A060475 -id:A060475 - cp's OEIS frontend

A143413 Apéry-like numbers for the constant e: a(n) = 1/(n-1)!Sum_{k = 0..n+1} (-1)^kC(n+1,k)(2n-k)! for n >= 1.

-1, 1, 11, 181, 3539, 81901, 2203319, 67741129, 2346167879, 90449857081, 3843107102339, 178468044946621, 8994348275804891, 488964835817842021, 28523735794360301039, 1777328098986754744081, 117817961601577138782479, 8279178465722546926265329

Offset: 0

Peter Bala, Aug 14 2008

This sequence satisfies the recursion (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1) *(2*n^2 - 2*n+1)*a(n-1), which leads to a rapidly converging series for Napier's constant: e = 2 * Sum_{n >= 1} (-1)^n * n^2/(a(n)* a(n-1)).

Notice the striking parallels with the theory of the Apéry numbers A(n) = A005258(n), which satisfy a similar recurrence relation n^2*A(n) - (n-1)^2*A(n-2) = (11*n^2-11*n+3)*A(n-1) and which appear in the series acceleration formula zeta(2) = 5*Sum {n >= 1} 1/(n^2*A(n)*A(n-1)) = 5*[1/(1*3) + 1/(2^2*3*19) + 1/(3^2*19*147) + ...].

Seiichi Manyama, Table of n, a(n) for n = 0..365
A. van der Poorten, A proof that Euler missed ... Apery's proof of the irrationality of zeta(3). An informal report., Math. Intelligencer 1 (1978/79), no 4, 195-203.

Cf. A005258, A060475, A086764, A143414, A143415.

The Apéry-like numbers [or Apéry-like sequences, Apery-like numbers, Apery-like sequences] include A000172, A000984, A002893, A002895, A005258, A005259, A005260, A006077, A036917, A063007, A081085, A093388, A125143 (apart from signs), A143003, A143007, A143413, A143414, A143415, A143583, A183204, A214262, A219692,A226535, A227216, A227454, A229111 (apart from signs), A260667, A260832, A262177, A264541, A264542, A279619, A290575, A290576. (The term "Apery-like" is not well-defined.)

a := n -> 1/(n-1)!*add((-1)^k*binomial(n+1,k)*(2*n-k)!, k = 0..n+1):
seq(a(n), n = 1..19);
# Alternative
a := n -> `if`(n<2, 2*n-1, (2*n)!/(n-1)!*hypergeom([-n-1], [-2*n], -1)):
seq(simplify(a(n)), n=0..17); # Peter Luschny, Nov 14 2018

Join[{-1}, Table[(1/(n-1)!)*Sum[(-1)^k*Binomial[n+1,k]*(2*n-k)!, {k, 0, n+1}], {n, 1, 50}]] (* G. C. Greubel, Oct 24 2017 *)

concat([-1], for(n=1,25, print1((1/(n-1)!)*sum(k=0,n+1, (-1)^k*binomial(n+1,k)*(2*n-k)!), ", "))) \\ G. C. Greubel, Oct 24 2017

a(0):= -1, a(n) = 1/(n-1)!*sum {k = 0..n+1} (-1)^k*C(n+1,k)*(2*n-k)! for n >= 1.
Apart from the initial term, this sequence is the second superdiagonal of the square array A060475; equivalently, the second subdiagonal of the square array A086764.
Recurrence relation: a(0) = -1, a(1) = 1, (n-1)^2*a(n) - n^2*a(n-2) = (2*n-1)*(2*n^2-2*n+1)*a(n-1), n >= 2.
Let b(n) denote the solution to this recurrence with initial conditions b(0) = 0, b(1) = 2. Then b(n) = A143414(n) = 1/(n-1)!*sum {k = 0..n-1} C(n-1,k)*(2*n-k)!. The rational number b(n)/a(n) is equal to the Padé approximation to exp(x) of degree (n-1,n+1) evaluated at x = 1 and b(n)/a(n) -> e very rapidly.
For example, b(100)/a(100) - e is approximately 1.934 * 10^(-436). The identity b(n)*a(n-1) - b(n-1)*a(n) = (-1)^n *2*n^2 leads to rapidly converging series for e and 1/e: e = 2 * Sum_{n >= 1} (-1)^n * n^2/(a(n)*a(n-1)) = 2*[1 + 2^2/(1*11) - 3^2/(11*181) + 4^2/(181*3539) - ...]; 1/e = 1/2 - 2*Sum_{n >= 2} (-1)^n * n^2/(b(n)*b(n-1)) = 1/2 - 2*[2^2/(2*30) - 3^2/(30*492) + 4^2/(492*9620) - ...].
Conjectural congruences: for r >= 0 and odd prime p, calculation suggests that a(p^r*(p+1)) + a(p^r) == 0 (mod p^(r+1)).
a(n) = ((2*n)!/(n-1)!)*hypergeom([-n-1], [-2*n], -1) for n >= 2. - Peter Luschny, Nov 14 2018
a(n) ~ 2^(2*n + 1/2) * n^(n+1) / exp(n + 1/2). - Vaclav Kotesovec, Jul 11 2021

A047920 Triangular array formed from successive differences of factorial numbers.

Original entry on oeis.org

1, 1, 0, 2, 1, 1, 6, 4, 3, 2, 24, 18, 14, 11, 9, 120, 96, 78, 64, 53, 44, 720, 600, 504, 426, 362, 309, 265, 5040, 4320, 3720, 3216, 2790, 2428, 2119, 1854, 40320, 35280, 30960, 27240, 24024, 21234, 18806, 16687, 14833, 362880, 322560

Offset: 0

N. J. A. Sloane

Number of permutations of 1,2,...,k,n+1,n+2,...,2n-k that have no agreements with 1,...,n. For example, consider 1234 and 1256, then n=4 and k=2, so T(4,2)=14. Compare A000255 for the case k=1. - Jon Perry, Jan 23 2004
From Emeric Deutsch, Apr 21 2009: (Start)
T(n-1,k-1) is the number of non-derangements of {1,2,...,n} having smallest fixed point equal to k. Example: T(3,1)=4 because we have 4213, 4231, 3214, and 3241 (the permutations of {1,2,3,4} having smallest fixed equal to 2).
Row sums give the number of non-derangement permutations of {1,2,...,n} (A002467).
Mirror image of A068106.
Closely related to A134830, where each row has an extra term (see the Charalambides reference).
(End)
T(n,k) is the number of permutations of {1..n} that don't fix the points 1..k. - Robert FERREOL, Aug 04 2016

			Triangle begins:
    1;
    1,  0;
    2,  1,  1;
    6,  4,  3,  2;
   24, 18, 14, 11,  9;
  120, 96, 78, 64, 53, 44;
  ...
The left-hand column is the factorial numbers (A000142). The other numbers in the row are calculated by subtracting the numbers in the previous row. For example, row 4 is 6, 4, 3, 2, so row 5 is 4! = 24, 24 - 6 = 18, 18 - 4 = 14, 14 - 3 = 11, 11 - 2 = 9. - _Michael B. Porter_, Aug 05 2016

Ch. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, Boca Raton, Florida, 2002, p. 176, Table 5.3. [From Emeric Deutsch, Apr 21 2009]

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
E. Deutsch and S. Elizalde, The largest and the smallest fixed points of permutations, arXiv:0904.2792 [math.CO], 2009.
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122. [Annotated scanned copy]
J. D. H. Dickson, Discussion of two double series arising from the number of terms in determinants of certain forms, Proc. London Math. Soc., 10 (1879), 120-122.
Ira M. Gessel, Symmetric inclusion-exclusion, Séminaire Lotharingien de Combinatoire, B54b (2005).
Peter Kagey, Ranking and Unranking Restricted Permutations, arXiv:2210.17021 [math.CO], 2022.
Index entries for sequences related to factorial numbers

Columns give A000142, A001563, A001564, etc. Cf. A047922.
See A068106 for another version of this triangle.
Orthogonal columns: A000166, A000255, A055790. Main diagonal A033815.
Cf. A002467, A068106, A134830. - Emeric Deutsch, Apr 21 2009
Cf. A155521.
T(n+2,n) = 2*A000153(n+1). T(n+3,n) = 6*A000261(n+2). T(n+4,n) = 24*A001909(n+3). T(n+5, n) = 120*A001910(n+4). T(n+6,n) = 720*A176732(n).
T(n+7,n) = 5040*A176733(n) - Richard R. Forberg, Dec 29 2013

a047920 n k = a047920_tabl !! n !! k
a047920_row n = a047920_tabl !! n
a047920_tabl = map fst $ iterate e ([1], 1) where
   e (row, n) = (scanl (-) (n * head row) row, n + 1)
-- Reinhard Zumkeller, Mar 05 2012

d[0] := 1: for n to 15 do d[n] := n*d[n-1]+(-1)^n end do: T := proc (n, k) if k <= n then sum(binomial(n-k, j)*d[n-j], j = 0 .. n-k) else 0 end if end proc: for n from 0 to 9 do seq(T(n, k), k = 0 .. n) end do; # yields sequence in triangular form - Emeric Deutsch, Jul 17 2009
# second Maple program:
T:= proc(n, k) option remember;
     `if`(k=0, n!, T(n, k-1)-T(n-1, k-1))
    end:
seq(seq(T(n, k), k=0..n), n=0..12);  # Alois P. Heinz, Sep 01 2021

t[n_, k_] = Sum[(-1)^j*Binomial[k, j]*(n-j)!, {j, 0, n}]; Flatten[Table[t[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, May 17 2011, after Philippe Deléham *)
T[n_, k_] := n! Hypergeometric1F1[-k, -n, -1];
Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten  (* Peter Luschny, Jul 28 2024 *)

row(n) = vector(n+1, k, k--; sum(j=0, n, (-1)^j * binomial(k, j)*(n-j)!)); \\ Michel Marcus, Sep 04 2021

t(n, k) = t(n, k-1) - t(n-1, k-1) = t(n, k+1) - t(n-1, k) = n*t(n-1, k) + k*t(n-2, k-1) = (n-1)*t(n-1, k-1) + (k-1)*t(n-2, k-2) = A060475(n, k)*(n-k)!. - Henry Bottomley, Mar 16 2001
T(n, k) = Sum_{j>=0} (-1)^j * binomial(k, j)*(n-j)!. - Philippe Deléham, May 29 2005
T(n,k) = Sum_{j=0..n-k} d(n-j)*binomial(n-k,j), where d(i)=A000166(i) are the derangement numbers. - Emeric Deutsch, Jul 17 2009
Sum_{k=0..n} (k+1)*T(n,k) = A155521(n+1). - Emeric Deutsch, Jul 18 2009
T(n, k) = n!*hypergeom([-k], [-n], -1). - Peter Luschny, Jul 28 2024

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

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 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

			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

D. Hanson, K. Seyffarth and J. H. Weston, Matchings, Derangements, Rencontres, Mathematics Magazine, Vol. 56, No. 4, September 1983.
StackExchange, How many injective functions f:[1,...,m]->[1,...,n] have no fixed point?

Cf. A076732, A086764.

Similar to A060475.

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

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 *)

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

Additional comments from Zerinvary Lajos, Mar 30 2006

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

cp's OEIS Frontend

A143413 Apéry-like numbers for the constant e: a(n) = 1/(n-1)!*Sum_{k = 0..n+1} (-1)^k*C(n+1,k)*(2*n-k)! for n >= 1.

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A047920 Triangular array formed from successive differences of factorial numbers.

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

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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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A143413 Apéry-like numbers for the constant e: a(n) = 1/(n-1)!Sum_{k = 0..n+1} (-1)^kC(n+1,k)(2n-k)! for n >= 1.