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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A058368 Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 5 sites wide.

Original entry on oeis.org

1, 1, 1, 1, 6, 7, 8, 9, 10, 16, 23, 31, 40, 50, 66, 89, 120, 160, 210, 276, 365, 485, 645, 855, 1131, 1496, 1981, 2626, 3481, 4612, 6108, 8089, 10715, 14196, 18808, 24916, 33005, 43720, 57916, 76724, 101640, 134645, 178365, 236281, 313005, 414645
Offset: 1

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 17 2000

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.

Examples

			a(5) = 6 because there is one way to put zero molecule to the necklace and 5 ways to put one molecule.
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{1,0,0,0,1},{1,1,1,1,6},50] (* Harvey P. Dale, Aug 14 2020 *)

Formula

a(n) = 1 + n*Sum_{i=1..n/5} binomial(n-4*i-1, i-1)/i.
a(n) = a(n-1) + a(n-5) for n >= 6.
G.f.: (x+5*x^5)/(1-x-x^5).

A058364 Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 9 sites wide.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 28, 39, 51, 64, 78, 93, 109, 126, 144, 172, 211, 262, 326, 404, 497, 606, 732, 876, 1048, 1259, 1521, 1847, 2251, 2748, 3354, 4086, 4962, 6010, 7269, 8790, 10637, 12888, 15636, 18990, 23076, 28038
Offset: 1

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 17 2000

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.

Examples

			a(9) = 10 because there is one way to put zero molecule to the necklace and 9 ways to put one molecule.
		

References

  • E. Di Cera and Y. Kong, Theory of multivalent binding in one and two-dimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107-124.
  • Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), pp. 4790-4799.

Crossrefs

Formula

a(n) = 1 + n*sum(binomial(n-1-8*i, i-1)/i, i=1..n/9). a(n) = a(n-1) + a(n-9), a(n) = 1 for n = 1..8, a(9) = 10. generating function = (x+9*x^9)/(1-x-x^9).

A058365 Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 8 sites wide.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 9, 10, 11, 12, 13, 14, 15, 16, 25, 35, 46, 58, 71, 85, 100, 116, 141, 176, 222, 280, 351, 436, 536, 652, 793, 969, 1191, 1471, 1822, 2258, 2794, 3446, 4239, 5208, 6399, 7870, 9692, 11950, 14744, 18190, 22429, 27637, 34036, 41906
Offset: 1

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 17 2000

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.

Examples

			a(8) = 9 because there is one way to put zero molecule to the necklace and 8 ways to put one molecule.
		

References

  • E. Di Cera and Y. Kong, Theory of multivalent binding in one and two-dimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107-124.
  • Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), pp. 4790-4799.

Crossrefs

Formula

a(n) = 1 + n*sum(binomial(n-1-7*i, i-1)/i, i=1..n/8). a(n) = a(n-1) + a(n-8), a(n) = 1 for n = 1..7, a(8) = 9. generating function = (x+8*x^8)/(1-x-x^8).

A058366 Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 7 sites wide.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 8, 9, 10, 11, 12, 13, 14, 22, 31, 41, 52, 64, 77, 91, 113, 144, 185, 237, 301, 378, 469, 582, 726, 911, 1148, 1449, 1827, 2296, 2878, 3604, 4515, 5663, 7112, 8939, 11235, 14113, 17717, 22232, 27895, 35007, 43946, 55181, 69294, 87011
Offset: 1

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 17 2000

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.

Examples

			a(7) = 8 because there is one way to put zero molecule to the necklace and 7 ways to put one molecule.
		

References

  • E. Di Cera and Y. Kong, Theory of multivalent binding in one and two-dimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107-124.
  • Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), pp. 4790-4799.

Crossrefs

Formula

a(n) = 1 + n*sum(binomial(n-1-6*i, i-1)/i, i=1..n/7). a(n) = a(n-1) + a(n-7), a(n) = 1 for n = 1..6, a(7) = 8. generating function = (x+7*x^7)/(1-x-x^7).

A058367 Number of ways to cover (without overlapping) a ring lattice (necklace) of n sites with molecules that are 6 sites wide.

Original entry on oeis.org

1, 1, 1, 1, 1, 7, 8, 9, 10, 11, 12, 19, 27, 36, 46, 57, 69, 88, 115, 151, 197, 254, 323, 411, 526, 677, 874, 1128, 1451, 1862, 2388, 3065, 3939, 5067, 6518, 8380, 10768, 13833, 17772, 22839, 29357, 37737, 48505, 62338, 80110, 102949, 132306, 170043, 218548
Offset: 1

Views

Author

Yong Kong (ykong(AT)curagen.com), Dec 17 2000

Keywords

Comments

This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 1...m-1, a(m) = m+1. The generating function is (x+m*x^m)/(1-x-x^m). Also a(n) = 1 + n*Sum_{i=1..n/m} binomial(n-1-(m-1)*i, i-1)/i. This gives the number of ways to cover (without overlapping) a ring lattice (or necklace) of n sites with molecules that are m sites wide. Special cases: m=2: A000204, m=3: A001609, m=4: A014097, m=5: A058368, m=6: A058367, m=7: A058366, m=8: A058365, m=9: A058364.

Examples

			a(6) = 7 because there is one way to put zero molecule to the necklace and 6 ways to put one molecule.
		

References

  • E. Di Cera and Y. Kong, Theory of multivalent binding in one and two-dimensional lattices, Biophysical Chemistry, Vol. 61 (1996), pp. 107-124.
  • Y. Kong, General recurrence theory of ligand binding on a three-dimensional lattice, J. Chem. Phys. Vol. 111 (1999), pp. 4790-4799.

Crossrefs

Formula

a(n) = 1 + n*sum(binomial(n-1-5*i, i-1)/i, i=1..n/6). a(n) = a(n-1) + a(n-6), a(n) = 1 for n = 1..5, a(6) = 7. generating function = (x+6*x^6)/(1-x-x^6).

A141539 Square array A(n,k) of numbers of length n binary words with at least k "0" between any two "1" digits (n,k >= 0), read by antidiagonals.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 3, 8, 1, 2, 3, 5, 16, 1, 2, 3, 4, 8, 32, 1, 2, 3, 4, 6, 13, 64, 1, 2, 3, 4, 5, 9, 21, 128, 1, 2, 3, 4, 5, 7, 13, 34, 256, 1, 2, 3, 4, 5, 6, 10, 19, 55, 512, 1, 2, 3, 4, 5, 6, 8, 14, 28, 89, 1024, 1, 2, 3, 4, 5, 6, 7, 11, 19, 41, 144, 2048, 1, 2, 3, 4, 5, 6, 7, 9, 15, 26, 60, 233, 4096
Offset: 0

Views

Author

Alois P. Heinz, Aug 15 2008

Keywords

Comments

A(n,k+1) = A(n,k) - A143291(n,k).
From Gary W. Adamson, Dec 19 2009: (Start)
Alternative method generated from variants of an infinite lower triangle T(n) = A000012 = (1; 1,1; 1,1,1; ...) such that T(n) has the leftmost column shifted up n times. Then take lim_{k->infinity} T(n)^k, obtaining a left-shifted vector considered as rows of an array (deleting the first 1) as follows:
1, 2, 4, 8, 16, 32, 64, 128, 256, ... = powers of 2
1, 1, 2, 3, 5, 8, 13, 21, 34, ... = Fibonacci numbers
1, 1, 1, 2, 3, 4, 6, 9, 13, ... = A000930
1, 1, 1, 1, 2, 3, 4, 5, 7, ... = A003269
... with the next rows A003520, A005708, A005709, ... such that beginning with the Fibonacci row, the succession of rows are recursive sequences generated from a(n) = a(n-1) + a(n-2); a(n) = a(n-1) + a(n-3), ... a(n) = a(n-1) + a(n-k); k = 2,3,4,... Last, columns going up from the topmost 1 become rows of triangle A141539. (End)

Examples

			A(4,2) = 6, because 6 binary words of length 4 have at least 2 "0" between any two "1" digits: 0000, 0001, 0010, 0100, 1000, 1001.
Square array A(n,k) begins:
    1,  1,  1,  1,  1,  1,  1,  1, ...
    2,  2,  2,  2,  2,  2,  2,  2, ...
    4,  3,  3,  3,  3,  3,  3,  3, ...
    8,  5,  4,  4,  4,  4,  4,  4, ...
   16,  8,  6,  5,  5,  5,  5,  5, ...
   32, 13,  9,  7,  6,  6,  6,  6, ...
   64, 21, 13, 10,  8,  7,  7,  7, ...
  128, 34, 19, 14, 11,  9,  8,  8, ...
		

Crossrefs

Cf. column k=0: A000079, k=1: A000045(n+2), k=2: A000930(n+2), A068921, A078012(n+5), k=3: A003269(n+4), A017898(n+7), k=4: A003520(n+4), A017899(n+9), k=5: A005708(n+5), A017900(n+11), k=6: A005709(n+6), A017901(n+13), k=7: A005710(n+7), A017902(n+15), k=8: A005711(n+7), A017903(n+17), k=9: A017904(n+19), k=10: A017905(n+21), k=11: A017906(n+23), k=12: A017907(n+25), k=13: A017908(n+27), k=14: A017909(n+29).
Main diagonal gives A000027(n+1).
A(2n,n) gives A000217(n+1)
A(3n,n) gives A008778.
A(3n,2n) gives A034856(n+1).
A(2n,3n) gives A005408.
A(2^n-1,n) gives A376697.
See also A143291.

Programs

  • Maple
    A:= proc(n, k) option remember;
          if k=0 then 2^n
        elif n<=k and n>=0 then n+1
        elif n>0 then A(n-1, k) +A(n-k-1, k)
        else          A(n+1+k, k) -A(n+k, k)
          fi
        end:
    seq(seq(A(n, d-n), n=0..d), d=0..15);
  • Mathematica
    a[n_, k_] := a[n, k] = Which[k == 0, 2^n, n <= k && n >= 0, n+1, n > 0, a[n-1, k] + a[n-k-1, k], True, a[n+1+k, k] - a[n+k, k]]; Table[Table[a[n, d-n], {n, 0, d}], {d, 0, 15}] // Flatten (* Jean-François Alcover, Dec 17 2013, translated from Maple *)

Formula

G.f. of column k: x^(-k)/(1-x-x^(k+1)).
A(n,k) = 2^n if k=0, otherwise A(n,k) = n+1 if n<=k, otherwise A(n,k) = A(n-1,k) + A(n-k-1,k).

A369813 Expansion of 1/(1 - x^2 - x^7).

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 1, 1, 2, 1, 3, 1, 4, 2, 5, 4, 6, 7, 7, 11, 9, 16, 13, 22, 20, 29, 31, 38, 47, 51, 69, 71, 98, 102, 136, 149, 187, 218, 258, 316, 360, 452, 509, 639, 727, 897, 1043, 1257, 1495, 1766, 2134, 2493, 3031, 3536, 4288, 5031, 6054, 7165, 8547, 10196, 12083, 14484, 17114, 20538, 24279, 29085, 34475
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2024

Keywords

Comments

Number of compositions of n into parts 2 and 7.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/(1-x^2-x^7),{x,0,80}],x] (* or *) LinearRecurrence[{0,1,0,0,0,0,1},{1,0,1,0,1,0,1},80] (* Harvey P. Dale, Dec 04 2024 *)
  • PARI
    my(N=70, x='x+O('x^N)); Vec(1/(1-x^2-x^7))
    
  • PARI
    a(n) = sum(k=0, n\7, ((n-5*k)%2==0)*binomial((n-5*k)/2, k));

Formula

a(n) = a(n-2) + a(n-7).
a(n) = A007380(n-7) for n >= 8.

A369814 Expansion of 1/(1 - x^3 - x^7).

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 2, 0, 1, 3, 1, 1, 4, 3, 1, 5, 6, 2, 6, 10, 5, 7, 15, 11, 9, 21, 21, 14, 28, 36, 25, 37, 57, 46, 51, 85, 82, 76, 122, 139, 122, 173, 224, 204, 249, 346, 343, 371, 519, 567, 575, 768, 913, 918, 1139, 1432, 1485, 1714, 2200, 2398, 2632, 3339, 3830, 4117, 5053, 6030, 6515, 7685
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2024

Keywords

Comments

Number of compositions of n into parts 3 and 7.

Crossrefs

Programs

  • PARI
    my(N=80, x='x+O('x^N)); Vec(1/(1-x^3-x^7))
    
  • PARI
    a(n) = sum(k=0, n\7, ((n-4*k)%3==0)*binomial((n-4*k)/3, k));

Formula

a(n) = a(n-3) + a(n-7).

A369815 Expansion of 1/(1 - x^4 - x^7).

Original entry on oeis.org

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 2, 1, 0, 1, 3, 1, 0, 3, 4, 1, 1, 6, 5, 1, 4, 10, 6, 2, 10, 15, 7, 6, 20, 21, 9, 16, 35, 28, 15, 36, 56, 37, 31, 71, 84, 52, 67, 127, 121, 83, 138, 211, 173, 150, 265, 332, 256, 288, 476, 505, 406, 553, 808, 761, 694, 1029, 1313, 1167, 1247, 1837, 2074, 1861, 2276, 3150, 3241
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2024

Keywords

Comments

Number of compositions of n into parts 4 and 7.

Crossrefs

Programs

  • PARI
    my(N=80, x='x+O('x^N)); Vec(1/(1-x^4-x^7))
    
  • PARI
    a(n) = sum(k=0, n\7, ((n-3*k)%4==0)*binomial((n-3*k)/4, k));

Formula

a(n) = a(n-4) + a(n-7).

A369816 Expansion of 1/(1 - x^5 - x^7).

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 2, 0, 1, 1, 0, 3, 0, 3, 1, 1, 4, 0, 6, 1, 4, 5, 1, 10, 1, 10, 6, 5, 15, 2, 20, 7, 15, 21, 7, 35, 9, 35, 28, 22, 56, 16, 70, 37, 57, 84, 38, 126, 53, 127, 121, 95, 210, 91, 253, 174, 222, 331, 186, 463, 265, 475, 505, 408, 794, 451, 938, 770, 883, 1299, 859, 1732, 1221
Offset: 0

Views

Author

Seiichi Manyama, Feb 02 2024

Keywords

Comments

Number of compositions of n into parts 5 and 7.

Crossrefs

Programs

  • PARI
    my(N=80, x='x+O('x^N)); Vec(1/(1-x^5-x^7))
    
  • PARI
    a(n) = sum(k=0, n\7, ((n-2*k)%5==0)*binomial((n-2*k)/5, k));

Formula

a(n) = a(n-5) + a(n-7).
G.f.: 1/((1-x+x^2)*(1+x-x^3-x^4-x^5)). - R. J. Mathar, Jul 03 2024
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