A060961 -id:A060961 - cp's OEIS frontend

A013983 Expansion of 1/(1-x^2-x^3-x^4-x^5-x^6).

1, 0, 1, 1, 2, 3, 5, 7, 12, 18, 29, 45, 71, 111, 175, 274, 431, 676, 1062, 1667, 2618, 4110, 6454, 10133, 15911, 24982, 39226, 61590, 96706, 151842, 238415, 374346, 587779, 922899, 1449088, 2275281, 3572527

Offset: 0

N. J. A. Sloane

Number of compositions of n into parts p where 2 <= p < = 6. [Joerg Arndt, Jun 24 2013]

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
R. Mullen, On Determining Paint by Numbers Puzzles with Nonunique Solutions, JIS 12 (2009) 09.6.5.
Index entries for linear recurrences with constant coefficients, signature (0,1,1,1,1,1).

First differences of A023437.

m:=40; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(1/(1-x^2-x^3-x^4-x^5-x^6))); // Vincenzo Librandi, Jun 24 2013

CoefficientList[Series[1 / (1 - x^2 - x^3 - x^4 - x^5 - x^6), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 23 2013 *)
LinearRecurrence[{0,1,1,1,1,1},{1,0,1,1,2,3},50] (* Harvey P. Dale, Dec 31 2013 *)

Vec(1/(1-x^2-x^3-x^4-x^5-x^6)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = a(n-6) + a(n-5) + a(n-4) + a(n-3) + a(n-2). - Jon E. Schoenfield, Aug 07 2006

G.f.: 1 / ( (1+x)*(1-x^5-x^3-x)). a(n)+a(n+1) = A060961(n). - R. J. Mathar, Mar 22 2011

A078802 Triangular array T given by T(n,k) = number of 01-words of length n containing k 1's, no three of which are consecutive.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 3, 0, 1, 4, 6, 2, 0, 1, 5, 10, 7, 1, 0, 1, 6, 15, 16, 6, 0, 0, 1, 7, 21, 30, 19, 3, 0, 0, 1, 8, 28, 50, 45, 16, 1, 0, 0, 1, 9, 36, 77, 90, 51, 10, 0, 0, 0, 1, 10, 45, 112, 161, 126, 45, 4, 0, 0, 0, 1, 11, 55, 156, 266, 266, 141, 30, 1, 0, 0, 0, 1, 12, 66, 210, 414

Offset: 0

Clark Kimberling, Dec 06 2002

The rows of T are essentially the antidiagonals of A027907 (trinomial coefficients). Reversing the rows produces A078803. Row sums: A000073.

Also, the diagonals of T are essentially the rows of A027907, so diagonal sums = 3^n. Antidiagonal sums are essentially A060961 (number of ordered partitions of n into 1's, 3's and 5's). - Gerald McGarvey, May 13 2005

			T(4,3) = 2 counts 1+0+1+1 and 1+1+0+1. Top of triangle T:
  1;
  1, 1;
  1, 2, 1;
  1, 3, 3, 0;
  1, 4, 6, 2, 0;

Clark Kimberling, Binary words with restricted repetitions and associated compositions of integers, in Applications of Fibonacci Numbers, vol. 10, Proceedings of the Eleventh International Conference on Fibonacci Numbers and Their Applications, William Webb, editor, Congressus Numerantium, Winnipeg, Manitoba 194 (2009) 141-151.

Cf. A027907, A078803. See A082601 for another version.

seq(seq(sum(binomial(n+1-k,k-j)*binomial(k-j,j),j=0..ceil((k-1)/2)),k=0..n),n=0..20); # Dennis P. Walsh, Apr 04 2012

nn=15; a=1+y x+y^2 x^2;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[a/(1-x a),{x,0,nn}],{x,y}]]//Grid (* Geoffrey Critzer, Sep 15 2012 *)

T(n, k) = T(n-1, k) + T(n-2, k-1) + T(n-3, k-2) with initial values as in first 3 rows.
T(n,k) = Sum_{j=0..ceiling((k-1)/2)} C(n+1-k, k-j)*C(k-j, j). - Dennis P. Walsh, Apr 04 2012
G.f.: (1 + y*x + y^2*x^2)/(1 - (x*(1 + y*x + y^2*x^2))).  - Geoffrey Critzer, Sep 15 2012

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

Alois P. Heinz, Sep 01 2014

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], ... .

			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, ...

Alois P. Heinz, Antidiagonals n = 0..140

Columns k=0-21, 23, 25-28 give: A000007, A000012, A059841, A000045(n+1), A079978, A000930, A121262, A003269(n+1), A182097, A079998, A000073(n+2), A003520, A079977, A079979, A060945, A005708, A001687(n+1), A017817, A082784, A079971, A006498, A005709, A052920, A120400, A060961, A005710, A013979.
Main diagonal gives A246691.
Cf. A246688, A246720 (the same for partitions).

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);

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 *)

A191238 Triangle T(n,k) = coefficient of x^n in expansion of (x+x^3+x^5)^k.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 0, 3, 0, 4, 0, 1, 0, 0, 6, 0, 5, 0, 1, 0, 2, 0, 10, 0, 6, 0, 1, 0, 0, 7, 0, 15, 0, 7, 0, 1, 0, 1, 0, 16, 0, 21, 0, 8, 0, 1, 0, 0, 6, 0, 30, 0, 28, 0, 9, 0, 1, 0, 0, 0, 19, 0, 50, 0, 36, 0, 10, 0, 1, 0, 0, 3, 0, 45, 0, 77, 0, 45, 0, 11, 0, 1, 0, 0, 0, 16, 0, 90, 0, 112, 0, 55, 0, 12, 0, 1

Offset: 1

Vladimir Kruchinin, May 27 2011

Riordan Array (1,x+x^3+x^5) without first column.
Riordan Array (1+x^2+x^4,x+x^3+x^5) numbering triangle (0,0).
For the g.f. 1/(1-x-x^3-x^5) we have a(n)=sum(k=1..n, T(n,k)) (see A060961).
For the e.g.f. exp(1-x-x^3-x^5) we have a(n)=n!*sum(k=1..n, T(n,k)/k!) (see A191237).
Bell Polynomial of second kind B(n,k){1,0,6,0,120,0,0,...,0}=n!/k!*T(n,k).
For more formulas see preprints.

			Triangle begins:
  1,
  0,1,
  1,0,1,
  0,2,0,1,
  1,0,3,0,1,
  0,3,0,4,0,1,
  0,0,6,0,5,0,1,
  0,2,0,10,0,6,0,1,
  0,0,7,0,15,0,7,0,1,
  0,1,0,16,0,21,0,8,0,1

Milan Janjic, Binomial Coefficients and Enumeration of Restricted Words, Journal of Integer Sequences, 2016, Vol 19, #16.7.3.
Vladimir Kruchinin, Derivation of Bell Polynomials of the Second Kind, arXiv:1104.5065 [math.CO], 2011.
Vladimir Kruchinin and D. V. Kruchinin, Composita and their properties, arXiv:1103.2582 [math.CO], 2011-2013.

Cf. A060961 (row sums).

A191238 := proc(n,k)
    add(binomial(j,((n-k-2*j)/2))*binomial(k,j)*((-1)^(n-k)+1),j=0..k)/2 ;
end proc:
seq(seq(A191238(n,m),m=1..n),n=1..10) ;# R. J. Mathar, Dec 16 2015

T(n,k):=sum(binomial(j,((n-k-2*j)/2))*binomial(k,j)*((-1)^(n-k)+1),j,0,k)/2;

T(n,k) = Sum_{j=0..k} binomial(j,((n-k-2*j)/2))*binomial(k,j)*((-1)^(n-k)+1)/2.

A271970 Linear recurrence, with both signature and original terms = 1,0,1,0,1.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 3, 5, 8, 12, 20, 31, 48, 76, 119, 187, 294, 461, 724, 1137, 1785, 2803, 4401, 6910, 10850, 17036, 26749, 42000, 65946, 103545, 162581, 255276, 400821, 629348, 988169, 1551571, 2436195, 3825185, 6006104, 9430468, 14807224, 23249523, 36505176

Offset: 1

Harvey P. Dale, Apr 17 2016

Index entries for linear recurrences with constant coefficients, signature (1, 0, 1, 0, 1).

With[{c={1,0,1,0,1}},LinearRecurrence[c,c,100]]

a(1)=1, a(2)=0, a(3)=1, a(4)=0, a(5)=1, a(n)=a(n-1)+a(n-3)+a(n-5).
G.f.: ( -x*(x-1)*(x^3-x^2-1) ) / ( -1+x+x^3+x^5 ). - R. J. Mathar, Apr 17 2016
a(n) = A060961(n) -A060961(n-2) -2*A060961(n-4). - R. J. Mathar, Apr 17 2016

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A013983 Expansion of 1/(1-x^2-x^3-x^4-x^5-x^6).

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A078802 Triangular array T given by T(n,k) = number of 01-words of length n containing k 1's, no three of which are consecutive.

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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.

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A191238 Triangle T(n,k) = coefficient of x^n in expansion of (x+x^3+x^5)^k.

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A271970 Linear recurrence, with both signature and original terms = 1,0,1,0,1.

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