A094795 -id:A094795 - cp's OEIS frontend

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

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

Benoit Cloitre, Jun 11 2004

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

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A001563, A001564, A001565, A001688, A001689, A023043.

Cf. A094791, A094792, A094794, A094795.

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

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

Benoit Cloitre, Jun 11 2004

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Cf. A001563, A001564, A001565, A001688, A001689, A023043, A096307.

Cf. A094791, A094792, A094793, A094795.

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]

a(n)=n^5+10*n^4+45*n^3+100*n^2+109*n+44 \\ Charles R Greathouse IV, Oct 16 2015

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

A096341 E.g.f.: exp(x)/(1-x)^7.

Original entry on oeis.org

1, 8, 71, 694, 7421, 86276, 1084483, 14665106, 212385209, 3280842496, 53862855551, 936722974958, 17205245113141, 332864226563324, 6766480571358971, 144202473398010826, 3215159679583864433

Offset: 0

Philippe Deléham, Jun 28 2004

nonn

Sum_{k=0..n} A094816(n,k)*x^k give A000522(n), A001339(n), A082030(n), A095000(n), A095177(n), A096307(n) for x = 1, 2, 3, 4, 5, 6 respectively.

G. C. Greubel, Table of n, a(n) for n = 0..250

Cf. E.g.f. exp(x)/(1-x)^k: A000522 (k = 1), A001339 (k = 2), A082030 (k = 3), A095000 (k = 4), A095177 (k = 5), A096307 (k = 6).

Cf. A094795, A094816.

Table[HypergeometricPFQ[{7, -n}, {}, -1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)
With[{nn = 250}, CoefficientList[Series[Exp[x]/(1 - x)^7, {x, 0, nn}], x] Range[0, nn]!] (* G. C. Greubel, May 27 2016 *)

a(n) = Sum_{k = 0..n} A094816(n, k)*7^k.
a(n) = Sum_{k = 0..n} binomial(n, k)*(k+6)!/6!.
a(n) = 2F0(7,-n;;-1). - Benedict W. J. Irwin, May 27 2016
From Peter Bala, Jul 26 2021: (Start)
a(n) = (n+7)*a(n-1) - (n-1)*a(n-2) with a(0) = 1 and a(1) = 8.
First-order recurrence: P(n-1)*a(n) = n*P(n)*a(n-1) + 1 with a(0) = 1, where P(n) = n^6 + 15*n^5 + 100*n^4 + 355*n^3 + 694*n^2 + 689*n + 265 = A094795(n).
(End)

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

Benoit Cloitre, Jun 11 2004

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

Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A001563, A001564, A001565, A001688, A001689, A023043.

Cf. A094791, A094793, A094794, A094795.

with(combinat):seq(fibonacci(4, i)-1,i=1..40); # Zerinvary Lajos, Mar 20 2008

Table[n^3+3*n^2+5*n+2,{n,0,70}] (* Vladimir Joseph Stephan Orlovsky, May 04 2011 *)

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

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

Benoit Cloitre, Jun 11 2004

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

			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
  ...

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.
Cf. A094792, A094793, A094794, A094795.

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

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 *)

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
The n-th row polynomial R(n,x) = Sum {k = 0..n} T(n,k)*x^k is P-recursive in the variable x: x*R(n,x) = (x+n+1)*R(n,x-1) - R(n,x-2). - _Peter Bala, Jul 25 2021

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

Mohammad K. Azarian, Oct 28 2002

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.

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

D. Hanson, K. Seyffarth and J. H. Weston, Matchings, Derangements, Rencontres, Mathematics Magazine, Vol. 56, No. 4, September 1983.

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

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

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 *)

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)

Edited and information added by Johannes W. Meijer, Jul 27 2011

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

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

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A096341 E.g.f.: exp(x)/(1-x)^7.

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

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

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

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