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 1-7 of 7 results.

A086764 Triangle T(n, k), read by row, related to Euler's difference table A068106 (divide column k of A068106 by k!).

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 9, 11, 7, 3, 1, 44, 53, 32, 13, 4, 1, 265, 309, 181, 71, 21, 5, 1, 1854, 2119, 1214, 465, 134, 31, 6, 1, 14833, 16687, 9403, 3539, 1001, 227, 43, 7, 1, 133496, 148329, 82508, 30637, 8544, 1909, 356, 57, 8, 1
Offset: 0

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Author

Philippe Deléham, Aug 02 2003

Keywords

Comments

The k-th column sequence, k >= 0, without leading zeros, enumerates the ways to distribute n beads, n >= 1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k+1 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords each contribute a factor 1, hence for n=0 one has 1. See A000255 for the description of a fixed cord with beads. This comment derives from a family of recurrences found by Malin Sjodahl for a combinatorial problem for certain quark and gluon diagrams (Feb 27 2010). - Wolfdieter Lang, Jun 02 2010

Examples

			Formatted as a square array:
      1      3     7    13   21   31  43 57 ... A002061;
      2     11    32    71  134  227 356    ... A094792;
      9     53   181   465 1001 1909        ... A094793;
     44    309  1214  3539 8544             ... A094794;
    265   2119  9403 30637                  ... A023043;
   1854  16687 82508                        ... A023044;
  14833 148329                              ... A023045;
Formatted as a triangular array (mirror of A076731):
       1;
       0      1;
       1      1     1;
       2      3     2     1;
       9     11     7     3    1;
      44     53    32    13    4    1;
     265    309   181    71   21    5    1;
    1854   2119  1214   465  134   31    6   1;
   14833  16687  9403  3539 1001  227   43   7   1;
  133496 148329 82508 30637 8544 1909  356  57   8   1;
		

Crossrefs

Programs

  • Magma
    A086764:= func< n,k | (&+[(-1)^j*Binomial(n-k,j)*Factorial(n-j): j in [0..n]])/Factorial(k) >;
    [A086764(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023
    
  • Mathematica
    T[n_,k_]:=(1/k!)*Sum[(-1)^j*Binomial[n-k,j]*(n-j)!,{j,0,n}];Flatten[Table[T[n,k],{n,0,11},{k,0,n}]] (* Indranil Ghosh, Feb 20 2017 *)
    T[n_, k_] := (n!/k!) HypergeometricPFQ[{k-n},{-n},-1];
    Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)
  • SageMath
    def A086764(n,k): return sum((-1)^j*binomial(n-k,j)*factorial(n-j) for j in range(n+1))//factorial(k)
    flatten([[A086764(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

Formula

T(n, n) = 1; T(n+1, n) = n.
T(n+2, n) = A002061(n+1) = n^2 + n + 1; T(n+3, n) = n^3 + 3*n^2 + 5*n + 2.
T(n, k) = (k + 1)*T(n, k + 1) - T(n-1, k); T(n, n) = 1; T(n, k) = 0, if k > n.
T(n, k) = (n-1)*T(n-1, k) + (n-k-1)*T(n-2, k).
k!*T(n, k) = A068106(n, k). [corrected by Georg Fischer, Aug 13 2022]
Sum_{k>=0} T(n, k) = A003470(n+1).
T(n, k) = (1/k!) * Sum_{j>=0} (-1)^j*binomial(n-k, j)*(n-j)!. - Philippe Deléham, Jun 13 2005
From Peter Bala, Aug 14 2008: (Start)
The following remarks all relate to the array read as a square array: e.g.f for column k: exp(-y)/(1-y)^(k+1); e.g.f. for array: exp(-y)/(1-x-y) = (1 + x + x^2 + x^3 + ...) + (x + 2*x^2 + 3*x^3 + 4*x^4 + ...)*y + (1 + 3*x + 7*x^2 + 13*x^3 + ...)*y^2/2! + ... .
This table is closely connected to the constant e. The row, column and diagonal entries of this table occur in series formulas for e.
Row n for n >= 2: e = n!*(1/T(n,0) + (-1)^n*[1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) + 1/(3!*T(n,2)*T(n,3)) + ...]). For example, row 3 gives e = 6*(1/2 - 1/(1!*2*11) - 1/(2!*11*32) - 1/(3!*32*71) - ...). See A095000.
Column 0: e = 2 + Sum_{n>=2} (-1)^n*n!/(T(n,0)*T(n+1,0)) = 2 + 2!/(1*2) - 3 !/(2*9) + 4!/(9*44) - ... .
Column k, k >= 1: e = (1 + 1/1! + 1/2! + ... + 1/k!) + 1/k!*Sum_{n >= 0} (-1)^n*n!/(T(n,k)*T(n+1,k)). For example, column 3 gives e = 8/3 + 1/6*(1/(1*3) - 1/(3*13) + 2/(13*71) - 6/(71*465) + ...).
Main diagonal: e = 1 + 2*(1/(1*1) - 1/(1*7) + 1/(7*71) - 1/(71*1001) + ...).
First subdiagonal: e = 8/3 + 5/(3*32) - 7/(32*465) + 9/(465*8544) - ... .
Second subdiagonal: e = 2*(1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...). See A143413.
Third subdiagonal: e = 3 - (2*3*5)/(2*53) + (3*4*7)/(53*1214) - (4*5*9)/(1214*30637) + ... .
For the corresponding results for the constants 1/e, sqrt(e) and 1/sqrt(e) see A143409, A143410 and A143411 respectively. For other arrays similarly related to constants see A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)). (End)
G.f. for column k is hypergeom([1,k+1],[],x/(x+1))/(x+1). - Mark van Hoeij, Nov 07 2011
T(n, k) = (n!/k!)*hypergeom([k-n], [-n], -1). - Peter Luschny, Oct 05 2017

Extensions

More terms from David Wasserman, Mar 28 2005
Additional comments from Zerinvary Lajos, Mar 30 2006
Edited by N. J. A. Sloane, Sep 24 2011

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

A076731 Table T(n,k) giving number of ways of obtaining exactly 0 correct answers on an (n,k)-matching problem (1 <= k <= n).

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 3, 7, 11, 9, 4, 13, 32, 53, 44, 5, 21, 71, 181, 309, 265, 6, 31, 134, 465, 1214, 2119, 1854, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496, 9, 73, 527, 3333, 18089, 81901, 296967, 808393
Offset: 1

Views

Author

Mohammad K. Azarian, Oct 28 2002

Keywords

Comments

Hanson et al. define the (n,k)-matching problem in the following realistic way. A matching question on an exam has k questions with n possible answers to choose from, each question having a unique answer. If a student guesses the answers at random, using each answer at most once, what is the probability of obtaining r of the k correct answers?
The T(n,k) represent the number of ways of obtaining exactly zero correct answers, i.e., r=0, given k questions and n possible answers, 1 <= k <= n.
T(n,k) is the number of injections from [1,...,k] into [1,...,n] with no fixed points. - David Bevan, Apr 29 2013

Examples

			0; 1,1; 2,3,2; 3,7,11,9; ...
Formatted as a square array:
0 1 2 3 4 5 6 7 8
1 3 7 13 21 31 43 57 which equals A002061
2 11 32 71 134 227 356 which equals A094792
9 53 181 465 1001 1909 which equals A094793
44 309 1214 3539 8544 which equals A094794
265 2119 9403 30637 which equals A023043
1854 16687 82508 which equals A023044
14833 148329 which equals A023045
Columns give A000255 A000153 A000261 A001909 A001910
Formatted as a triangular array (mirror image of A086764):
0
1 1
2 3 2
3 7 11 9
4 13 32 53 44
5 21 71 181 309 265
6 31 134 465 1214 2119 1854
7 43 227 1001 3539 9403 16687 14833
8 57 356 1909 8544 30637 82508 148329 133496
		

Crossrefs

Similar to A060475.

Programs

  • Maple
    A076731 := proc(n,k): (1/(n-k)!)*A061312(n-1,k-1) end: A061312:=proc(n,k): add(((-1)^j)*binomial(k+1,j)*(n+1-j)!, j=0..k+1) end: for n from 1 to 7 do seq(A076731(n,k), k=1..n) od; seq(seq(A076731(n,k), k=1..n), n=1..9); # Johannes W. Meijer, Jul 27 2011
  • Mathematica
    t[n_,k_] := k!(n - k)! SeriesCoefficient[Exp[z(1-u+u^2z)/(1-z u)]/(1-z u), {z,0,n}, {u,0,k}]; Table[t[n,k], {n,9}, {k,n}] //TableForm (* David Bevan, Apr 29 2013 *)
    t[n_, k_] := Pochhammer[n-k+1, k]*Hypergeometric1F1[-k, -n, -1]; Table[t[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 29 2013 *)

Formula

T(n,k) = F(n,k)*Sum{((-1)^j)*C(k, j)*(n-j)! (j=0 to k)}, where F(n,k) = 1/(n-k)! for 1 <= k <= n.
From Johannes W. Meijer, Jul 27 2011: (Start)
T(n,k) = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) with T(n,1) = (n-1) and T(n,n) = A000166(n) [Hanson et al.]
T(n,k) = (1/(n-k)!)*A061312(n-1,k-1)
sum(T(n,k), k=1..n) = A193464(n); row sums. (End)
T(n,k) = k!(n-k)![z^n*u^k]J(z,u) where J(z,u) = exp(z(1-u+z*u^2)/(1-z*u))/(1-z*u) is the exponential generating function of labeled digraphs consisting just of directed paths and oriented cycles (of length at least 2), z marking the vertices and u the edges; [z^n*u^k]J(z,u) is the coefficient of z^n*u^k in J(z,u). - David Bevan, Apr 29 2013

Extensions

Additional comments from Zerinvary Lajos, Mar 30 2006

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

A094791 Triangle read by rows giving coefficients of polynomials arising in successive differences of (n!)_{n>=0}.

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, 1, 36, 582, 5460, 32403, 124908, 309602, 470328, 391641, 133496
Offset: 0

Views

Author

Benoit Cloitre, Jun 11 2004

Keywords

Comments

Let D_0(n)=n! and D_{k+1}(n)=D_{k}(n+1)-D_{k}(n), then D_{k}(n)=n!*P_{k}(n) where P_{k} is a polynomial with integer coefficients of degree k.
The horizontal reversal of this triangle arises as a binomial convolution of the derangements coefficients der(n,i) (numbers of permutations of size n with i derangements = A098825(n,i) = number of permutations of size n with n-i rencontres = A008290(n,n-i), see formula section). - Olivier Gérard, Jul 31 2011

Examples

			D_3(n) = n!*(n^3 + 3*n^2 + 5*n + 2).
D_4(n) = n!*(n^4 + 6*n^3 + 17*n^2 + 20*n + 9).
Table begins:
  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
  ...
		

Crossrefs

Successive differences of factorial numbers: A001563, A001564, A001565, A001688, A001689, A023043.
Rencontres numbers A008290. Partial derangements A098825.
Row sum is A000255. Signed version in A126353.

Programs

  • Maple
    with(LREtools): A094791_row := proc(n)
    delta(x!,x,n); simplify(%/x!); seq(coeff(%,x,n-j),j=0..n) end:
    seq(print(A094791_row(n)),n=0..9); # Peter Luschny, Jan 09 2015
  • Mathematica
    d[0][n_] := n!; d[k_][n_] := d[k][n] = d[k - 1][n + 1] - d[k - 1][n] // FullSimplify;
    row[k_] := d[k][n]/n! // FullSimplify // CoefficientList[#, n]& // Reverse;
    Array[row, 10, 0] // Flatten (* Jean-François Alcover, Aug 02 2019 *)

Formula

T(n, n) = A000166(n).
T(2, k) = A000217(k).
Sum_{k=0..n} T(n,n-k)*x^k = Sum_{i=0..n} der(n,i)*binomial( n+x, i) (an analog of Worpitzky's identity). - Olivier Gérard, Jul 31 2011

Extensions

Edited and T(0,0) corrected according to the author's definition by Olivier Gérard, Jul 31 2011

A076732 Table T(n,k) giving number of ways of obtaining exactly one correct answer on an (n,k)-matching problem (1 <= k <= n).

Original entry on oeis.org

1, 1, 0, 1, 2, 3, 1, 4, 9, 8, 1, 6, 21, 44, 45, 1, 8, 39, 128, 265, 264, 1, 10, 63, 284, 905, 1854, 1855, 1, 12, 93, 536, 2325, 7284, 14833, 14832, 1, 14, 129, 908, 5005, 21234, 65821, 133496, 133497, 1, 16, 171, 1424, 9545, 51264, 214459, 660064, 1334961, 1334960
Offset: 1

Views

Author

Mohammad K. Azarian, Oct 28 2002

Keywords

Comments

Hanson et al. define the (n,k)-matching problem in the following realistic way. A matching question on an exam has k questions with n possible answers to choose from, each question having a unique answer. If a student guesses the answers at random, using each answer at most once, what is the probability of obtaining r of the k correct answers?
The T(n,k) represent the number of ways of obtaining exactly one correct answer, i.e., r=1, given k questions and n possible answers, 1 <= k <= n.

Examples

			Triangle begins
  1;
  1,0;
  1,2,3;
  1,4,9,8;
  ...
		

Crossrefs

Columns: A000012(n), 2*A001477(n-2), 3*A002061(n-2), 4*A094792(n-4), 5*A094793(n-5), 6*A094794(n-6), 7*A094795(n-7); A000240(n), A000166(n). - Johannes W. Meijer, Jul 27 2011

Programs

  • Maple
    A076732:=proc(n,k): (k/(n-k)!)*A047920(n,k) end: A047920:=proc(n,k): add(((-1)^j)*binomial(k-1,j)*(n-1-j)!, j=0..k-1) end: seq(seq(A076732(n,k), k=1..n), n=1..10); # Johannes W. Meijer, Jul 27 2011
  • Mathematica
    A000240[n_] := Subfactorial[n] - (-1)^n;
    T[n_, k_] := T[n, k] = Switch[k, 1, 1, n, A000240[n], _, k*T[n-1, k-1] + T[n-1, k]];
    Table[T[n, k], {n, 1, 10}, {k, 1, n}] // Flatten (* Jean-François Alcover, Nov 14 2023 *)

Formula

T(n,k) = F(n,k)*Sum{((-1)^j)*C(k-1, j)*(n-1-j)! (j=0 to k-1)}, where F(n,k) = k/(n-k)!, for 1 <= k <= n.
From Johannes W. Meijer, Jul 27 2011: (Start)
T(n,k) = k*T(n-1,k-1) + T(n-1,k) with T(n,1) = 1 and T(n,n) = A000240(n). [Hanson et al.]
T(n,k) = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) + (1-k)*A076731(n-2,k-2) + A076731(n-1,k-1) with T(0,0) = T(n,0) = 0 and T(n,1) = 1. [Hanson et al.]
T(n,k) = k*A060475(n-1,k-1).
T(n,k) = (k/(n-k)!)*A047920(n-1,k-1).
Sum_{k=1..n} T(n,k) = A193463(n); row sums.
Sum_{k=1..n} T(n,k)/k = A003470(n-1). (End)

Extensions

Edited and information added by Johannes W. Meijer, Jul 27 2011
Showing 1-7 of 7 results.