A086764 -id:A086764 - cp's OEIS frontend

A143409 Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.

1, 2, 1, 5, 3, 1, 16, 11, 4, 1, 65, 49, 19, 5, 1, 326, 261, 106, 29, 6, 1, 1957, 1631, 685, 193, 41, 7, 1, 13700, 11743, 5056, 1457, 316, 55, 8, 1, 109601, 95901, 42079, 12341, 2721, 481, 71, 9, 1, 986410, 876809, 390454, 116125, 25946, 4645, 694, 89, 10, 1

Offset: 0

Peter Bala, Aug 14 2008

The Euler-Seidel matrix for the sequence {k!} is array A076571 read as a square, whose k-th column entries have a common factor of k!. Removing these common factors gives the current table.
This table is closely connected to the constant 1/e. The row, column and diagonal entries of this table occur in series acceleration formulas for 1/e.
For a similar table based on the differences of the sequence {k!} and related to the constant e, see A086764. For other arrays similarly related to constants see A143410 (for sqrt(e)), A143411 (for 1/sqrt(e)), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

			The Euler-Seidel matrix for the sequence {k!} begins
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....2.....6....24...120...720
1..|.....2.....3.....8....30...144...840
2..|.....5....11....38...174...984
3..|....16....49...212..1158
4..|....65...261..1370
5..|...326..1631
6..|..1957
...
Dividing the k-th column by k! gives
==============================================
n\k|.....0.....1.....2.....3.....4.....5.....6
==============================================
0..|.....1.....1.....1.....1.....1.....1.....1
1..|.....2.....3.....4.....5.....6.....7
2..|.....5....11....19....29....41
3..|....16....49...106...193
4..|....65...261...685
5..|...326..1631
6..|..1957
...
Examples of series formula for 1/e:
Row 2: 1/e = 2*(1/5 - 1/(1!*5*11) + 1/(2!*11*19) - 1/(3!*19*29) + ...).
Column 4: 24/e = 9 - (0!/(1*6) + 1!/(6*41) + 2!/(41*316) + ...).
...
Displayed as a triangle:
0 |     1
1 |     2,     1
2 |     5,     3,    1
3 |    16,    11,    4,    1
4 |    65,    49,   19,    5,   1
5 |   326,   261,  106,   29,   6,  1
6 |  1957,  1631,  685,  193,  41,  7, 1
7 | 13700, 11743, 5056, 1457, 316, 55, 8, 1

Robert Israel, Table of n, a(n) for n = 0..10010 (antidiagonals 0 to 140, flattened)
D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
Eric Weisstein's World of Mathematics Poisson-Charlier polynomial

Cf. A008288, A076571, A086764, A108625, A143007, A143410, A143411, A143413, A001517 (main diagonal), A028387 (row 2), A000522 (column 0), A001339 (column 1), A082030 (column 2), A095000 (column 3), A095177 (column 4).

Cf. A273596, A003470.

T := (n, k) -> 1/k!*add(binomial(n,j)*(k+j)!, j = 0..n):
for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;
# Alternate:
T:= proc(n,k) option remember;
  if n = 0 then return 1 fi;
  (n+k)*procname(n-1,k) + procname(n-1,k-1);
end proc:
seq(seq(T(s-n,n),n=0..s),s=0..10); # Robert Israel, Jul 07 2017
# Or:
A143409 := (n,k) -> hypergeom([k+1, k-n], [], -1):
seq(seq(simplify(A143409(n,k)),k=0..n),n=0..9); # Peter Luschny, Oct 05 2017

T[n_, k_] := HypergeometricPFQ[{k+1,k-n}, {}, -1];
Table[T[n,k], {n,0,9}, {k,0,n}] // Flatten (* Peter Luschny, Oct 05 2017 *)

T(n,k) = (1/k!)*Sum_{j = 0..n} binomial(n,j)*(k+j)!.
T(n,k) = ((n+k)!/k!)*Num_Pade(n,k), where Num_Pade(n,k) denotes the numerator of the Padé approximation for the function exp(x) of degree (n,k) evaluated at x = 1.
Recurrence relations:
T(n,k) = T(n-1,k) + (k+1)*T(n-1,k+1);
T(n,k) = (n+k)*T(n-1,k) + T(n-1,k-1).
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 + ...) + (2 + 3*x + 4*x^2 + ...)*y + (5 + 11*x + 19*x^2 + ...)*y^2/2! + ... .
Row n lists the values of the Poisson-Charlier polynomial x^(n) + C(n,1)*x^(n-1) + C(n,2)*x^(n-2) + ... + C(n,n) for x = 1,2,3,..., where x^(m) denotes the rising factorial x*(x+1)*...*(x+m-1).
Main diagonal is A001517.
Series formulas for 1/e:
Row n: 1/e = n!*[1/T(n,0) - 1/(1!*T(n,0)*T(n,1)) + 1/(2!*T(n,1)*T(n,2)) - 1/(3!*T(n,2)*T(n,3)) + ...].
Column k: k!/e = A000166(k) + (-1)^(k+1)*[0!/(T(0,k)*T(1,k)) + 1!/(T(1,k)*T(2,k)) + 2!/(T(2,k)*T(3,k)) + ...].
Main diagonal: 1/e = 1 - 2*Sum_{n>=0} (-1)^n/(T(n,n)*T(n+1,n+1)) = 1 - 2*[1/(1*3) - 1/(3*19) + 1/(19*193) - ...].
Second subdiagonal: 1/e = 2*(1^2/(1*5) - 2^2/(5*49) + 3^2/(49*685) - ...).
Compare with A143413.
From Peter Luschny, Oct 05 2017: (Start)
T(n, k) = hypergeom([k+1, k-n], [], -1).
When seen as a triangular array then the row sums are A273596 and the alternating row sums are A003470. (End)

A176732 a(n) = (n+5)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 6, 43, 356, 3333, 34754, 398959, 4996032, 67741129, 988344062, 15434831091, 256840738076, 4536075689293, 84731451264186, 1668866557980343, 34563571477305464, 750867999393119889, 17072113130285524982, 405423357986250112699, 10037458628015142154452, 258639509502117306002581

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=6 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A001725(n+5) = (n+5)!/5!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).

			Necklaces and 6 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c6(1), (binomial(4,2)*2)*c6(2), and 1*c6(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c6(n):=A001725(n+5) numbers for the pure 6-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=6: 1/(1-x)^6). This adds up as 9 + 4*2*6 + (6*1)*42 + 3024 = 3333 = a(4).

Cf. A000153, A000261, A001909, A001910 (necklaces and k=5 cords), A176732.

a := n -> hypergeom([-n,7],[],1)*(-1)^n:
seq(simplify(a(n)),n=0..9); # Peter Luschny, Apr 25 2015

Rest[RecurrenceTable[{a[0]==1,a[-1]==0,a[n]==(n+5)a[n-1]+(n-1)a[n-2]},a,{n,20}]] (* Harvey P. Dale, Oct 01 2012 *)

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^6) = exp(-x)/(1-x)^7, equivalent to the recurrence.
a(n) = A086764(n+6,6).
a(n) = A090010(n), n>0. - R. J. Mathar, Jul 22 2010
a(n) = (-1)^n*hypergeom([-n,7],[],1). - Peter Luschny, Apr 25 2015

A176733 a(n) = (n+6)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 7, 57, 527, 5441, 61959, 770713, 10391023, 150869313, 2346167879, 38896509881, 684702346767, 12752503850497, 250514001320647, 5176062576469401, 112204510124346479, 2546140161382663553, 60356495873790805383, 1491840283714484609593, 38382424018590349736719

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=7 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A001730(n+6) = (n+6)!/6!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).

			Necklaces and 7 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c7(1), (binomial(4,2)*!2)*c7(2), and 1*c7(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c7(n):=A001730(n+6) numbers for the pure 7-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=7: 1/(1-x)^7). This adds up as 9 + 4*2*7 + (6*1)*56 + 5040 = 5441 = a(4).

Robert Israel, Table of n, a(n) for n = 0..442

Cf. A176732 (necklaces and k=6 cords).

f:= proc(n) option remember; (n+6)*procname(n-1) + (n-1)*procname(n-2) end proc:
f(-1):= 0: f(0):= 1:
map(f, [$0..30]); # Robert Israel, Dec 01 2024

Table[(-1)^n HypergeometricPFQ[{8, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^7) = exp(-x)/(1-x)^8, equivalent to the given recurrence.
a(n) = A086764(n+7,7).
a(n) = (-1)^n*2F0(8,-n;;1). - Benedict W. J. Irwin, May 29 2016

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

A143411 Square array, read by antidiagonals: form the Euler-Seidel matrix for the sequence {2^kk!} and then divide column k by 2^kk!.

Original entry on oeis.org

1, 3, 1, 13, 5, 1, 79, 33, 7, 1, 633, 277, 61, 9, 1, 6331, 2849, 643, 97, 11, 1, 75973, 34821, 7993, 1225, 141, 13, 1, 1063623, 493825, 114751, 17793, 2071, 193, 15, 1, 17017969, 7977173, 1870837, 292681, 34361, 3229, 253, 17, 1

Offset: 0

Peter Bala, Aug 19 2008

This table is closely connected to the constant 1/sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for 1/sqrt(e). For a similar table based on the differences of the sequence {2^k*k!} and related to the constant sqrt(e), see A143410. For other arrays similarly related to constants see A086764 (for e), A143409 (for 1/e), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

			The Euler-Seidel matrix for the sequence {2^k*k!} begins
  ========================================
  n\k|     0     1     2     3     4     5
  ========================================
  0  |     1     2     8    48   384  3840
  1  |     3    10    56   432  4224
  2  |    13    66   488  4656
  3  |    79   554  5144
  4  |   633  5698
  5  |  6331
  ...
.
  Dividing the k-th column by 2^k*k! gives
  ========================================
  n\k|     0     1     2     3     4     5
  ========================================
  0  |     1     1     1     1     1     1
  1  |     3     5     7     9    11
  2  |    13    33    61    97
  3  |    79   277   643
  4  |   633  2849
  5  |  6331
  ...

G. C. Greubel, Antidiagonals n = 0..50, flattened
D. Dumont, Matrices d'Euler-Seidel, Sem. Loth. Comb. B05c (1981) 59-78.
Eric Weisstein's World of Mathematics, Poisson-Charlier polynomial.

Cf. A008288, A065919 (main diagonal), A076571, A086764, A108625, A143007, A143409, A143410.

A:= func< n,k | (&+[Binomial(n,j)*Factorial(k+j)*2^j/Factorial(k): j in [0..n]]) >; // Array
A143411:= func< n,k | A(n-k,k) >; // antidiagonal triangle
[A143411(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Oct 05 2023

with combinat: T := (n, k) -> 1/k!*add(2^j*binomial(n,j)*(k+j)!, j = 0..n): for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;

A[n_, k_]:= (1/k!)*Sum[Binomial[n,j]*(k+j)!*2^j, {j,0,n}]; (* array *)
A143411[n_, k_]:= A[n-k,k]; (* antidiagonals *)
Table[A143411[n,k], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Oct 05 2023 *)

def A(n,k): return sum(binomial(n,j)*factorial(j+k)*2^j/factorial(k) for j in range(n+1)) # array
def A143411(n,k): return A(n-k,k) # antidiagonal triangle
flatten([[A143411(n,k) for k in range(n+1)] for n in range(13)]) # G. C. Greubel, Oct 05 2023

T(n,k) = (1/k!)*Sum_{j = 0..n} 2^j*binomial(n,j)*(k+j)!.
Relation with Poisson-Charlier polynomials c_n(x,a):
T(n,k) = (-1)^n*c_n(-(k+1),1/2).
Recurrence relations:
T(n,k) = 2*n*T(n-1,k) + T(n,k-1);
T(n,k) = 2*(n+k)*T(n-1,k) + T(n-1,k-1);
T(n,k) = 2*(k+1)*T(n-1,k+1) + T(n-1,k).
Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k-1)*T(n,k-1) + T(n,k-2).
E.g.f. for column k: exp(y)/(1 - 2*y)^(k+1).
E.g.f. for array: exp(y)/(1 - x - 2*y) = (1 + x + x^2 + ...) + (3 + 5*x + 7*x^2 + ...)*y + (13 + 37*x + 61*x^2 + ...)*y^2/2! + ... .
Series acceleration formulas for 1/sqrt(e):
Row n: 1/sqrt(e) = 2^n*n!*(1/T(n,0) - 1/(2*1!*T(n,0)*T(n,1)) + 1/(2^2*2!*T(n,1)*T(n,2)) - 1/(2^3*3!*T(n,2)*T(n,3)) + ...). For example, row 3 gives 1/sqrt(e) = 48*(1/79 - 1/(2*79*277) + 1/(8*277*643) - 1/(48*643*1225) + ...).
Column k: 1/sqrt(e) = (1 - (1/2)/1! + (1/2)^2/2! - ... + (-1/2)^k/k!) + (-1)^(k+1)/(2^k*k!)*( Sum_{n = 0..inf} 2^n*n!/(T(n,k)*T(n+1,k)) ). For example, column 3 gives 1/sqrt(e) = 29/48 + 1/48*( 1/(1*9) + 2/(9*97) + 8/(97*1225) + 48/(1225*17793) + ... ).
Main diagonal: 1/sqrt(e) = 1 - 2*( 1/(1*5) - 1/(5*61) + 1/(61*1225) - ... ). See A065919.

A176735 a(n) = (n+8)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 9, 91, 1019, 12501, 166589, 2394751, 36920799, 607496041, 10622799089, 196677847971, 3843107102339, 79025598374461, 1705654851091749, 38551739502886471, 910569176481673319, 22431936328103456721, 575367515026293191129, 15340898308261381733611, 424560869593530584247819

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=9 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049389(n) = (n+8)!/8!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).

			Necklaces and 9 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c9(1), (binomial(4,2)*! 2)*c9(2), and 1*c9(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c9(n):=A049389(n) numbers for the pure 9-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=9: 1/(1-x)^9). This adds up as 9 + 4*2*9 + (6*1)*90 + 11880 = 12501 = a(4).

Harvey P. Dale, Table of n, a(n) for n = 0..400

Cf. A176734 (necklaces and k=8 cords).

RecurrenceTable[{a[0]==1,a[1]==9,a[n]==(n+8)a[n-1]+(n-1)a[n-2]},a[n],{n,20}] (* Harvey P. Dale, Oct 20 2011 *)
Table[(-1)^n HypergeometricPFQ[{10, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 27 2016 *)

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^9) = exp(-x)/(1-x)^10, equivalent to the given recurrence.
a(n) = A086764(n+9,9).
a(n) = (-1)^n*2F0(10,-n;;1). - Benedict W. J. Irwin, May 27 2016

A143410 Form the difference table of the sequence {2^kk!}, then divide k-th column entries by 2^kk!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.

Original entry on oeis.org

1, 1, 1, 5, 3, 1, 29, 17, 5, 1, 233, 131, 37, 7, 1, 2329, 1281, 353, 65, 9, 1, 27949, 15139, 4105, 743, 101, 11, 1, 391285, 209617, 56189, 10049, 1349, 145, 13, 1, 6260561, 3325923, 883885, 156679, 20841, 2219, 197, 15, 1, 112690097, 59475329, 15700313

Offset: 0

Peter Bala, Aug 19 2008

This table is closely connected to the constant sqrt(e). The row, column and diagonal entries of this table occur in series acceleration formulas for sqrt(e). For a similar table based on the Euler-Seidel matrix of the sequence {2^k*k!} and related to the constant 1/sqrt(e), see A143411. For other arrays similarly related to constants see A086764 (for e), A143409 (for 1/e), A008288 (for log(2)), A108625 (for zeta(2)) and A143007 (for zeta(3)).

			Table of differences of {2^k*k!}
  =====================================================
  Column                0     1     2     3     4     5
  =====================================================
  Sequence 2^k*k!       1     2     8    48   384  3840
  First differences     1     6    40   336  3456
  Second differences    5    34   296  3120
  Third differences    29   262  2824
  Fourth differences  233  2562
  ...
Remove the common factor 2^k*k! from k-th column entries:
  ====================================
  n\k|   0      1      2      3      4
  ====================================
  0  |   1      1      1      1      1
  1  |   1      3      5      7      9
  2  |   5     17     37     65    101
  3  |  29    131    353    743   1349
  4  | 233   1281   4105  10049  20841

Eric Weisstein's World of Mathematics, Poisson-Charlier polynomial.

Cf. A008288, A076571, A086764, A108625, A143007, A143409, A143411, A143412.

T := (n, k) -> (-1)^n/k!*add((-2)^j*binomial(n,j)*(k+j)!, j = 0..n):
for n from 0 to 9 do seq(T(n, k), k = 0..9) end do;

T(n,k) = ((-1)^n/k!)*Sum {j = 0..n} (-2)^j*C(n,j)*(k+j)!.
Relation with Poisson-Charlier polynomials c_n(x,a): T(n,k) = c_n(-(k+1),-1/2).
Recurrence relations: T(n,k) = 2*n*T(n-1,k) + T(n,k-1); T(n,k) = 2*(n+k)*T(n-1,k) - T(n-1,k-1); T(n,k) = 2*(k+1)*T(n-1,k+1) - T(n-1,k);
Recurrence for row n entries: 2*k*T(n,k) = (2*n+2*k+1)*T(n,k-1) - T(n,k-2).
E.g.f. for column k: exp(-y)/(1-2*y)^(k+1).
E.g.f. for array: exp(-y)/(1-x-2*y) = (1 + x + x^2 + ...) + (1 + 3*x + 5*x^2 + ...)*y + (5 + 17*x + 37*x^2 + ...)*y^2/2! + ... .
Series acceleration formulas for sqrt(e):
Row n: sqrt(e) = 2^n*n!*(1/T(n,0) + (-1)^n*(1/(2*1!*T(n,0)*T(n,1)) + 1/(2^2*2!*T(n,1)*T(n,2)) + 1/(2^3*3!*T(n,2)*T(n,3)) + ...)). For example, row 3 gives sqrt(e) = 48*(1/29 - 1/(2*29*131) - 1/(8*131*353) - 1/(48*353*743) - ...).
Column k: sqrt(e) = (1 + (1/2)/1! + (1/2)^2 / 2! + ... + (1/2)^k/k!) + 1/(2^k*k!) * Sum_{n>= 0} ((-2)^n *n!/(T(n,k)*T(n+1,k))). For example, column 3 gives sqrt(e) = 79/48 + (1/48)*(1/(1*7) - 2/(7*65) + 8/(65*743) - 48/(743*10049) + ...).
Main diagonal: sqrt(e) = 1 + 2*(1/(1*3) - 1/(3*37) + 1/(37*743) - ...). See A143412.
T(n, k) = (-1)^n*(-1/2)^(k + 1)*KummerU(k + 1, k + n + 2, -1/2). - Peter Luschny, Jan 02 2020

A176734 a(n) = (n+7)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 8, 73, 746, 8425, 104084, 1395217, 20157542, 312129649, 5155334720, 90449857081, 1679650774658, 32908313146393, 678322072223756, 14672571587601985, 332293083938376254, 7862829504396683617, 194024597448534426872, 4984283037788104293289, 133083801736564331309210

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=8 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049388(n) = (n+7)!/7!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).

			Necklaces and 8 cords problem. For n=4 one considers the following weak 2 part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c8(1), (binomial(4,2)*!2)*c8(2), and 1*c8(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c8(n):=A049388(n) numbers for the pure 8-cord problem (see the remark on the e.g.f. for the k cords problem in A000153; here for k=8: 1/(1-x)^8). This adds up as 9 + 4*2*8 + (6*1)*72 + 7920 = 8425 = a(4).

Cf. A176733 (necklaces and k=7 cords).

nxt[{n_,a_,b_}]:={n+1,b,(n+8)b+n*a}; Transpose[NestList[nxt,{1,1,8},20]][[2]] (* Harvey P. Dale, Mar 19 2013 *)
Table[(-1)^n HypergeometricPFQ[{9, -n}, {}, 1], {n, 0, 20}] (* Benedict W. J. Irwin, May 29 2016 *)

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^8) = exp(-x)/(1-x)^9, equivalent to the given recurrence.
a(n) = A086764(n+8,8).
a(n) = (-1)^n*2F0(9,-n;;1). - Benedict W. J. Irwin, May 29 2016

A176736 a(n) = (n+9)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

Original entry on oeis.org

1, 10, 111, 1352, 17909, 256134, 3931555, 64441684, 1123029513, 20730064706, 403978495031, 8286870547680, 178468044946621, 4025739435397822, 94912091598455979, 2334250550458513004, 59779945135439664785, 1591626582328767492474, 43990176790179196598143, 1260374228606935319612536

Offset: 0

Wolfdieter Lang, Jul 14 2010

a(n) enumerates the possibilities for distributing n beads, n>=1, labeled differently from 1 to n, over a set of (unordered) necklaces, excluding necklaces with exactly one bead, and k=10 indistinguishable, ordered, fixed cords, each allowed to have any number of beads. Beadless necklaces as well as beadless cords contribute a factor 1 in the counting, e.g., a(0):= 1*1 =1. See A000255 for the description of a fixed cord with beads. This produces for a(n) the exponential (aka binomial) convolution of the subfactorial sequence {A000166(n)} and the sequence {A049398(n) = (n+9)!/9!}. See the necklaces and cords problem comment in A000153. Therefore the recurrence with inputs holds. 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).

			Necklaces and 10 cords problem. For n=4 one considers the following weak 2-part compositions of 4: (4,0), (3,1), (2,2), and (0,4), where (1,3) does not appear because there are no necklaces with 1 bead. These compositions contribute respectively !4*1,binomial(4,3)*!3*c10(1), (binomial(4,2)*! 2)*c10(2), and 1*c10(4) with the subfactorials !n:=A000166(n) (see the necklace comment there) and the c10(n):=A049398(n) numbers for the pure 10-cord problem (see the remark on the e.g.f. for the k-cord problem in A000153; here for k=10: 1/(1-x)^10). This adds up as 9 + 4*2*10 + (6*1)*110 + 17160 = 17909 = a(4).

Cf. A176735 (necklaces and k=9 cords).

E.g.f. (exp(-x)/(1-x))*(1/(1-x)^10) = exp(-x)/(1-x)^11, equivalent to the given recurrence.

a(n) = A086764(n+10,10).

cp's OEIS Frontend

A143409 Square array read by antidiagonals: form the Euler-Seidel matrix for the sequence {k!} and then divide column k by k!.

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A176732 a(n) = (n+5)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

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A176733 a(n) = (n+6)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

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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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A143411 Square array, read by antidiagonals: form the Euler-Seidel matrix for the sequence {2^k*k!} and then divide column k by 2^k*k!.

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A176735 a(n) = (n+8)*a(n-1) + (n-1)*a(n-2), a(-1)=0, a(0)=1.

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A143410 Form the difference table of the sequence {2^k*k!}, then divide k-th column entries by 2^k*k!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.

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A176732 a(n) = (n+5)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

A176733 a(n) = (n+6)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

A143411 Square array, read by antidiagonals: form the Euler-Seidel matrix for the sequence {2^kk!} and then divide column k by 2^kk!.

A176735 a(n) = (n+8)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

A143410 Form the difference table of the sequence {2^kk!}, then divide k-th column entries by 2^kk!. Array read by ascending antidiagonals, T(n, k) for n, k >= 0.

A176734 a(n) = (n+7)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.

A176736 a(n) = (n+9)a(n-1) + (n-1)a(n-2), a(-1)=0, a(0)=1.