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.

A193464 Row sums of triangle A076731.

Original entry on oeis.org

0, 2, 7, 30, 146, 852, 5823, 45740, 405844, 4012710, 43733975, 520795002, 6726601062, 93651619880, 1398047697151, 22275111534552, 377278848390248, 6768744159489930, 128228860181918439, 2557808459478878870, 53585748788874537850, 1176328664895760953852
Offset: 1

Views

Author

Johannes W. Meijer, Jul 27 2011

Keywords

Programs

  • Maple
    A193464 := proc(n): add(A076731(n,k), k=1..n) end: 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: seq(A193464(n), n=1..22);

Formula

a(n) = sum(A076731(n,k), k=1..n).
a(n) = sum((1/(n-k)!)*A061312(n-1,k-1), k=1..n).
a(n) = sum((1/(n-k)!)*sum(((-1)^j)*binomial(k,j)*(n-j)!, j=0..k), k=1..n).

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

Views

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

A060475 Triangular array formed from successive differences of factorial numbers, then with factorials removed.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 3, 2, 1, 3, 7, 11, 9, 1, 4, 13, 32, 53, 44, 1, 5, 21, 71, 181, 309, 265, 1, 6, 31, 134, 465, 1214, 2119, 1854, 1, 7, 43, 227, 1001, 3539, 9403, 16687, 14833, 1, 8, 57, 356, 1909, 8544, 30637, 82508, 148329, 133496, 1, 9, 73, 527, 3333, 18089, 81901, 296967, 808393, 1468457, 1334961
Offset: 0

Views

Author

Henry Bottomley, Mar 16 2001

Keywords

Comments

T(n,k) is also the number of partial bijections (of an n-element set) with a fixed domain of size k and without fixed points. Equivalently, T(n,k) is the number of partial derangements with a fixed domain of size k in the symmetric inverse semigroup (monoid), I sub n. - Abdullahi Umar, Sep 14 2008

Examples

			Triangle begins
  1,
  1,  0,
  1,  1,  1,
  1,  2,  3,  2,
  1,  3,  7, 11,  9,
  1,  4, 13, 32, 53, 44,
  ...
		

Crossrefs

Columns include A000012, A001477, A002061.
Diagonals include A000166, A000255, A000153, A000261, A001909, A001910.
Main diagonal is abs of A002119.
Similar to A076731.
Row sums equal A003470. - Johannes W. Meijer, Jul 27 2011

Programs

  • Magma
    [[Factorial(k)*(&+[(-1)^j*Binomial(n-j, k-j)/Factorial(j): j in [0..k]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Mar 04 2019
    
  • Maple
    A060475 := proc(n,k): k! * add(binomial(n-j,k-j)*(-1)^j/j!, j=0..k) end:
    seq(seq(A060475(n,k), k=0..n), n=0..7); # Johannes W. Meijer, Jul 27 2011
    T := (n,k) -> KummerU(-k, -n, -1):
    seq(seq(simplify(T(n, k)), k = 0..n), n = 0..10); # Peter Luschny, Jul 07 2022
  • Mathematica
    t[n_, k_] := k!*Sum[Binomial[n - j, k - j]*(-1)^j/j!, {j, 0, k}]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Robert G. Wilson v, Aug 08 2011 *)
  • PARI
    {T(n,k) = k!*sum(j=0,k, (-1)^j*binomial(n-j, k-j)/j!)};
    for(n=0,10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Mar 04 2019
    
  • Sage
    [[factorial(k)*sum((-1)^j*binomial(n-j, k-j)/factorial(j) for j in (0..k)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 04 2019

Formula

T(n,k) = A047920(n,k)/(n-k)! = (n-1)*T(n-1,k-1) + (k-1)*T(n-2,k-2) = (n-k+1)*T(n, k-1) - T(n-1,k-1).
From Abdullahi Umar, Sep 14 2008: (Start)
T(n,k) = k! * Sum_{j=0..k} C(n-j,k-j)*(-1)^j/j!.
C(n,k)*T(n,k) = A144089(n, k). (End)
T(n,k) = A076732(n+1,k+1)/(k+1). - Johannes W. Meijer, Jul 27 2011
E.g.f. as a square array: A(x,y) = exp(-x)/(1 - x - y) = (1 + y + y^2 + y^3 + ...) + (y + 2*y^2 + 3*y^3 + 4*y^4 + ...)*x + (1 + 3*y + 7*y^2 + 13*y^3 + ...)*x^2/2! + (2 + 11*y + 32*y^2 + 71*y^3 + ...)*x^3/3! + .... Observe that (1 - y)*A(x*(1 - y),y) = exp(x*(y - 1))/(1 - x) is the e.g.f. for A008290. - Peter Bala, Sep 25 2013
T(n, k) = KummerU(-k, -n, -1). - Peter Luschny, Jul 07 2022

A061312 Triangle T[n,m]: T[n,-1] = 0; T[0,0] = 0; T[n,0] = n*n!; T[n,m] = T[n,m-1] - T[n-1,m-1].

Original entry on oeis.org

0, 1, 1, 4, 3, 2, 18, 14, 11, 9, 96, 78, 64, 53, 44, 600, 504, 426, 362, 309, 265, 4320, 3720, 3216, 2790, 2428, 2119, 1854, 35280, 30960, 27240, 24024, 21234, 18806, 16687, 14833, 322560, 287280, 256320, 229080, 205056, 183822, 165016, 148329
Offset: 0

Views

Author

Wouter Meeussen, Jun 06 2001

Keywords

Comments

Appears in the (n,k)-matching problem A076731. [Johannes W. Meijer, Jul 27 2011]

Examples

			0,
1, 1,
4, 3, 2,
18, 14, 11, 9,
96, 78, 64, 53, 44,
600, 504, 426, 362, 309, 265,
4320, 3720, 3216, 2790, 2428, 2119, 1854,
35280, 30960, 27240, 24024, 21234, 18806, 16687, 14833,
		

Crossrefs

Cf. A061018.
From Johannes W. Meijer, Jul 27 2011: (Start)
The row sums equal A193465. (End)

Programs

  • Magma
    [[(&+[(-1)^j*Binomial(k+1,j)*Factorial(n-j+1): j in [0..k+1]]): k in [0..n]]: n in [0..20]]; // G. C. Greubel, Aug 13 2018
  • Maple
    A061312 := proc(n,m): add(((-1)^j)*binomial(m+1,j)*(n+1-j)!, j=0..m+1) end: seq(seq(A061312(n,m), m=0..n), n=0..7); # Johannes W. Meijer, Jul 27 2011
  • Mathematica
    T[n_, k_]:= Sum[(-1)^j*Binomial[k + 1, j]*(n + 1 - j)!, {j, 0, k + 1}]; Table[T[n, k], {n, 0, 100}, {k, 0, n}] // Flatten  (* G. C. Greubel, Aug 13 2018 *)
  • PARI
    for(n=0,20, for(k=0,n, print1(sum(j=0,k+1, (-1)^j*binomial(k+1,j) *(n-j+1)!), ", "))) \\ G. C. Greubel, Aug 13 2018
    

Formula

T[n,m] = T[n,m-1]-T[n-1,m-1] with T[n,-1] = 0 and T[n,0] = A001563(n) = n*n!
T(n,m) = sum(((-1)^j)*binomial(m+1,j)*(n+1-j)!, j=0..m+1) [Johannes W. Meijer, Jul 27 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 1-5 of 5 results.