-th bit is set in the binary representation of k. A(6,3) = 13: {}, {1}, {2}, {3}, {4}, {5}, {6}, {1,4}, {1,5}, {1,6}, {2,5}, {2,6}, {3,6}.
Each column sequence satisfies a linear recurrence with constant coefficients.
The sequence of row n is periodic with period A011782(n) = ceiling(2^(n-1)).

Examples

			A(6,6) = 17: 000000, 000001, 000010, 000011, 000100, 000110, 001000, 001100, 010000, 010001, 011000, 100000, 100001, 100010, 100011, 110000, 110001 because 6 = 110_2 and no two "1" digits have distance 2 or 3.
A(6,7) = 10: 000000, 000001, 000010, 000100, 001000, 010000, 010001, 100000, 100001, 100010.
A(7,7) = 14: 0000000, 0000001, 0000010, 0000100, 0001000, 0010000, 0010001, 0100000, 0100001, 0100010, 1000000, 1000001, 1000010, 1000100.
Square array A(n,k) begins:
     1,  1,   1,  1,   1,  1,  1,  1,   1,  1, ...
     2,  2,   2,  2,   2,  2,  2,  2,   2,  2, ...
     4,  3,   4,  3,   4,  3,  4,  3,   4,  3, ...
     8,  5,   6,  4,   8,  5,  6,  4,   8,  5, ...
    16,  8,   9,  6,  12,  7,  8,  5,  16,  8, ...
    32, 13,  15,  9,  18, 11, 11,  7,  24, 11, ...
    64, 21,  25, 13,  27, 16, 17, 10,  36, 17, ...
   128, 34,  40, 19,  45, 25, 27, 14,  54, 25, ...
   256, 55,  64, 28,  75, 37, 41, 19,  81, 37, ...
   512, 89, 104, 41, 125, 57, 60, 26, 135, 57, ...
		

Crossrefs

Columns k=0-20 give: A000079, A000045(n+2), A006498(n+2), A000930(n+2), A006500, A130137, A079972(n+3), A003269(n+4), A031923(n+1), A263710(n+1), A224809(n+4), A317669(n+4), A351873, A351874, A121832(n+4), A003520(n+4), A208742, A374737, A375977, A375980, A375978.
Rows n=0-2 give: A000012, A007395(k+1), A010702(k+1).
Main diagonal gives A376091.
A(n,2^k-1) gives A141539.
A(2^n-1,2^n-1) gives A376697.
A(n,2^k) gives A209435.

Programs

  • Maple
    h:= proc(n) option remember; `if`(n=0, 1, 2^(1+ilog2(n))) end:
    b:= proc(n, k, t) option remember; `if`(n=0, 1, add(`if`(j=1 and
          Bits[And](t, k)>0, 0, b(n-1, k, irem(2*t+j, h(k)))), j=0..1))
        end:
    A:= (n, k)-> b(n, k, 0):
    seq(seq(A(n, d-n), n=0..d), d=0..12);
  • PARI
    step(v,b)={vector(#v, i, my(j=(i-1)>>1); if(bittest(i-1,0), if(bitand(b,j)==0, v[1+j], 0), v[1+j] + v[1+#v/2+j]));}
    col(n,k)={my(v=vector(2^(1+logint(k,2))), r=vector(1+n)); v[1]=r[1]=1; for(i=1, n, v=step(v,k); r[1+i]=vecsum(v)); r}
    A(n,k)=if(k==0, 2^n, col(n,k)[n+1]) \\ Andrew Howroyd, Oct 03 2024

Formula

A(n,k) = A(n,k+ceiling(2^(n-1))).
A(n,ceiling(2^(n-1))-1) = n+1.
A(n,ceiling(2^(n-2))) = ceiling(3*2^(n-2)) = A098011(n+2).

A350111 Triangle read by rows: T(n,k) is the number of tilings of an (n+k)-board using k (1,3)-fences and n-k squares.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 4, 2, 0, 1, 3, 6, 7, 4, 0, 0, 1, 4, 9, 12, 8, 0, 0, 0, 1, 5, 13, 20, 16, 8, 4, 2, 1, 1, 6, 18, 32, 36, 28, 19, 12, 3, 0, 1, 7, 24, 50, 69, 69, 58, 31, 9, 0, 0, 1, 8, 31, 74, 120, 144, 127, 78, 27, 0, 0, 0
Offset: 0

Views

Author

Michael A. Allen, Dec 22 2021

Keywords

Comments

This is the m=4 member in the sequence of triangles A007318, A059259, A350110, A350111, A350112 which give the number of tilings of an (n+k) X 1 board using k (1,m-1)-fences and n-k unit square tiles. A (1,g)-fence is composed of two unit square tiles separated by a gap of width g.
It is also the m=4, t=2 member of a two-parameter family of triangles such that T(n,k) is the number of tilings of an (n+(t-1)*k) X 1 board using k (1,m-1;t)-combs and n-k unit square tiles. A (1,g;t)-comb is composed of a line of t unit square tiles separated from each other by gaps of width g.
T(4*j+r-k,k) is the coefficient of x^k in (f(j,x))^(4-r)*(f(j+1,x))^r for r=0,1,2,3 where f(n,x) is one form of a Fibonacci polynomial defined by f(n+1,x)=f(n,x)+x*f(n-1,x) where f(0,x)=1 and f(n<0,x)=0.
T(n+4-k,k) is the number of subsets of {1,2,...,n} of size k such that no two elements in a subset differ by 4.
Sum of (n+3)-th antidiagonal (counting initial 1 as the 0th) is A031923(n).

Examples

			Triangle begins:
  1;
  1,   0;
  1,   0,   0;
  1,   0,   0,   0;
  1,   1,   1,   1,   1;
  1,   2,   3,   4,   2,   0;
  1,   3,   6,   7,   4,   0,   0;
  1,   4,   9,  12,   8,   0,   0,   0;
  1,   5,  13,  20,  16,   8,   4,   2,   1;
  1,   6,  18,  32,  36,  28,  19,  12,   3,   0;
  1,   7,  24,  50,  69,  69,  58,  31,   9,   0,   0;
  1,   8,  31,  74, 120, 144, 127,  78,  27,   0,   0,   0;
  1,   9,  39, 105, 195, 264, 265, 189,  81,  27,   9,   3,   1;
  1,  10,  48, 144, 300, 458, 522, 432, 270, 132,  58,  24,   4,   0;
		

Crossrefs

Sums of antidiagonals: A031923
Other members of the two-parameter family of triangles: A007318 (m=1,t=2), A059259 (m=2,t=2), A350110 (m=3,t=2), A350112 (m=5,t=2), A354665 (m=2,t=3), A354666 (m=2,t=4), A354667 (m=2,t=5), A354668 (m=3,t=3).
Other triangles related to tiling using fences: A123521, A157897, A335964.

Programs

  • Mathematica
    f[n_]:=If[n<0,0,f[n-1]+x*f[n-2]+KroneckerDelta[n,0]];
    T[n_, k_]:=Module[{j=Floor[(n+k)/4],r=Mod[n+k,4]},
      Coefficient[f[j]^(4-r)*f[j+1]^r,x,k]];
    Flatten@Table[T[n,k], {n, 0, 13}, {k, 0, n}]
    (* or *)
    T[n_,k_]:=If[k<0 || n
    				

Formula

T(n,k) = T(n-1,k) + T(n-2,k-1) - T(n-3,k-1) + T(n-3,k-2) + T(n-4,k-1) + T(n-4,k-3) + 2*T(n-4,k-4) + T(n-5,k-2) + 2*T(n-5,k-3) - T(n-5,k-4) - T(n-6,k-3)-T(n-6,k-5) - T(n-7,k-4)-T(n-7,k-5) - T(n-7,k-6) - T(n-8,k-7)-T(n-8,k-8) + delta(n,0)*delta(k,0) - delta(n,2)*delta(k,1) - delta(n,3)*delta(k,2) - delta(n,4)*delta(k,4) with T(n
T(n,0) = 1.
T(n,n) = delta(n mod 4,0).
T(n,1) = n-3 for n>2.
T(4*j-r,4*j-p) = 0 for j>0, p=1,2,3, and r=1,...,p.
T(4*(j-1)+p,4*(j-1)) = T(4*j,4*j-p) = j^p for j>0 and p=0,1,2,3,4.
T(4*j+1,4*j-1) = 4*j(j+1)/2 for j>0.
T(4*j+2,4*j-2) = 4*C(j+2,4) + 6*C(j+1,2)^2 for j>1.
G.f. of row sums: (1-x-x^3)/((1-2*x)*(1-x^2)*(1+2*x^2+x^3+x^4)).
G.f. of antidiagonal sums: (1-x^2-x^3+x^4-x^6)/((1-x-x^2)*(1-x^4)*(1+3*x^4+x^8)).
T(n,k) = T(n-1,k) + T(n-1,k-1) for n>=3*k+1 if k>=0.

A209434 Table T(n,m), read by antidiagonals, is the number of subsets of {1,...,n} which do not contain two elements whose difference is m+1.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 5, 4, 2, 1, 8, 6, 4, 2, 1, 13, 9, 8, 4, 2, 1, 21, 15, 12, 8, 4, 2, 1, 34, 25, 18, 16, 8, 4, 2, 1, 55, 40, 27, 24, 16, 8, 4, 2, 1, 89, 64, 45, 36, 32, 16, 8, 4, 2, 1, 144, 104, 75, 54, 48, 32, 16, 8, 4, 2, 1, 233, 169, 125, 81, 72, 64, 32
Offset: 0

Author

David Nacin, Mar 09 2012

Keywords

Comments

1st column is the Fibonacci sequence.

Examples

			Table begins:
1,   1,   1,   1,   1,   1,   1,   1,   1,   1,    1,    ...
2,   2,   2,   2,   2,   2,   2,   2,   2,   2,    2,    ...
3,   4,   4,   4,   4,   4,   4,   4,   4,   4,    4,    ...
5,   6,   8,   8,   8,   8,   8,   8,   8,   8,    8,    ...
8,   9,   12,  16,  16,  16,  16,  16,  16,  16,   16,   ...
13,  15,  18,  24,  32,  32,  32,  32,  32,  32,   32,   ...
21,  25,  27,  36,  48,  64,  64,  64,  64,  64,   64,   ...
34,  40,  45,  54,  72,  96,  128, 128, 128, 128,  128,  ...
55,  64,  75,  81,  108, 144, 192, 256, 256, 256,  256,  ...
89,  104, 125, 135, 162, 216, 288, 384, 512, 512,  512,  ...
144, 169, 200, 225, 243, 324, 432, 576, 768, 1024, 1024, ...
............................................................
		

References

  • M. El-Mikkawy, T. Sogabe, A new family of k-Fibonacci numbers, Appl. Math. Comput. 215 (2010) 4456-4461 doi:10.1016/j.amc.2009.12.069, Table 1.

Programs

  • Mathematica
    a[n_, m_] := Product[Fibonacci[Floor[(n + i)/(m + 1) + 2]], {i, 0, m}]; Flatten[Table[a[j - i, i], {j, 0, 30}, {i, 0, j}]]

Formula

T(n,m) = Product_{i=0 to m} F(floor[(n + i)/(m + 1) + 2]) where F(n) is the n-th Fibonacci number.

A376743 Number of permutations (p(1),p(2),...,p(n)) of (1,2,...,n) such that p(i)-i is in {-2,-1,4} for all i=1,...,n.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 5, 5, 6, 8, 11, 15, 25, 35, 46, 61, 85, 125, 175, 245, 341, 470, 650, 925, 1300, 1810, 2521, 3520, 4915, 6880, 9640, 13476, 18801, 26251, 36721, 51346, 71776, 100335, 140210, 195886, 273813, 382821, 535105, 747850, 1045220
Offset: 0

Author

Michael A. Allen, Oct 03 2024

Keywords

Comments

Other sequences related to strongly restricted permutations pi(i) of i in {1,..,n} along with the sets of allowed p(i)-i (containing at least 3 elements): A000045 {-1,0,1}, A189593 {-1,0,2,3,4,5,6}, A189600 {-1,0,2,3,4,5,6,7}, A006498 {-2,0,2}, A080013 {-2,1,2}, A080014 {-2,0,1,2}, A033305 {-2,-1,1,2}, A002524 {-2,-1,0,1,2}, A080000 {-2,0,3}, A080001 {-2,1,3}, A080004 {-2,0,1,3}, A080002 {-2,2,3}, A080005 {-2,0,2,3}, A080008 {-2,1,2,3}, A080011 {-2,0,1,2,3}, A079999 {-2,-1,3}, A080003 {-2,-1,0,3}, A080006 {-2,-1,1,3}, A080009 {-2,-1,0,1,3}, A080007 {-2,-1,2,3}, A080010 {-2,-1,0,2,3}, A080012 {-2,-1,1,2,3}, A072827 {-2,-1,0,1,2,3}, A224809 {-2,0,4}, A189585 {-2,0,1,3,4}, A189581 {-2,-1,0,3,4}, A072850 {-2,-1,0,1,2,3,4}, A189587 {-2,0,1,3,4,5}, A189588 {-2,-1,0,3,4,5}, A189594 {-2,0,1,3,4,5,6}, A189595 {-2,-1,0,3,4,5,6}, A189601 {-2,0,1,3,4,5,6,7}, A189602 {-2,-1,0,3,4,5,6,7}, A224811 {-2,0,8}, A224812 {-2,0,10}, A224813 {-2,0,12}, A006500 {-3,0,3}, A079981 {-3,1,3}, A079983 {-3,0,1,3}, A079982 {-3,2,3}, A079984 {-3,0,2,3}, A079988 {-3,1,2,3}, A079989 {-3,0,1,2,3}, A079986 {-3,-1,1,3}, A079992 {-3,-1,0,1,3}, A079987 {-3,-1,2,3}, A079990 {-3,-1,0,2,3}, A079993 {-3,-1,1,2,3}, A079985 {-3,-2,2,3}, A079991 {-3,-2,0,2,3}, A079996 {-3,-2,0,1,2,3}, A079994 {-3,-2,1,2,3}, A079997 {-3,-2, -1,1,2,3}, A002526 {-3,-2,-1,0,1,2,3}, A189586 {-3,0,1,2,4}, A189583 {-3,-1,0,2,4}, A189582 {-3,-2,0,1,4}, A189584 {-3,-2,-1,0,4}, A189589 {-3,0,1,2,4,5}, A189590 {-3,-1,0,2,4,5}, A189591 {-3,-2,1,4,5}, A189592 {-3,-2,-1,0,4,5}, A224810 {-3,0,6}, A189596 {-3,0,1,2,4,5,6}, A189597 {-3,-1,0,2,4,5,6}, A189598 {-3,-2,0,1,4,5,6}, A189599 {-3,-2,-1,0,4,5,6}, A224814 {-3,0,9}, A031923 {-4,0,4}, A072856 {-4,-3, -2,-1,0,1,2,3,4}, A224815 {-4,0,8}, A154654 {-5,-4,-3,-2,-1,0,1,2,3,4,5}, A154655 {-6,-5,-4,-3, -2,-1,0,1,2,3,4,5,6}.
[Keyword "less", because this comment should be moved to the Index to the OEIS, it is not appropriate here. - N. J. A. Sloane, Oct 25 2024]

References

  • D. H. Lehmer, Permutations with strongly restricted displacements. Combinatorial theory and its applications, II (Proc. Colloq., Balatonfured, 1969), North-Holland, Amsterdam, 1970, pp. 755-770.

Crossrefs

See comments for other sequences related to strongly restricted permutations.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x^3 - x^4 - x^6 + x^9)/(1 - x^3 - x^4 - x^5 - 2*x^6 - x^7 + 2*x^9 + 2*x^10 + x^12 - x^15),{x,0,49}],x]
    LinearRecurrence[{0, 0, 1, 1, 1, 2, 1, 0, -2, -2, 0, -1, 0, 0, 1}, {1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 5, 5, 6, 8}, 50]

Formula

a(n) = a(n-3) + a(n-4) + a(n-5) + 2*a(n-6) + a(n-7) - 2*a(n-9) - 2*a(n-10) - a(n-12) + a(n-15).
G.f.: (1 - x^3 - x^4 - x^6 + x^9)/(1 - x^3 - x^4 - x^5 - 2*x^6 - x^7 + 2*x^9 + 2*x^10 + x^12 - x^15).

A209408 Number of subsets of {1,...,n} containing {a,a+4} for some a.

Original entry on oeis.org

0, 0, 0, 0, 0, 8, 28, 74, 175, 377, 799, 1673, 3471, 7192, 14784, 30208, 61440, 124416, 251328, 506712, 1020015, 2051015, 4119775, 8268215, 16582735, 33239558, 66599068, 133392344, 267099120, 534709192, 1070244924, 2141826898, 4285816671, 8575127217
Offset: 0

Author

David Nacin, Mar 08 2012

Keywords

Comments

For n=5, subsets containing {a,a+4} occur only when a=1. There are 2^3 subsets including {1,5}, thus a(5) = 8.

Crossrefs

Programs

  • Magma
    [2^n - Fibonacci(Floor(n/4) + 2)*Fibonacci(Floor((n + 1)/4) + 2)*Fibonacci(Floor((n + 2)/4) + 2)*Fibonacci(Floor((n + 3)/4) + 2): n in [0..30]]; // G. C. Greubel, Jan 03 2018
  • Mathematica
    Table[2^n - Product[Fibonacci[Floor[(n + i)/4] + 2], {i, 0, 3}], {n, 0, 30}]
    LinearRecurrence[{3, -1, -2, -2, 6, -2, -4, 2, -6, 2, 4, 1, -3, 1, 2}, {0, 0, 0, 0, 0, 8, 28, 74, 175, 377, 799, 1673, 3471, 7192, 14784}, 30]
  • PARI
    for(n=0,20, print1(2^n - fibonacci(floor(n/4) + 2)*fibonacci( floor((n + 1)/4) + 2)*fibonacci(floor((n + 2)/4) + 2)*fibonacci( floor((n + 3)/4) + 2), ", ")) \\ G. C. Greubel, Jan 03 2018
    
  • Python
    #Returns the actual list of valid subsets
    def contains10001(n):
     patterns=list()
     for start in range (1,n-3):
      s=set()
      for i in range(5):
       if (1,0,0,0,1)[i]:
        s.add(start+i)
      patterns.append(s)
     s=list()
     for i in range(2,n+1):
      for temptuple in comb(range(1,n+1),i):
       tempset=set(temptuple)
       for sub in patterns:
        if sub <= tempset:
         s.append(tempset)
         break
     return s
    #Counts all such sets
    def countcontains10001(n):
     return len(contains10001(n))
    #From recurrence
    def a(n, adict={0:0, 1:0, 2:0, 3:0, 4:0, 5:8, 6:28, 7:74, 8:175, 9:377, 10:799, 11:1673, 12:3471, 13:7192, 14:14784}):
     if n in adict:
      return adict[n]
     adict[n]=3*a(n-1)-a(n-2)-2*a(n-3)-2*a(n-4)+6*a(n-5)-2*a(n-6)-4*a(n-7)+2*a(n-8)-6*a(n-9)+2*a(n-10)+4*a(n-11)+a(n-12)-3*a(n-13)+a(n-14)+2*a(n-15)
     return adict[n]
    

Formula

a(n) = 2^n - A208741(n-1).
a(n) = 2^n - Product_{i=0..3} Fibonacci(floor((n + i)/4) + 2).
a(n) = 3*a(n-1) - a(n-2) -2*a(n-3) -2*a(n-4) + 6*a(n-5) - 2*a(n-6) - 4*a(n-7) + 2*a(n-8) - 6*a(n-9) + 2*a(n-10) + 4*a(n-11) + a(n-12) - 3*a(n-13) + a(n-14) + 2*a(n-15).
G.f.: x^5*(8 + 4 x - 2 x^2 - 3 x^3 - 2 x^4 - x^5 - x^6 - x^7 - 2 x^8 - x^9) / ((1 - x) (1 + x) (1 - 2 x) (1 + x^2) (1 - x - x^2) (1 + 3 x^4 + x^8)).

A356639 Number of integer sequences b with b(1) = 1, b(m) > 0 and b(m+1) - b(m) > 0, of length n which transform under the map S into a nonnegative integer sequence. The transform c = S(b) is defined by c(m) = Product_{k=1..m} b(k) / Product_{k=2..m} (b(k) - b(k-1)).

Original entry on oeis.org

1, 1, 3, 17, 155, 2677, 73327, 3578339, 329652351
Offset: 1

Author

Thomas Scheuerle, Aug 19 2022

Keywords

Comments

This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;
A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;
A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;
A279312 <--> A------; A031923 <--> A------.
These transformation pairs are conjectured:
A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;
A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;
A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.

Examples

			a(4) = 17. The 17 transformation pairs of length 4 are:
  {1, 2, 3, 4}  = S({1, 2, 6, 24}).
  {1, 2, 3, 5}  = S({1, 2, 6, 15}).
  {1, 2, 3, 6}  = S({1, 2, 6, 12}).
  {1, 2, 3, 9}  = S({1, 2, 6, 9}).
  {1, 2, 3, 12} = S({1, 2, 6, 8}).
  {1, 2, 3, 21} = S({1, 2, 6, 7}).
  {1, 2, 4, 5}  = S({1, 2, 4, 20}).
  {1, 2, 4, 6}  = S({1, 2, 4, 12}).
  {1, 2, 4, 8}  = S({1, 2, 4, 8}).
  {1, 2, 4, 12} = S({1, 2, 4, 6}).
  {1, 2, 4, 20} = S({1, 2, 4, 5}).
  {1, 2, 6, 7}  = S({1, 2, 3, 21}).
  {1, 2, 6, 8}  = S({1, 2, 3, 12}).
  {1, 2, 6, 9}  = S({1, 2, 3, 9}).
  {1, 2, 6, 12} = S({1, 2, 3, 6}).
  {1, 2, 6, 15} = S({1, 2, 3, 5}).
  {1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
a(4) = A000005((b(2)+1)*(b(2)+4)) + A000005((b(2)+2)*(b(2)+2)) + A000005((b(2)+4)*(b(2)+1)) = 17.
		
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