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.

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A073003 Decimal expansion of -exp(1)*Ei(-1), also called Gompertz's constant, or the Euler-Gompertz constant.

Original entry on oeis.org

5, 9, 6, 3, 4, 7, 3, 6, 2, 3, 2, 3, 1, 9, 4, 0, 7, 4, 3, 4, 1, 0, 7, 8, 4, 9, 9, 3, 6, 9, 2, 7, 9, 3, 7, 6, 0, 7, 4, 1, 7, 7, 8, 6, 0, 1, 5, 2, 5, 4, 8, 7, 8, 1, 5, 7, 3, 4, 8, 4, 9, 1, 0, 4, 8, 2, 3, 2, 7, 2, 1, 9, 1, 1, 4, 8, 7, 4, 4, 1, 7, 4, 7, 0, 4, 3, 0, 4, 9, 7, 0, 9, 3, 6, 1, 2, 7, 6, 0, 3, 4, 4, 2, 3, 7
Offset: 0

Views

Author

Robert G. Wilson v, Aug 03 2002

Keywords

Comments

0! - 1! + 2! - 3! + 4! - 5! + ... = (Borel) Sum_{n>=0} (-y)^n n! = KummerU(1,1,1/y)/y.
Decimal expansion of phi(1) where phi(x) = Integral_{t>=0} e^-t/(x+t) dt. - Benoit Cloitre, Apr 11 2003
The divergent series g(x=1,m) = 1^m*1! - 2^m*2! + 3^m*3! - 4^m*4! + ..., m => -1, is intimately related to Gompertz's constant. We discovered that g(x=1,m) = (-1)^m * (A040027(m) - A000110(m+1) * A073003) with A000110 the Bell numbers and A040027 a sequence that was published by Gould, see for more information A163940. - Johannes W. Meijer, Oct 16 2009
Named by Le Lionnais (1983) after the English self-educated mathematician and actuary Benjamin Gompertz (1779 - 1865). It was named the Euler-Gompertz constant by Finch (2003). Lagarias (2013) noted that he has not located this constant in Gompertz's writings. - Amiram Eldar, Aug 15 2020

Examples

			0.59634736232319407434107849936927937607417786015254878157348491...
With n := 10^5, Sum_{k >= 0} (n/(n + 1))^k/(n + k) = 0.5963(51...). - _Peter Bala_, Jun 19 2024
		

References

  • Bruce C. Berndt, Ramanujan's notebooks Part II, Springer, p. 171
  • Bruce C. Berndt, Ramanujan's notebooks Part I, Springer, p. 144-145.
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, pp. 303, 424-425.
  • Francois Le Lionnais, Les nombres remarquables, Paris: Hermann, 1983. See p. 29.
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 44, page 426.
  • H. S. Wall, Analytic Theory of Continued Fractions, Van Nostrand, New York, 1948, p. 356.

Crossrefs

Cf. A000522 (arrangements), A001620, A000262, A002720, A002793, A058006 (alternating factorial sums), A091725, A099285, A153229, A201203, A245780, A283743 (Ei(1)/e), A321942, A369883.

Programs

  • Magma
    SetDefaultRealField(RealField(100)); ExponentialIntegralE1(1)*Exp(1); // G. C. Greubel, Dec 04 2018
    
  • Mathematica
    RealDigits[N[-Exp[1]*ExpIntegralEi[-1], 105]][[1]]
    (* Second program: *)
    G = 1/Fold[Function[2*#2 - #2^2/#1], 2, Reverse[Range[10^4]]] // N[#, 105]&; RealDigits[G] // First (* Jean-François Alcover, Sep 19 2014 *)
  • PARI
    eint1(1)*exp(1) \\ Charles R Greathouse IV, Apr 23 2013
    
  • Sage
    numerical_approx(exp_integral_e(1,1)*exp(1), digits=100) # G. C. Greubel, Dec 04 2018

Formula

phi(1) = e*(Sum_{k>=1} (-1)^(k-1)/(k*k!) - Gamma) = 0.596347362323194... where Gamma is the Euler constant.
G = 0.596347... = 1/(1+1/(1+1/(1+2/(1+2/(1+3/(1+3/(1+4/(1+4/(1+5/(1+5/(1+6/(... - Philippe Deléham, Aug 14 2005
Equals A001113*A099285. - Johannes W. Meijer, Oct 16 2009
From Peter Bala, Oct 11 2012: (Start)
Stieltjes found the continued fraction representation G = 1/(2 - 1^2/(4 - 2^2/(6 - 3^2/(8 - ...)))). See [Wall, Chapter 18, (92.7) with a = 1]. The sequence of convergents to the continued fraction begins [1/2, 4/7, 20/34, 124/209, ...]. The numerators are in A002793 and the denominators in A002720.
Also, 1 - G has the continued fraction representation 1/(3 - 2/(5 - 6/(7 - ... -n*(n+1)/((2*n+3) - ...)))) with convergents beginning [1/3, 5/13, 29/73, 201/501, ...]. The numerators are in A201203 (unsigned) and the denominators are in A000262.
(End)
G = f(1) with f solution to the o.d.e. x^2*f'(x) + (x+1)*f(x)=1 such that f(0)=1. - Jean-François Alcover, May 28 2013
From Amiram Eldar, Aug 15 2020: (Start)
Equals Integral_{x=0..1} 1/(1-log(x)) dx.
Equals Integral_{x=1..oo} exp(1-x)/x dx.
Equals Integral_{x=0..oo} exp(-x)*log(x+1) dx.
Equals Integral_{x=0..oo} exp(-x)/(x+1) dx. (End)
From Gleb Koloskov, May 01 2021: (Start)
Equals Integral_{x=0..1} LambertW(e/x)-1 dx.
Equals Integral_{x=0..1} 1+1/LambertW(-1,-x/e) dx. (End)
Equals lim_{n->oo} A040027(n)/A000110(n+1). - Vaclav Kotesovec, Feb 22 2021
G = lim_{n->oo} A321942(n)/A000262(n). - Peter Bala, Mar 21 2022
Equals Sum_{n >= 1} 1/(n*L(n, -1)*L(n-1, -1)), where L(n, x) denotes the n-th Laguerre polynomial. This is the case x = 1 of the identity Integral_{t >= 0} exp(-t)/(x + t) dt = Sum_{n >= 1} 1/(n*L(n, -x)*L(n-1, -x)) valid for Re(x) > 0. - Peter Bala, Mar 21 2024
Equals lim_{n->oo} Sum_{k >= 0} (n/(n + 1))^k/(n + k). Cf. A099285. - Peter Bala, Jun 18 2024

Extensions

Additional references from Gerald McGarvey, Oct 10 2005
Link corrected by Johannes W. Meijer, Aug 01 2009

A055209 a(n) = Product_{i=0..n} i!^2.

Original entry on oeis.org

1, 1, 4, 144, 82944, 1194393600, 619173642240000, 15728001190723584000000, 25569049282962188245401600000000, 3366980847587422591723894776791040000000000, 44337041641882947649156022595410930014617600000000000000
Offset: 0

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Author

N. J. A. Sloane, Jul 18 2000

Keywords

Comments

a(n) is the discriminant of the polynomial x(x+1)(x+2)...(x+n). - Yuval Dekel (dekelyuval(AT)hotmail.com), Nov 13 2003
This is the Hankel transform (see A001906 for definition) of the sequence: 1, 0, 1, 0, 5, 0, 61, 0, 1385, 0, 50521, ... (see A000364: Euler numbers). - Philippe Deléham, Apr 06 2005
Also, for n>0, the quotient of (-1)^(n-1)S(u)^(n^2)/S(un) and the determinant of the (n-1) X (n-1) square matrix [P'(u), P''(u), ..., P^(n-1)(u); P''(u), P'''(u), ..., P^(n)(u); P'''(u), P^(4)(u), ..., P^(n+1)(u); ...; P^(n-1)(u), P^(n)(u), ..., P^(2n-3)(u)] where S and P are the Weierstrass Sigma and The Weierstrass P-function, respectively and f^(n) is the n-th derivative of f. See the King and Schwarz & Weierstrass references. - Balarka Sen, Jul 31 2013
a(n) is the number of idempotent monotonic labeled magmas. That is, prod(i,j) >= max(i,j) and prod(i,i) = i. - Chad Brewbaker, Nov 03 2013
Ramanujan's infinite nested radical sqrt(1+2*sqrt(1+3*sqrt(1+...))) = 3 can be written sqrt(1+sqrt(4+sqrt(144+...))) = sqrt(a(1)+sqrt(a(2)+sqrt(a(3)+...))). Vijayaraghavan used that to prove convergence of Ramanujan's formula. - Petros Hadjicostas and Jonathan Sondow, Mar 22 2014
a(n) is the determinant of the (n+1)-th order Hankel matrix whose (i,j)-entry is equal to A000142(i+j), i,j = 0,1,...,n. - Michael Shmoish, Sep 02 2020

References

  • R. Bruce King, Beyond The Quartic Equation, Birkhauser Boston, Berlin, 1996, p. 72.
  • Srinivasa Ramanujan, J. Indian Math. Soc., III (1911), 90 and IV (1912), 226.
  • T. Vijayaraghavan, in Collected Papers of Srinivasa Ramanujan, G.H. Hardy, P.V. Seshu Aiyar and B.M. Wilson, eds., Cambridge Univ. Press, 1927, p. 348; reprinted by Chelsea, 1962.

Crossrefs

Cf. A055209 is the Hankel transform (see A001906 for definition) of A000023, A000142, A000166, A000522, A003701, A010842, A010843, A051295, A052186, A053486, A053487.

Programs

  • Magma
    [1] cat [(&*[(Factorial(k))^2: k in [1..n]]): n in [1..10]]; // G. C. Greubel, Oct 14 2018
  • Maple
    seq(mul(mul(j^2,j=1..k), k=0..n), n=0..10); # Zerinvary Lajos, Sep 21 2007
  • Mathematica
    Table[Product[(i!)^2,{i,n}],{n,0,11}] (* Harvey P. Dale, Jul 06 2011 *)
    Table[BarnesG[n + 2]^2, {n, 0, 11}] (* Jan Mangaldan, May 07 2014 *)
  • PARI
    a(n)=prod(i=1,n,i!)^2 \\ Charles R Greathouse IV, Jan 12 2012
    
  • Sage
    def A055209(n) :
       return prod(factorial(i)^(2) for i in (0..n))
    [A055209(n) for n in (0..11)] # Jani Melik, Jun 06 2015
    

Formula

a(n) = A000178(n)^2. - Philippe Deléham, Mar 06 2004
a(n) = Product_{i=0..n} i^(2*n - 2*i + 2). - Charles R Greathouse IV, Jan 12 2012
Asymptotic: a(n) ~ exp(2*zeta'(-1)-3/2*(1+n^2)-3*n)*(2*Pi)^(n+1)*(n+1)^ (n^2+2*n+5/6). - Peter Luschny, Jun 23 2012
lim_{n->infinity} a(n)^(2^(-(n+1))) = 1. - Vaclav Kotesovec, Jun 06 2015
Sum_{n>=0} 1/a(n) = A258619. - Amiram Eldar, Nov 17 2020

A093101 Cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 20, 1, 10, 1, 8, 5, 2, 5, 4, 1, 130, 1, 4000, 1, 2, 5, 52, 5, 494, 1, 40, 1, 10, 13, 4, 25, 38, 5, 16, 13, 230, 13, 20, 1, 46, 5, 104, 475, 62, 1, 20, 1, 130, 31, 832, 2755, 74, 5, 4, 13, 50, 1, 40, 23, 2, 2795, 76, 34385, 2, 1, 80, 1, 650, 1, 2812, 5, 74, 5
Offset: 0

Views

Author

Jonathan Sondow, May 10 2004, Oct 18 2006

Keywords

Comments

Same as n!/A061355(n) and (1+n+n(n-1)+n(n-1)(n-2)+...+n!)/A061354(n).
a(n) is relatively prime to n.
gcd(a(n),a(n+1)) = 1.

Examples

			E.g. 1/0!+1/1!+1/2!+1/3!=16/6=(2*8)/(2*3) so a(3)=2.
		

Crossrefs

(n+1)!/(a(n)*a(n+1)) = A123899(n).
(n+3)!/(a(n)*a(n+1)*a(n+2)) = A123900(n).
(n+3)/GCD(a(n), a(n+2)) = A123901(n).
Cf. also A000522, A061354, A061355.

Programs

  • Mathematica
    f[n_] := n! / Denominator[ Sum[1/k!, {k, 0, n}]]; Table[ f[n], {n, 0, 74}] (* Robert G. Wilson v *)
    (* Second program: *)
    A[n_] := If[n==0,1,n*A[n-1]+1]; Table[GCD[A[n],n! ], {n, 0, 74}]
  • PARI
    A000522(n) = sum(k=0, n, binomial(n, k)*k!); \\ This function from Joerg Arndt, Dec 14 2014
    A093101(n) = gcd(n!,A000522(n)); \\ Antti Karttunen, Jul 12 2017

Formula

a(n) = gcd(n!, 1+n+n(n-1)+n(n-1)(n-2)+...+n!).
a(n) = gcd(n!, A(n)) where A(0) = 1, A(n) = n*A(n-1)+1.

Extensions

More terms from Robert G. Wilson v, May 14 2004

A002807 a(n) = Sum_{k=3..n} (k-1)!*C(n,k)/2.

Original entry on oeis.org

0, 0, 0, 1, 7, 37, 197, 1172, 8018, 62814, 556014, 5488059, 59740609, 710771275, 9174170011, 127661752406, 1904975488436, 30341995265036, 513771331467372, 9215499383109573, 174548332364311563, 3481204991988351553, 72920994844093191553, 1600596371590399671784
Offset: 0

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Author

Keywords

Comments

Number of cycles in the complete graph on n nodes K_{n}. - Erich Friedman
Number of equations that must be checked to verify reversibility of an n state Markov chain using the Kolmogorov criterion. - Qian Jiang (jiang1h(AT)uwindsor.ca), Jun 08 2009
Also the number of paths in the (n-1)-triangular honeycomb rook graph. - Eric W. Weisstein, Jul 14 2017

References

  • E.P.C. Kao, An Introduction to Stochastic Processes, Duxbury Press, 1997, 209-210. [From Qian Jiang (jiang1h(AT)uwindsor.ca), Jun 08 2009]
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A284947 (triangle of k-cycle counts in K_n). - Eric W. Weisstein, Apr 06 2017

Programs

  • Magma
    [&+[Factorial(k-1)*Binomial(n,k) div 2: k in [3..n]]: n in [3..30]]; // Vincenzo Librandi, Mar 06 2016
    
  • Mathematica
    Table[Sum[((k-1)!Binomial[n,k])/2,{k,3,n}],{n,0,25}] (* Harvey P. Dale, Jun 24 2011 *)
    a[n_] := n/4*(2*HypergeometricPFQ[{1, 1, 1 - n}, {2}, -1] - n - 1); a[0] = 0; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Oct 05 2012 *)
  • PARI
    a(n)=sum(k=3,n, (k-1)!*binomial(n,k)/2) \\ Charles R Greathouse IV, Feb 08 2017

Formula

E.g.f.: (-1/4)*exp(x)*(2*log(1-x)+2*x+x^2). - Vladeta Jovovic, Oct 26 2004
a(n) = (n-1)*(n-2)/2 + n*a(n-1) - (n-1)*a(n-2). - Vladeta Jovovic, Jan 22 2005
a(n) ~ exp(1)/2 * (n-1)! * (1 + 1/n + 2/n^2 + 5/n^3 + 15/n^4 + 52/n^5 + 203/n^6 + 877/n^7 + 4140/n^8 + 21147/n^9 + ...). Coefficients are the Bell numbers (A000110). - Vaclav Kotesovec, Mar 08 2016
For n>2, a(n) = Sum_{k=1..n-2} A000522(k-1)*A000217(k). - Vaclav Kotesovec, Mar 08 2016

A038156 a(n) = n! * Sum_{k=1..n-1} 1/k!.

Original entry on oeis.org

0, 0, 2, 9, 40, 205, 1236, 8659, 69280, 623529, 6235300, 68588311, 823059744, 10699776685, 149796873604, 2246953104075, 35951249665216, 611171244308689, 11001082397556420, 209020565553571999, 4180411311071440000, 87788637532500240021, 1931350025715005280484
Offset: 0

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Author

Keywords

Comments

Related to number of operations of addition and multiplication to evaluate a determinant of order n by cofactor expansion - see A026243.
Also number of operations needed to create all permutations of n distinct elements using Algorithm L (lexicographic permutation generation) from Knuth's The Art of Computer Programming, Vol. 4, chapter 7.2.1.2. Sequence gives number of comparisons required to find j in step L2 (see answer to exercise 5). - Hugo Pfoertner, Jan 24 2003
For n>1, the number of possible ballots where there are n candidates and voters may identify their top m most preferred ones, where 0 < m < n. - Shaye Horwitz, Jun 28 2011
For n > 1, a(n) is the expected number of comparisons required to sort a random list of n distinct elements using the "bogosort" algorithm. - Andrew Slattery, Jun 02 2022
The number of permutations of all proper nonempty subsets of an n element set. - P. Christopher Staecker, May 09 2024

Examples

			a(2) = floor((2.718... - 1)*2) - 1 = 3 - 1 = 2,
a(3) = floor((2.718... - 1)*6) - 1 = 10 - 1 = 9.
		

References

  • D. E. Knuth: The Art of Computer Programming, Volume 4, Fascicle 2. Generating All Tuples and Permutations, Addison-Wesley, 2005.

Crossrefs

Programs

Formula

a(n) = floor((e-1)*n!) - 1.
a(0) = a(1) = 0, a(n) = n*(a(n-1) + 1) for n>1. - Philippe Deléham, Oct 16 2009
E.g.f.: (exp(x) - 1)*x/(1 - x). - Ilya Gutkovskiy, Jan 26 2017
a(n) = A002627(n)-1, n>=1. - R. J. Mathar, Jan 03 2018
a(n) = A000522(n)-n!-1, n>=1. - P. Christopher Staecker, May 09 2024

Extensions

a(28) ff. corrected by Georg Fischer, Apr 11 2020

A132382 Lower triangular array T(n,k) generator for group of arrays related to A001147 and A102625.

Original entry on oeis.org

1, -1, 1, -1, -2, 1, -3, -3, -3, 1, -15, -12, -6, -4, 1, -105, -75, -30, -10, -5, 1, -945, -630, -225, -60, -15, -6, 1, -10395, -6615, -2205, -525, -105, -21, -7, 1, -135135, -83160, -26460, -5880, -1050, -168, -28, -8, 1, -2027025, -1216215, -374220, -79380, -13230, -1890, -252, -36, -9, 1
Offset: 0

Views

Author

Tom Copeland, Nov 11 2007, Nov 12 2007, Nov 19 2007, Dec 04 2007, Dec 06 2007

Keywords

Comments

Let b(n) = LPT[ A001147 ] = -A001147(n-1) for n > 0 and 1 for n=0, where LPT represents the action of the list partition transform described in A133314.
Then T(n,k) = binomial(n,k) * b(n-k) .
Form the matrix of polynomials TB(n,k,t) = T(n,k) * t^(n-k) = binomial(n,k) * b(n-k) * t^(n-k) = binomial(n,k) * Pb(n-k,t),
beginning as
1;
-1, 1;
-1*t, -2, 1;
-3*t^2, -3*t, -3, 1;
-15*t^3, -12*t^2, -6*t, -4, 1;
-105*t^4, -75*t^3, -30*t^2, -10*t, -5, 1;
Let Pc(n,t) = LPT(Pb(.,t)).
Then [TB(t)]^(-1) = TC(t) = [ binomial(n,k) * Pc(n-k,t) ] = LPT(TB),
whose first column is
Pc(0,t) = 1
Pc(1,t) = 1
Pc(2,t) = 2 + t
Pc(3,t) = 6 + 6*t + 3*t^2
Pc(4,t) = 24 + 36*t + 30*t^2 + 15*t^3
Pc(5,t) = 120 + 240*t + 270*t^2 + 210*t^3 + 105*t^4.
The coefficients of these polynomials are given by the reverse of A102625 with the highest order coefficients given by A001147 with an additional leading 1.
Note this is not the complete matrix TC. The complete matrix is formed by multiplying along the diagonal of the lower triangular Pascal matrix by these polynomials, embedding trees of coefficients in the matrix.
exp[Pb(.,t)*x] = 1 + [(1-2t*x)^(1/2) - 1] / (t-0) = [1 + a finite diff. of [(1-2t*x)^(1/2)] with step t] = e.g.f. of the first column of TB.
exp[Pc(.,t)*x] = 1 / { 1 + [(1-2t*x)^(1/2) - 1] / t } = 1 / exp[Pb(.,t)*x) = e.g.f. of the first column of TC.
TB(t) and TC(t), being inverse to each other, are the generators of an Abelian group.
TB(0) and TC(0) are generators for a subgroup representing the iterated Laguerre operator described in A132013 and A132014.
Let sb(t,m) and sc(t,m) be the associated sequences under the LPT to TB(t)^m = B(t,m) and TC(t)^m = C(t,m).
Let Esb(t,m) and Esc(t,m) be e.g.f.'s for sb(t,m) and sc(t,m), rB(t,m) and rC(t,m) be the row sums of B(t,m) and C(t,m) and aB(t,m) and aC(t,m) be the alternating row sums.
Then B(t,m) is the inverse of C(t,m), Esb(t,m) is the reciprocal of Esc(t,m) and sb(t,m) and sc(t,m) form a reciprocal pair under the LPT. Similar relations hold among the row sums and the alternating sign row sums and associated quantities.
All the group members have the form B(t,m) * C(u,p) = TB(t)^m * TC(u)^p = [ binomial(n,k) * s(n-k) ]
with associated e.g.f. Es(x) = exp[m * Pb(.,t) * x] * exp[p * Pc(.,u) * x] for the first column of the matrix, with terms s(n), so group multiplication is isomorphic to matrix multiplication and to multiplication of the e.g.f.'s for the associated sequences (see examples).
These results can be extended to other groups of integer-valued arrays by replacing the 2 by any natural number in the expression for exp[Pb(.,t)*x].
More generally,
[ G.f. for M = Product_{i=0..j} B[s(i),m(i)] * C[t(i),n(i)] ]
= exp(u*x) * Product_{i=0..j} { exp[m(i) * Pb(.,s(i)) * x] * exp[n(i) * Pc(.,t(i)) * x] }
= exp(u*x) * Product_{i=0..j} { 1 + [ (1 - 2*s(i)*x)^(1/2) - 1 ] / s(i) }^m(i) / { 1 + [ (1 - 2*t(i)*x)^(1/2) - 1 ] / t(i) }^n(i)
= exp(u*x) * H(x)
[ E.g.f. for M ] = I_o[2*(u*x)^(1/2)] * H(x).
M is an integer-valued matrix for m(i) and n(i) positive integers and s(i) and t(i) integers. To invert M, change B to C in Product for M.
H(x) is the e.g.f. for the first column of M and diagonally multiplying the Pascal matrix by the terms of this column generates M. See examples.
The G.f. for M, i.e., the e.g.f. for the row polynomials of M, implies that the row polynomials form an Appell sequence (see Wikipedia and Mathworld). - Tom Copeland, Dec 03 2013

Examples

			Some group members and associated arrays are
(t,m) :: Array :: Asc. Matrix :: Asc. Sequence :: E.g.f. for sequence
..............................................................................
(0,1).::.B..::..A132013.::.(1,-1,0,0,0,0,...).....::.s(x).=.1-x
(0,1).::.C..::..A094587.::.(0!,1!,2!,3!,...)......::.1./.s(x)
(0,1).::.rB.::.~A055137.::.(1,0,-1,-2,-3,-4,...)..::.exp(x).*.s(x)
(0,1).::.rC.::....-.....::..A000522...............::.exp(x)./.s(x)
(0,1).::.aB.::....-.....::.(1,-2,3,-4,5,-6,...)...::.exp(-x).*.s(x)
(0,1).::.aC.::..A008290.::..A000166...............::.exp(-x)./.s(x)
..............................................................................
(0,2).::.B..::..A132014.::.(1,-2,2,0,0,0,0...)....::.s(x).=.(1-x)^2
(0,2).::.C..::..A132159.::.(1!,2!,3!,4!,...)......::..1./.s(x).
(0,2).::.rB.::...-......::.(1,-1,-1,1,5,11,19,29,)::.exp(x).*.s(x).
(0,2).::.rC.::...-......::..A001339...............::.exp(x)./.s(x).
(0,2).::.aB.::...-......::.(-1)^n.A002061(n+1)....::.exp(-x).*.s(x).
(0,2).::.aC.::...-......::..A000255...............::.exp(-x)./.s(x).
..............................................................................
(1,1).::.B..::..T.......::.(1,-A001147(n-1))......::.s(x).=.(1-2x)^(1/2)
(1,1).::.C..::.~A113278.::..A001147...............::.1./.s(x)...
(1,1).::.rB.::...-......::..A055142...............::.exp(x).*.s(x).
(1,1).::.rC.::...-......::..A084262...............::.exp(x)./.s(x).
(1,1).::.aB.::...-......::.(1,-2,2,-4,-4,-56,...).::.exp(-x).*.s(x).
(1,1).::.aC.::...-......::..A053871...............::.exp(-x)./.s(x).
..............................................................................
(2,1).::.B..::...-......::.(1,-A001813)...........::.s=[1+(1-4x)^(1/2)]/2....
(2,1).::.C..::...-......::..A001761...............::.1./.s(x)..
(2,1).::.rB.::...-......::.(1,0,-3,-20,-183,...)..::.exp(x).*.s(x)..
(2,1).::.rC.::...-......::.(1,2,7,46,485,...).....::.exp(x)./.s(x).
(2,1).::.aB.::...-......::.(1,-2,1,-10,-79,...)...::.exp(-x).*.s(x).
(2,1).::.aC.::...-......::.(1,0,3,20,237,...).....::.exp(-x)./.s(x)
..............................................................................
(1,2).::.B..::.~A134082.::.(1,-2,0,0,0,0,...).....::.s(x).=.1.-.2x
(1,2).::.C..::....-.....::..A000165...............::.1./.s(x)..
(1,2).::.rB.::....-.....::.(1,-1,-3,-5,-7,-9,...).::.exp(x).*.s(x).
(1,2).::.rC.::....-.....::..A010844...............::.exp(x)./.s(x)..
(1,2).::.aB.::....-.....::.(1,-3,5,-7,9,-11,...)..::.exp(-x).*.s(x).
(1,2).::.aC.::....-.....::..A000354...............::.exp(-x)./.s(x).
..............................................................................
(The tilde indicates the match is not exact--specifically, there are differences in signs from the true matrices.)
Note the row sums correspond to binomial transforms of s(x) and the alternating row sums, to inverse binomial transforms, or, finite differences.
Some additional examples:
C(1,2)*B(0,1) = B(1,-2)*C(0,-1) = [ binomial(n,k)*A002866(n-k) ] with asc. e.g.f. (1-x) / (1-2x).
B(1,2)*C(0,1) = C(1,-2)*B(0,-1) = 2I - A094587 with asc. e.g.f. (1-2x) / (1-x).
		

Formula

[G.f. for TB(n,k,t)] = GTB(u,x,t) = exp(u*x) * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pb(.,t))*x] where TB(n,k,t) = (D_x)^n (D_u)^k /k! GTB(u,x,t) eval. at u=x=0.
[G.f. for TC(n,k,t)] = GTC(u,x,t) = exp(u*x) / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } = exp[(u+Pc(.,t))*x] where TC(n,k,t) = (D_x)^n (D_u)^k /k! GTC(u,x,t) eval. at u=x=0.
[E.g.f. for TB(n,k,t)] = I_o[2*(u*x)^(1/2)] * { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t } and
[E.g.f. for TC(n,k,t)] = I_o[2*(u*x)^(1/2)] / { 1 + [ (1 - 2t*x)^(1/2) - 1 ] / t }
where I_o is the zeroth modified Bessel function of the first kind, i.e.,
I_o[2*(u*x)^(1/2)] = Sum_{j>=0} (u^j/j!) * (x^j/j!).
So [e.g.f. for TB(n,k)] = I_o[2*(u*x)^(1/2)] * (1 - 2x)^(1/2).

Extensions

More terms from Tom Copeland, Dec 05 2007

A056542 a(n) = n*a(n-1) + 1, a(1) = 0.

Original entry on oeis.org

0, 1, 4, 17, 86, 517, 3620, 28961, 260650, 2606501, 28671512, 344058145, 4472755886, 62618582405, 939278736076, 15028459777217, 255483816212690, 4598708691828421, 87375465144740000, 1747509302894800001, 36697695360790800022, 807349297937397600485
Offset: 1

Views

Author

Henry Bottomley, Jun 20 2000

Keywords

Comments

For n >= 2 also operation count to create all permutations of n distinct elements using Algorithm L (lexicographic permutation generation) from Knuth's The Art of Computer Programming, Vol. 4, chapter 7.2.1.2. Sequence gives number of loop repetitions of the j search loop in step L2. - Hugo Pfoertner, Feb 06 2003
More directly: sum over all permutations of length n-1 of the product of the length of the first increasing run by the value of the first position. The recurrence follows from this definition. - Olivier Gérard, Jul 07 2011
This sequence shares divisibility properties with A000522; each of the primes in A072456 divide only a finite number of terms of this sequence. - T. D. Noe, Jul 07 2005
This sequence also represents the number of subdeterminant evaluations when calculation a determinant by Laplace recursive method. - Reinhard Muehlfeld, Sep 14 2010
Also, a(n) equals the number of non-isomorphic directed graphs of n+1 vertices with 1 component, where each vertex has exactly one outgoing edge, excluding loops and cycle graphs. - Stephen Dunn, Nov 30 2019

Examples

			a(4) = 4*a(3) + 1 = 4*4 + 1 = 17.
Permutations of order 3 .. Length of first run * First position
123..3*1
132..2*1
213..1*2
231..2*2
312..1*3
321..1*3
a(4) = 3+2+2+4+3+3 = 17. - _Olivier Gérard_, Jul 07 2011
		

References

  • D. E. Knuth: The Art of Computer Programming, Volume 4, Combinatorial Algorithms, Volume 4A, Enumeration and Backtracking. Pre-fascicle 2B, A draft of section 7.2.1.2: Generating all permutations. Available online; see link.

Crossrefs

Cf. A079751 (same recursion formula, but starting from a(3)=0), A038155, A038156, A080047, A080048, A080049.
Equals the row sums of A162995 triangle (n>=2). - Johannes W. Meijer, Jul 21 2009
Cf. A070213 (indices of primes).

Programs

  • Haskell
    a056542 n = a056542_list !! (n-1)
    a056542_list = 0 : map (+ 1) (zipWith (*) [2..] a056542_list)
    -- Reinhard Zumkeller, Mar 24 2013
    
  • Magma
    [n le 2 select n-1 else n*Self(n-1)+1: n in [1..20]]; // Bruno Berselli, Dec 13 2013
  • Mathematica
    tmp=0; Join[{tmp}, Table[tmp=n*tmp+1, {n, 2, 100}]] (* T. D. Noe, Jul 12 2005 *)
    FoldList[ #1*#2 + 1 &, 0, Range[2, 21]] (* Robert G. Wilson v, Oct 11 2005 *)

Formula

a(n) = floor((e-2)*n!).
a(n) = A002627(n) - n!.
a(n) = A000522(n) - 2*n!.
a(n) = n! - A056543(n).
a(n) = (n-1)*(a(n-1) + a(n-2)) + 2, n > 2. - Gary Detlefs, Jun 22 2010
1/(e - 2) = 2! - 2!/(1*4) - 3!/(4*17) - 4!/(17*86) - 5!/(86*517) - ... (see A002627 and A185108). - Peter Bala, Oct 09 2013
E.g.f.: (exp(x) - 1 - x) / (1 - x). - Ilya Gutkovskiy, Jun 26 2022

Extensions

More terms from James Sellers, Jul 04 2000

A068996 Decimal expansion of 1 - 1/e.

Original entry on oeis.org

6, 3, 2, 1, 2, 0, 5, 5, 8, 8, 2, 8, 5, 5, 7, 6, 7, 8, 4, 0, 4, 4, 7, 6, 2, 2, 9, 8, 3, 8, 5, 3, 9, 1, 3, 2, 5, 5, 4, 1, 8, 8, 8, 6, 8, 9, 6, 8, 2, 3, 2, 1, 6, 5, 4, 9, 2, 1, 6, 3, 1, 9, 8, 3, 0, 2, 5, 3, 8, 5, 0, 4, 2, 5, 5, 1, 0, 0, 1, 9, 6, 6, 4, 2, 8, 5, 2, 7, 2, 5, 6, 5, 4, 0, 8, 0, 3, 5, 6
Offset: 0

Views

Author

N. J. A. Sloane, Apr 08 2002

Keywords

Comments

From the "derangements" problem: this is the probability that if a large number of people are given their hats at random, at least one person gets their own hat.
1-1/e is the limit to which (1 - !n/n!) {= 1 - A000166(n)/A000142(n) = A002467(n)/A000142(n)} converges as n tends to infinity. - Lekraj Beedassy, Apr 14 2005
Also, this is lim_{n->inf} P(n), where P(n) is the probability that a random rooted forest on [n] is not connected. - Washington Bomfim, Nov 01 2010
Also equals the mode of a Gompertz distribution when the shape parameter is less than 1. - Jean-François Alcover, Apr 17 2013
The asymptotic density of numbers with an even number of trailing zeros in their factorial base representation (A232744). - Amiram Eldar, Feb 26 2021

Examples

			0.6321205588285576784044762...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, Section 1.3, pp. 12-17.
  • Anders Hald, A History of Probability and Statistics and Their Applications before 1750, Wiley, NY, 1990 (Chapter 19).
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 65.

Crossrefs

Programs

Formula

Equals Integral_{x = 0 .. 1} exp(-x) dx. - Alonso del Arte, Jul 06 2012
Equals -Sum_{k>=1} (-1)^k/k!. - Bruno Berselli, May 13 2013
Equals Sum_{k>=0} 1/((2*k+2)*(2*k)!). - Fred Daniel Kline, Mar 03 2016
From Peter Bala, Nov 27 2019: (Start)
1 - 1/e = Sum_{n >= 0} n!/(A(n)*A(n+1)), where A(n) = A000522(n).
Continued fraction expansion: [0; 1, 1, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, ...].
Related continued fraction expansions include
2*(1 - 1/e) = [1; 3, 1, 3, 1, 1, 1, 3, 3, 3, 1, 3, 1, 3, 5, 3, 1, 5, ..., 1, 3, 2*n + 1, 3, 1, 2*n + 1, ...];
(1/2)*(1 - 1/e) = [0; 3, 6, 10, 14, 18, ..., 4*n + 2, ...];
4*(1 - 1/e) = [2; 1, 1, 8, 3, 1, 1, 1, 1, 7, 1, 1, 2, 1, 1, 1, 2, 7, 1, 2, 2, 1, 1, 1, 3, ..., 7, 1, n, 2, 1, 1, 1, n+1, ...];
(1/4)*(1 - 1/e) = [0; 6, 3, 20, 7, 36, 11, 52, 15, ..., 16*n + 4, 4*n + 3, ...]. (End)
Equals Integral_{x=0..1} x * cosh(x) dx. - Amiram Eldar, Aug 14 2020
Equals A091131/e. - Hugo Pfoertner, Aug 20 2024

A142979 a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n+1)^2*a(n).

Original entry on oeis.org

1, 3, 13, 66, 406, 2868, 23220, 210192, 2116656, 23375520, 281792160, 3673814400, 51599514240, 775673176320, 12440524320000, 211848037632000, 3820318338816000, 72685037892096000, 1455838255452672000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 1 of the more general recurrence a(1) = 1, a(2) = 2*m + 1, a(n+2) = (2*m + 1)*a(n+1) + (n + 1)^2*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of Mercator's series for the constant log(2). For other cases see A024167 (m = 0), A142980 (m = 2), A142981 (m = 3) and A142982 (m = 4).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)), where p_m(x) = Sum_{k = 0..m} 2^k*C(m,k)*C(x,k) = Sum_{k = 0..m} C(m,k)*C(x+k,m), is the Ehrhart polynomial of the m-dimensional cross polytope (the hyperoctahedron).
The first few values are p_0(x) = 1, p_1(x) = 2*x + 1, p_2(x) = 2*x^2 + 2*x + 1 and p_3(x) = (4*x^3 + 6*x^2 + 8*x + 3)/3.
The sequence {p_m(k)},k>=0 is the crystal ball sequence for the product lattice A_1 x... x A_1 (m copies). The table of values [p_m(k)]m,k>=0 is the array of Delannoy numbers A008288.
The polynomial p_m(x) is the unique polynomial solution of the difference equation (x + 1)*f(x+1) - x*f(x-1) = (2*m + 1)*f(x), normalized so that f(0) = 1. These polynomials have their zeros on the vertical line Re x = -1/2 in the complex plane, that is, the polynomials p_m(x-1), m = 1,2,3,..., satisfy a Riemann hypothesis [BUMP et al., Theorems 4 and 6]. The o.g.f. for the p_m(x) is (1 + t)^x/(1 - t)^(x+1) = 1 + (2*x + 1)*t + (2*x^2 + 2*x + 1)*t^2 + ....
The general recurrence in the first paragraph above also has a second solution b(n) = n!*p_m(n) with initial conditions b(1) = 2*m + 1, b(2) = (2*m + 1)^2 + 1.
Hence the behavior of a(n) for large n is given by lim_{n -> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(k*p_m(k-1)*p_m(k)) = 1/((2*m + 1) + 1^2/((2*m + 1) + 2^2/((2*m + 1) + 3^2/((2*m + 1) + ... + n^2/((2*m + 1) + ...))))) = (-1)^m * (log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 29]).
For other sequences defined by similar recurrences and related to log(2) see A142983 and A142988. See also A142992 for the connection between log(2) and the C_n lattices. For corresponding results for the constants e, zeta(2) and zeta(3) see A000522, A142995 and A143003 respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Maple
    p := n -> 2*n+1: a := n -> n!*p(n)*sum ((-1)^(k+1)/(k*p(k-1)*p(k)), k = 1..n): seq(a(n), n = 1..20)
  • Mathematica
    RecurrenceTable[{a[1]==1,a[2]==3,a[n+2]==3a[n+1]+(n+1)^2 a[n]},a,{n,20}] (* Harvey P. Dale, May 20 2012 *)

Formula

a(n) = n!*p(n)*Sum_{k = 1..n} (-1)^(k+1)/(k*p(k-1)*p(k)), where p(n) = 2*n + 1.
Recurrence: a(1) = 1, a(2) = 3, a(n+2) = 3*a(n+1) + (n + 1)^2*a(n).
The sequence b(n):= n!*p(n) satisfies the same recurrence with b(1) = 3, b(2) = 10.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/3)))), for n >= 2.
Limit_{n -> oo} a(n)/b(n) = 1/(3 + 1^2/(3 + 2^2/(3 + 3^2/(3 + ... + (n-1)^2/(3 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(4k^2 - 1)) = 1 - log(2).
Thus a(n) ~ c*n*n! as n -> oo, where c = 2*(1 - log(2)).
From Peter Bala, Dec 09 2024: (Start)
E.g.f.: A(x) = (2*x - (1 + x)*log(1 + x))/(1 - x)^2 satisfies the differential equation 1 + (x + 3)*A(x) + (x^2 - 1)*A'(x) = 0 with A(0) = 0.
Sum_{k = 1..n} Stirling_2(n, k) * a(k) = A317057(n+1). (End)

A142983 a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).

Original entry on oeis.org

1, 2, 10, 44, 288, 1896, 15888, 137952, 1419840, 15255360, 186693120, 2387093760, 33898314240, 502247692800, 8123141376000, 136785729024000, 2483065912320000, 46822564905984000, 942853671825408000, 19678282007924736000, 435355106182520832000
Offset: 1

Views

Author

Peter Bala, Jul 17 2008

Keywords

Comments

This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series 1/2 + 1/2*Sum_{k > 1} (-1)^(k+1)/(k*(k + 1)) = log(2). For other cases see A142984 (m = 2), A142985 (m = 3), A142986 (m = 4) and A142987 (m = 5).
The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p_m(k)*p_m(k+1)), where p_m(x) = Sum_{k = 1..m} 2^(k-1)*C(m-1,k-1)*C(x,k) is the polynomial that gives the regular polytope numbers for the m-dimensional cross polytope as defined by [Kim] (see A142978). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_m(x) is the unique polynomial solution of the difference equation x*(f(x+1) - f(x-1)) = 2*m*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_m(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_m(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_m(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The general recurrence in the first paragraph above has a second solution b(n) = n!*p_m(n+1) with b(1) = 2*m, b(2) = m^2 + 2. Hence the behavior of a(n) for large n is given by Limit_{n-> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p_m(k)*p_m(k+1)) = 1/((2*m) + 1*2/((2*m) + 2*3/((2*m) + 3*4/((2*m) + ... + n*(n + 1)/((2*m) + ...))))) = 1 + (-1)^(m+1) * (2*m)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
See A142979, A142988 and A142992 for similar results. For corresponding results for Napier's constant e, the constant zeta(2) and Apery's constant zeta(3) refer to A000522, A142995 and A143003, respectively.

References

  • Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.

Crossrefs

Programs

  • Haskell
    a142983 n = a142983_list !! (n-1)
    a142983_list = 1 : 2 : zipWith (+)
                           (map (* 2) $ tail a142983_list)
                           (zipWith (*) (drop 2 a002378_list) a142983_list)
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    a := n -> (n+1)!*sum ((-1)^(k+1)/(k*(k+1)), k = 1..n): seq(a(n), n = 1..20);
  • Mathematica
    Rest[CoefficientList[Series[(-x+2*Log[x+1])/(x-1)^2,{x,0,20}],x]*Range[0,20]!] (* Vaclav Kotesovec, Oct 21 2012 *)

Formula

a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n.
Recurrence: a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 2 and b(2) = 6.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + (n - 1)*n/2)))), for n >= 2.
The behavior of a(n) for large n is given by Limit_{n -> oo} a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + n*(n+1)/(2 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)) = 2*log(2) - 1.
E.g.f.: (2*log(x+1)-x)/(x-1)^2. - Vaclav Kotesovec, Oct 21 2012
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