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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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).

A371125 Number of compositions of 6*n into parts 1 and 6.

Original entry on oeis.org

1, 2, 9, 43, 196, 882, 3970, 17887, 80608, 363254, 1636944, 7376591, 33241289, 149795989, 675029164, 3041899638, 13707783053, 61771701389, 278363253873, 1254394801761, 5652708454881, 25472931513057, 114789263420590, 517277526141329, 2331019740675071
Offset: 0

Views

Author

Seiichi Manyama, Jun 22 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+5*k, n-k));

Formula

a(n) = A005708(6*n).
a(n) = Sum_{k=0..n} binomial(n+5*k,n-k).
a(n) = 7*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
G.f.: 1/(1 - x - x/(1 - x)^5).

A017900 Expansion of 1/(1 - x^6 - x^7 - x^8 - ...).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196, 251, 322, 414, 533, 686, 882, 1133, 1455, 1869, 2402, 3088, 3970, 5103, 6558, 8427, 10829, 13917, 17887, 22990, 29548, 37975, 48804, 62721, 80608
Offset: 0

Views

Author

Keywords

Comments

A Lamé sequence of higher order.
Number of compositions of n into parts >= 6. - Milan Janjic, Jun 28 2010
a(n+6) equals the number of n-length binary words such that 0 appears only in a run which length is a multiple of 6. - Milan Janjic, Feb 17 2015
Same as sequence A005708 with 1, 0, 0, 0, 0, 0 prepended. - Linas Vepstas, Feb 06 2024

Crossrefs

For Lamé sequences of orders 1 through 9 see A000045, A000930, A017898-A017904.

Programs

  • Maple
    f := proc(r) local t1,i; t1 := []; for i from 1 to r do t1 := [op(t1),0]; od: for i from 1 to r+1 do t1 := [op(t1),1]; od: for i from 2*r+2 to 50 do t1 := [op(t1),t1[i-1]+t1[i-1-r]]; od: t1; end; # set r = order
    a:= n-> (Matrix(6, (i, j)-> `if`(i=j-1, 1, `if`(j=1, [1, 0$4, 1][i], 0)))^n)[6, 6]: seq(a(n), n=0..80); # Alois P. Heinz, Aug 04 2008
  • Mathematica
    f[n_] := If[n < 1, 1, Sum[ Binomial[ n - 5 k - 5, k], {k, 0, (n - 5)/6}]]; Array[f, 49, 0] (* Adi Dani, Robert G. Wilson v, Jul 04 2011 *)
    LinearRecurrence[{1, 0, 0, 0, 0, 1}, {1, 0, 0, 0, 0, 0}, 60] (* Jean-François Alcover, Feb 13 2016 *)
  • PARI
    Vec((1-x)/(1-x-x^6)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

Formula

G.f.: 1/(1-Sum_{k>=6} x^k).
G.f.: (1-x)/(1-x-x^6). - Alois P. Heinz, Aug 04 2008
For positive integers n and k such that k <= n <= 6*k, and 5 divides n-k, define c(n,k) = binomial(k,(n-k)/5), and c(n,k) = 0 otherwise. Then, for n >= 1, a(n+6) = Sum_{k=1..n} c(n,k). - Milan Janjic, Dec 09 2011

A246690 Number A(n,k) of compositions of n into parts of the k-th list of distinct parts in the order given by A246688; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 2, 0, 1, 0, 1, 1, 0, 3, 1, 1, 0, 1, 0, 1, 1, 5, 0, 1, 0, 1, 1, 0, 2, 0, 8, 1, 1, 0, 1, 0, 1, 0, 3, 0, 13, 0, 1, 0, 1, 0, 1, 1, 1, 4, 1, 21, 1, 1, 0, 1, 1, 0, 1, 2, 0, 6, 0, 34, 0, 1, 0, 1, 1, 2, 0, 1, 3, 0, 9, 0, 55, 1, 1, 0
Offset: 0

Views

Author

Alois P. Heinz, Sep 01 2014

Keywords

Comments

The first lists of distinct parts in the order given by A246688 are: 0:[], 1:[1], 2:[2], 3:[1,2], 4:[3], 5:[1,3], 6:[4], 7:[1,4], 8:[2,3], 9:[5], 10:[1,2,3], 11:[1,5], 12:[2,4], 13:[6], 14:[1,2,4], 15:[1,6], 16:[2,5], 17:[3,4], 18:[7], 19:[1,2,5], 20:[1,3,4], ... .

Examples

			Square array A(n,k) begins:
  1, 1, 1,  1, 1,  1, 1,  1, 1, 1,   1, 1, 1, 1,   1, ...
  0, 1, 0,  1, 0,  1, 0,  1, 0, 0,   1, 1, 0, 0,   1, ...
  0, 1, 1,  2, 0,  1, 0,  1, 1, 0,   2, 1, 1, 0,   2, ...
  0, 1, 0,  3, 1,  2, 0,  1, 1, 0,   4, 1, 0, 0,   3, ...
  0, 1, 1,  5, 0,  3, 1,  2, 1, 0,   7, 1, 2, 0,   6, ...
  0, 1, 0,  8, 0,  4, 0,  3, 2, 1,  13, 2, 0, 0,  10, ...
  0, 1, 1, 13, 1,  6, 0,  4, 2, 0,  24, 3, 3, 1,  18, ...
  0, 1, 0, 21, 0,  9, 0,  5, 3, 0,  44, 4, 0, 0,  31, ...
  0, 1, 1, 34, 0, 13, 1,  7, 4, 0,  81, 5, 5, 0,  55, ...
  0, 1, 0, 55, 1, 19, 0, 10, 5, 0, 149, 6, 0, 0,  96, ...
  0, 1, 1, 89, 0, 28, 0, 14, 7, 1, 274, 8, 8, 0, 169, ...
		

Crossrefs

Main diagonal gives A246691.
Cf. A246688, A246720 (the same for partitions).

Programs

  • Maple
    b:= proc(n, i) b(n, i):= `if`(n=0, [[]], `if`(i>n, [],
          [map(x->[i, x[]], b(n-i, i+1))[], b(n, i+1)[]]))
        end:
    f:= proc() local i, l; i, l:=0, [];
          proc(n) while n>=nops(l)
            do l:=[l[], b(i, 1)[]]; i:=i+1 od; l[n+1]
          end
        end():
    g:= proc(n, l) option remember; `if`(n=0, 1,
          add(`if`(i>n, 0, g(n-i, l)), i=l))
        end:
    A:= (n, k)-> g(n, f(k)):
    seq(seq(A(n, d-n), n=0..d), d=0..14);
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, {{}}, If[i>n, {}, Join[Prepend[#, i]& /@ b[n - i, i + 1], b[n, i + 1]]]];
    f = Module[{i = 0, l = {}}, Function[n, While[n >= Length[l], l = Join[l, b[i, 1]]; i++]; l[[n + 1]]]];
    g[n_, l_] := g[n, l] = If[n==0, 1, Sum[If[i>n, 0, g[n - i, l]], {i, l}]];
    A[n_, k_] := g[n, f[k]];
    Table[Table[A[n, d - n], {n, 0, d}], {d, 0, 14}] // Flatten (* Jean-François Alcover, Dec 07 2020, after Alois P. Heinz *)

A339088 Number of compositions (ordered partitions) of n into distinct parts congruent to 1 mod 6.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 1, 4, 6, 0, 0, 0, 1, 4, 6, 0, 0, 0, 1, 6, 12, 0, 0, 0, 1, 6, 18, 24, 0, 0, 1, 8, 24, 24, 0, 0, 1, 8, 30, 48, 0, 0, 1, 10, 42, 72, 0, 0, 1, 10, 48, 120, 120, 0, 1, 12, 60, 144, 120, 0, 1, 12, 72, 216, 240, 0, 1, 14, 84, 264, 360
Offset: 0

Views

Author

Ilya Gutkovskiy, Nov 23 2020

Keywords

Examples

			a(21) = 6 because we have [13, 7, 1], [13, 1, 7], [7, 13, 1], [7, 1, 13], [1, 13, 7] and [1, 7, 13].
		

Crossrefs

Programs

  • Mathematica
    nmax = 83; CoefficientList[Series[Sum[k! x^(k (3 k - 2))/Product[1 - x^(6 j), {j, 1, k}], {k, 0, nmax}], {x, 0, nmax}], x]

Formula

G.f.: Sum_{k>=0} k! * x^(k*(3*k - 2)) / Product_{j=1..k} (1 - x^(6*j)).

A367637 G.f. A(x) satisfies A(x) = 1 / (1 - x * A(x^6)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 3, 4, 5, 6, 7, 9, 12, 16, 21, 27, 34, 43, 55, 71, 92, 119, 153, 196, 251, 322, 414, 533, 686, 882, 1133, 1455, 1869, 2402, 3088, 3970, 5103, 6558, 8427, 10829, 13917, 17888, 22992, 29551, 37979, 48809, 62727, 80617, 103612, 133167
Offset: 0

Views

Author

Seiichi Manyama, Dec 01 2023

Keywords

Comments

This sequence is different from A005708.

Crossrefs

Programs

  • PARI
    a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=sum(j=0, (i-1)\6, v[j+1]*v[i-6*j])); v;

Formula

a(0) = 1; a(n) = Sum_{k=0..floor((n-1)/6)} a(k) * a(n-1-6*k).

A373302 Number of compositions of 6*n-2 into parts 1 and 6.

Original entry on oeis.org

1, 6, 27, 119, 533, 2402, 10829, 48804, 219925, 991044, 4465957, 20125051, 90690002, 408678475, 1841637299, 8299012941, 37398034921, 168527634148, 759439995404, 3422282105232, 15421909405056, 69496108849357, 313171930813206, 1411253951813003, 6359566489040219
Offset: 1

Views

Author

Seiichi Manyama, Jun 23 2024

Keywords

Crossrefs

Programs

  • PARI
    a(n) = sum(k=0, n, binomial(n+3+5*k, n-1-k));

Formula

a(n) = A005708(6*n-2).
a(n) = Sum_{k=0..n} binomial(n+3+5*k,n-1-k).
a(n) = 7*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
G.f.: x*(1-x)/((1-x)^6 - x).
a(n) = A099242(n-1) - A099242(n-2).
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