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.

Showing 1-5 of 5 results.

A094793 a(n) = (1/n!)*A001688(n).

Original entry on oeis.org

9, 53, 181, 465, 1001, 1909, 3333, 5441, 8425, 12501, 17909, 24913, 33801, 44885, 58501, 75009, 94793, 118261, 145845, 178001, 215209, 257973, 306821, 362305, 425001, 495509, 574453, 662481, 760265, 868501, 987909, 1119233, 1263241
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Number of injections from {1,2,3,4} to {1,2,...,n} with no fixed points. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
In general (cf. A094792, A094794, A094795, etc.), the number of injections [k] -> [n] with no fixed points is given by Sum_{i=0..k} (-1)^i*binomial(k,i)*(n-i)!/(n-k)!, which is equal to (1/n!)*f_k(n) where f_k(n) gives the k-th differences of factorial numbers. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{9,53,181,465,1001},40] (* Harvey P. Dale, May 23 2016 *)

Formula

a(n) = n^4 + 6*n^3 + 17*n^2 + 20*n + 9.
a(n) = Sum_{i=0..4} (-1)^i*binomial(4,i)*(n-i)!/(n-4)!. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
G.f.: -(x^4+6*x^2+8*x+9) / (x-1)^5. - Colin Barker, Jun 16 2013
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - Fung Lam, Apr 17 2014
P-recursive: n*a(n) = (n+5)*a(n-1) - a(n-2) with a(0) = 9 and a(1) = 53. Cf. A094791. - Peter Bala, Jul 25 2021

A094794 a(n) = (1/n!)*A001689(n).

Original entry on oeis.org

44, 309, 1214, 3539, 8544, 18089, 34754, 61959, 104084, 166589, 256134, 380699, 549704, 774129, 1066634, 1441679, 1915644, 2506949, 3236174, 4126179, 5202224, 6492089, 8026194, 9837719, 11962724, 14440269, 17312534, 20624939, 24426264
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Crossrefs

Programs

  • Mathematica
    Table[n^5+10n^4+45n^3+100n^2+109n+44,{n,0,30}] (* or *) LinearRecurrence[ {6,-15,20,-15,6,-1},{44,309,1214,3539,8544,18089},30]
  • PARI
    a(n)=n^5+10*n^4+45*n^3+100*n^2+109*n+44 \\ Charles R Greathouse IV, Oct 16 2015

Formula

a(n) = n^5 + 10*n^4 + 45*n^3 + 100*n^2 + 109*n + 44.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6), with a(0)=44, a(1)=309, a(2)=1214, a(3)=3539, a(4)=8544, a(5)=18089. - Harvey P. Dale, Jul 25 2012
G.f.: (x^5 + 10*x^3 + 20*x^2 + 45*x + 44) / (x-1)^6. - Colin Barker, Jun 15 2013
P-recursive: n*a(n) = (n+6)*a(n-1) - a(n-2) with a(0) = 44 and a(1) = 309. Cf. A094791 and A096307. - Peter Bala, Jul 25 2021

A094792 a(n) = (1/n!)*A001565(n).

Original entry on oeis.org

2, 11, 32, 71, 134, 227, 356, 527, 746, 1019, 1352, 1751, 2222, 2771, 3404, 4127, 4946, 5867, 6896, 8039, 9302, 10691, 12212, 13871, 15674, 17627, 19736, 22007, 24446, 27059, 29852, 32831, 36002, 39371, 42944, 46727, 50726, 54947, 59396, 64079
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Number of injections from {1,2,3} to {1,2,...,n} with no fixed points. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006

Crossrefs

Programs

Formula

a(n) = n^3 + 3*n^2 + 5*n + 2.
a(n) = Sum_{i=0..3} (-1)^i*binomial(3,i)*(n-i)!/(n-3)!. - Fiona T. Brunk (fbrunk(AT)mcs.st-and.ac.uk), May 23 2006
G.f.: (x^3+3*x+2) / (x-1)^4. - Colin Barker, Jun 15 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Fung Lam, Apr 17 2014
P-recursive: n*a(n) = (n+4)*a(n-1) - a(n-2) with a(0) = 2 and a(1) = 11. Cf. A094791. - Peter Bala, Jul 25 2021

A094795 a(n) = (1/n!)*A023043(n).

Original entry on oeis.org

265, 2119, 9403, 30637, 81901, 190435, 398959, 770713, 1395217, 2394751, 3931555, 6215749, 9513973, 14158747, 20558551, 29208625, 40702489, 55744183, 75161227, 99918301, 131131645, 170084179, 218241343, 277267657, 349044001, 435685615
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{7,-21,35,-35,21,-7,1},{265,2119,9403,30637,81901,190435,398959},30] (* Harvey P. Dale, Aug 29 2023 *)

Formula

a(n) = n^6 + 15*n^5 + 100*n^4 + 355*n^3 + 694*n^2 + 689*n + 265.
G.f.: -(265 + 264*x + 135*x^2 + 40*x^3 + 15*x^4 + x^6)/(x-1)^7. - R. J. Mathar, Nov 15 2019
P-recursive: n*a(n) = (n+7)*a(n-1) - a(n-2) with a(0) = 265 and a(1) = 2119. Cf. A094791. - Peter Bala, Jul 25 2021

A126353 Triangle read by rows: matrix product of the Stirling numbers of the first kind with the binomial coefficients.

Original entry on oeis.org

1, 1, 0, 1, -1, 1, 1, -3, 5, -2, 1, -6, 17, -20, 9, 1, -10, 45, -100, 109, -44, 1, -15, 100, -355, 694, -689, 265, 1, -21, 196, -1015, 3094, -5453, 5053, -1854, 1, -28, 350, -2492, 10899, -29596, 48082, -42048, 14833
Offset: 1

Views

Author

Thomas Wieder, Dec 29 2006

Keywords

Comments

Many well-known integer sequences arise from such a matrix product of combinatorial coefficients. In the present case we have as the first row A000166 = subfactorial or rencontres numbers, or derangements: number of permutations of n elements with no fixed points.

Examples

			Matrix begins:
1 0 1 -2 9 -44 265 -1854 14833
0 1 -1 5 -20 109 -689 5053 -42048
0 0 1 -3 17 -100 694 -5453 48082
0 0 0 1 -6 45 -355 3094 -29596
0 0 0 0 1 -10 100 -1015 10899
0 0 0 0 0 1 -15 196 -2492
0 0 0 0 0 0 1 -21 350
0 0 0 0 0 0 0 1 -28
0 0 0 0 0 0 0 0 1
		

Crossrefs

Signed version of A094791 [from Olivier Gérard, Jul 31 2011]

Formula

(In Maple notation:) Matrix product B.A of matrix A[i,j]:=binomial(j-1,i-1) with i = 1 to p+1, j = 1 to p+1, p=8 and of matrix B[i,j]:=stirling1(j,i) with i from 1 to d, j from 1 to d, d=9.
Showing 1-5 of 5 results.