A001147 -id:A001147 - cp's OEIS frontend

A249348 a(n) = (A001147(n+1)^2-1)/8, where A001147(n+1) = 35...*(2n+1).

0, 1, 28, 1378, 111628, 13507003, 2282683528, 513603793828, 148431496416328, 53583770206294453, 23630442660975853828, 12500504167656226675078, 7812815104785141671923828, 5695542211388368278832470703, 4789950999777617722498107861328

Offset: 0

M. F. Hasler, Oct 26 2014

nonn

These are the numerators of the partial sums S(n) = Sum_{k=1..n} A000217(k)/A001147(k+1)^2 before simplification, i.e., a(n) = S(n)*A001147(n+1)^2, where A000217(n) = n(n+1)/2. The series S(n) has sum 1/8, actually S(n) = 1/8 - 1/(8*A001147(n+1)^2). (Similarly, Sum_{n=1..oo} A249354(n)/A007559(n+1)^3 = 1/9, where A249354(n) = 3n^3+3n^2+n.)

This is a subsequence of the centered 9-gonal numbers A060544, which are a subsequence of the triangular numbers A000217.

S. Klein, A neat infinite sum ..., "Number Theory" group on LinkedIn, Oct. 2014.

A249348 := proc(n)
    (doublefactorial(2*n+1)^2-1)/8 ;
end proc:
seq(A249348(n),n=0..20) ;

PARI
```
a(n)=(prod(k=1,n,2*k+1)^2-1)/8
```

(-n+1)*a(n) +2*n*(2*n^2-1)*a(n-1) -(n+1)*(-1+2*n)^2*a(n-2)=0. - R. J. Mathar, Oct 28 2014

a(11)/a(12) corrected by Georg Fischer, Mar 12 2020

A166750 a(n) = (A001147(n))^3 = 2^(3n)GAMMA(n+1/2)^3/Pi^(3/2).

Original entry on oeis.org

1, 1, 27, 3375, 1157625, 843908625, 1123242379875, 2467763508585375, 8328701841475640625, 40918912147169822390625, 280662818417437811777296875, 2599218361363891574869546359375, 31624689802714468791437770554515625, 494135778167413574866215164914306640625

Offset: 0

Karol A. Penson, Oct 21 2009

nonn

Vincenzo Librandi, Table of n, a(n) for n = 0..160

DoubleFactorial:=func< n | &*[n..2 by -2] >; [DoubleFactorial((2*n-1))^3: n in [0..20]]; // Vincenzo Librandi, Jul 21 2017

seq((doublefactorial(2*n-1))^3, n=0..15);

Table[((2 n - 1)!!)^3, {n, 0, 30}] (* Vincenzo Librandi, Jul 21 2017 *)

G.f.: sum(a(n)*x^n/(n!)^3,n=0..infinity) = 4*EllipticK((1/2)*sqrt(2-2*sqrt(1-8*x)))^2/Pi^2, sum(a(n)*x^n/(n!)^4,n=0..infinity)=hypergeom([1/2,1/2,1/2],[1,1,1],8*x).
Asymptotics: a(n) = (2*sqrt(2)/((exp(-1/2))^3*(exp(1/2))^3)-(1/4)*sqrt(2)/((exp(-1/2))^3*(exp(1/2))^3*n)+(1/64)*sqrt(2)/((exp(-1/2))^3*(exp(1/2))^3*n^2)+O(1/n^3))*(2^n)^3/(((1/n)^n)^3*(exp(n))^3), n->infinity.
Integral representation as n-th moment of a positive function on a positive halfaxis (solution of the Stieltjes moment problem),in Maple notation:
a(n) = int(x^n*MeijerG([[],[]],[[ -1/2,-1/2,-1/2],[]],x/8)/(8*(Pi)^(3/2)), x=0..infinity), n=0,1... .
This solution is not unique.
a(n) -(2*n-1)^3*a(n-1) +a(n-2) -(2*n-5)^3*a(n-3) =0. - R. J. Mathar, Jul 24 2012

A248652 Union of the factorial numbers (A000142) and the double factorials of odd numbers (A001147).

Original entry on oeis.org

1, 2, 3, 6, 15, 24, 105, 120, 720, 945, 5040, 10395, 40320, 135135, 362880, 2027025, 3628800, 34459425, 39916800, 479001600, 654729075, 6227020800, 13749310575, 87178291200, 316234143225, 1307674368000, 7905853580625, 20922789888000, 213458046676875, 355687428096000

Offset: 1

Olivier Gérard, Oct 10 2014

nonn

Douglas Hoftstadter, Keynote lecture, DIMACS Workshop on Recognition of Integer Sequences, Oct. 10, 2014.

Abstract of D. Hoftstadter in the Workshop's program

Cf. A000142, A001147, A005214.

See A268645 for another version.

max = 10^15;
A000142 = Reap[For[k = 0, k!  <= max, k++, Sow[k!]]][[2, 1]];
A001147 = Reap[For[k = 1, k!! <= max, k = k+2, Sow[k!!]]][[2, 1]];
Union[A000142, A001147] (* Jean-François Alcover, Jul 18 2022 *)

Revised by N. J. A. Sloane, Feb 09 2016

A249349 (A001147(n+1)-1)/2, equals the index of A249348(n) within the triangular numbers A000217.

Original entry on oeis.org

0, 1, 7, 52, 472, 5197, 67567, 1013512, 17229712, 327364537, 6874655287, 158117071612, 3952926790312, 106729023338437, 3095141676814687, 95949391981255312, 3166329935381425312, 110821547738349885937, 4100397266318945779687, 159915493386438885407812

Offset: 0

M. F. Hasler, Oct 26 2014

nonn

Also a(n) = floor(sqrt(A249348(n)*2)).

The positive terms are of the form 3k-2; this k (= 1, 3, 18, 157, ...) is the index of A249348(n) within the centered 9-gonal numbers A060544.

PARI
```
a(n)=A001147(n+1)\2
```

vector(10,n,A001147(n)\2) \\ To get the initial term a(0) for n=1.

a(n) +(-2*n-3)*a(n-1) +(4*n-1)*a(n-2) +(-2*n+3)*a(n-3)=0. - R. J. Mathar, Oct 28 2014

A263801 Partial sums of odd double factorials (A001147) with alternating signs.

Original entry on oeis.org

1, 0, 3, -12, 93, -852, 9543, -125592, 1901433, -32557992, 622171083, -13127139492, 303107003733, -7602746576892, 205855300099983, -5984428053529392, 185914355908981233, -6146745514853869392, 215496349961845902483, -7985298182676045656892

Offset: 0

Vladimir Reshetnikov, Oct 26 2015

sign

			For n = 4, a(4) = Sum_{k=0..4} (-1)^k*(2*k-1)!! = (-1)!! - 1!! + 3!! - 5!! + 7!! = 1 - 1 + 3 - 15 + 105 = 93.
G.f. = 1 + 3*x^2 - 12*x^3 + 93*x^4 - 852*x^5 + 9543*x^6 - 125592*x^7 + ...

G. C. Greubel, Table of n, a(n) for n = 0..400
Eric Weisstein's MathWorld, Double Factorial.
Eric Weisstein's MathWorld, Incomplete Gamma Function.
Eric Weisstein's MathWorld, Erf.

Cf. A001147, A076795.

Table[Sum[(-1)^k (2k-1)!!, {k, 0, n}], {n, 0, 20}]
Round@Table[(Sqrt[Pi] Erfc[1/Sqrt[2]] - Gamma[-n-1/2, 1/2] (2n+1)!!/(-2)^(n+1)) Sqrt[E/2], {n, 0, 20}]

for(n=0,50, print1(sum(k=0,n, (-1)^k*(2*k)!/(2^k*k!)), ", ")) \\ G. C. Greubel, Apr 08 2017

a(n) = Sum_{k=0..n} (-1)^k*(2*k-1)!!.
a(n) = (sqrt(Pi)*erfc(1/sqrt(2))-Gamma(-n-1/2, 1/2)*(2*n+1)!!/(-2)^(n+1))*exp(1/2)/sqrt(2), where Gamma(a, x) is the upper incomplete Gamma function.
E.g.f.: 1/sqrt(2*x+1)+sqrt(Pi/2)*exp(x+1/2)*(erf(sqrt(x+1/2))-erf(1/sqrt(2))).
Recurrence: a(0) = 1, a(1) = 0, a(n+2) = (2*n+3)*a(n)-(2*n+2)*a(n+1).
0 = a(n)*(-2*a(n+1) + a(n+2) + a(n+3)) + a(n+1)*(+3*a(n+1) - 3*a(n+2) - a(n+3)) + a(n+2)*(+a(n+2)) if n>=0. - Michael Somos, Oct 30 2015

A350464 Table read by rows. Interpolating the swinging factorial (A056040) and the double factorial (A001147).

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 2, 15, 15, 0, 6, 91, 210, 105, 0, 6, 690, 2835, 3150, 945, 0, 30, 5214, 42405, 79695, 51975, 10395, 0, 20, 44772, 666666, 2057055, 2207205, 945945, 135135, 0, 140, 384756, 11274900, 54879825, 90090000, 62432370, 18918900, 2027025

Offset: 0

Peter Luschny, Mar 13 2022

			Triangle starts:
[0] 1;
[1] 0,  1;
[2] 0,  1,   3;
[3] 0,  2,   15,     15;
[4] 0,  6,   91,     210,     105;
[5] 0,  6,   690,    2835,    3150,     945;
[6] 0,  30,  5214,   42405,   79695,    51975,    10395;
[7] 0,  20,  44772,  666666,  2057055,  2207205,  945945,  135135;

Cf. A350465 (row sums), A350466 (alternating row sums).

Cf. A056040, A001147, A001880.

Swing[n_] := n! / Floor[n/2]!^2;
Z[n_] := Flatten[Table[{0, Swing[j]}, {j, 0, n}]];
T[n_, k_] := BellY[2 n, k, Z[n - k]];
Table[T[n, k], {n, 0, 8}, {k, 0, n}] // Flatten

The partial Bell polynomials Y_{2*n, k}(Z) applied to the list Z of the aerated swinging factorials (A056040).

A358987 Omit the trailing 5 from double factorial of odd numbers (A001147(n)).

Original entry on oeis.org

1, 1, 3, 1, 10, 94, 1039, 13513, 202702, 3445942, 65472907, 1374931057, 31623414322, 790585358062, 21345804667687, 619028335362937, 19189878396251062, 633265987076285062, 22164309547669977187, 820079453263789155937, 31983098677287777081562, 1311307045768798860344062

Offset: 0

Stefano Spezia, Dec 10 2022

A001147(n) has only a trailing five for n > 2.

Proof: being trivial to prove that A001147(n) ends with at least a digit 5 for n > 2, it remains to prove that the tenth digit of A001147(n) is not equal to 5. Considering the product A001147(n) = A001147(n-1)*(2*n - 1) for n > 2, it is easy to verify that the tenth digit of A001147(n) is congruent to 2 modulo 5 if the tenth digit of A001147(n-1) is congruent to 2 modulo 5. Since for n <= 8 the tenth digit of A001147(n) is not equal to 5, and it is equal to 2 for n = 8, it follows that the tenth digit of A001147(n) for n > 8 is congruent to 2 modulo 5, and therefore, not equal to 5. QED.

Stefano Spezia, Table of n, a(n) for n = 0..400

Cf. A001147.

Join[{1,1,3},Table[((2n-1)!!-5)/10,{n,3,21}]] (* or *)
CoefficientList[Series[(14-5Exp[x]+1/Sqrt[1-2x]+2x(7+8x))/10,{x,0,21}],x]Table[n!,{n,0,21}]

a(n) = (A001147(n) - 5)/10 for n > 2.
E.g.f.: (14 + 2*x*(7 + 8*x) - 5*exp(x) + 1/sqrt(1 - 2*x))/10.
D-finite with recurrence a(n) + (-2*n+1)*a(n-1) + (-n+1) = 0 for n > 3. - R. J. Mathar, Mar 25 2024

A000779 a(n) = 2(2n-1)!!-(n-1)!2^(n-1), where (2n-1)!! is A001147(n).

Original entry on oeis.org

1, 4, 22, 162, 1506, 16950, 224190, 3408930, 58596930, 1123663590, 23782729950, 550718680050, 13849716607650, 375904338960150, 10952237584237950, 340947694234397250, 11294123783425733250, 396665528378000631750

Offset: 1

N. J. A. Sloane

Nathaniel Johnston, Table of n, a(n) for n = 1..250
J. R. Stembridge, Some combinatorial aspects of reduced words in finite Coxeter groups, Trans. Amer. Math. Soc. 349 (1997), no. 4, 1285-1332.

Cf. A001147.

A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ 2*A001147(n)-Factorial(n-1)*2^(n-1): n in [1..20] ]; // Klaus Brockhaus, Jun 22 2011

seq(2*doublefactorial(2*n-1)-(n-1)!*2^(n-1), n=1..18); # Nathaniel Johnston, Jun 23 2011

Table[2*(2n-1)!! - (n-1)!*2^(n-1), {n, 1, 20}] (* Jean-François Alcover, Feb 11 2016 *)

More terms from Sean A. Irvine, Jun 13 2011

A034405 Let f(x) = (Pi - 2arctan(1/(sqrt(x)sqrt(x+2))))/(2sqrt(x)sqrt(x+2)), take (-1)^n*(n-th derivative from right at x=0) and multiply by A001147(n+1).

Original entry on oeis.org

1, 2, 14, 216, 5976, 262800, 16945200, 1511395200, 178458940800, 26959810348800, 5071861902240000, 1162523770531200000, 318880083535896960000, 103120648805872938240000, 38820554918130896951040000, 16829499728777665273344000000, 8323409867177396185818624000000, 4657912954052653582049258496000000

Offset: 0

James R. FitzSimons (cherry(AT)neta.com)

nonn

This is related to the solution of Problem 12150 of American Math. Monthly, vol. 126 (2019), page 946. - Stephen J. Herschkorn, Dec 14 2019

Robert Israel, Table of n, a(n) for n = 0..238

Cf. A001147.

f:= gfun:-rectoproc({-(3*n+5)*(n+2)*a(n+1)+a(n+2)+(2*n+3)*(n+1)^2*(n+2)*a(n),a(0)=1,a(1)=2},a(n),remember):
map(f, [$0..30]); # Robert Israel, Mar 14 2018

Table[FullSimplify[-(2*n + 1)! * Hypergeometric2F1[1, n + 3/2, n + 2, 2]/ ((n + 1)*2^n) - I*n!^2], {n, 0, 20}] (* Vaclav Kotesovec, Jan 02 2020 *)
Table[FullSimplify[-I*Gamma[1 + n]^2 + I*2^(-1 - 2 n) * Beta[2, 1 + n, 1/2] * Gamma[2 + 2 n]], {n, 0, 20}] (* Vaclav Kotesovec, Jan 02 2020 *)

a(n+2) = (3*n+5)*(n+2)*a(n+1) + (2*n+3)*(n+1)^2*a(n). - Robert Israel, Mar 14 2018
Empirical observation: a(n-1) = (2*n - 1)! / 4^(n-1) * Integral_{t=0..Pi/4} sec(t)^(2*n). - Stephen J. Herschkorn, Dec 14 2019
a(n) ~ sqrt(Pi) * 2^(n+2) * n^(2*n + 1/2) / exp(2*n). - Vaclav Kotesovec, Jan 02 2020
a(n) = (1/2^n)*Sum_{i=0..n} binomial(n,i) * (2*(n-i))! * (2*i)!. - Håvar Andre Melheim Salbu, May 22 2022

Edited by, and more terms from Robert Israel, Mar 14 2018

A103511 Smallest prime > double factorial number A001147(n).

Original entry on oeis.org

2, 5, 17, 107, 947, 10399, 135151, 2027033, 34459429, 654729139, 13749310577, 316234143227, 7905853580633, 213458046676877, 6190283353629379, 191898783962510743, 6332659870762850657, 221643095476699771957, 8200794532637891559553, 319830986772877770815629

Offset: 1

Lei Zhou, Feb 15 2005

nonn

			1!!=1, 2 is the first prime greater than 1;
6!!=10395, 10399 is the first prime greater than 10395.

Cf. A001147.

Table[NextPrime[(2n-1)!!],{n,20}] (* James C. McMahon, Jan 21 2024 *)

a(19)-a(20) from James C. McMahon, Jan 21 2024

cp's OEIS Frontend

A249348 a(n) = (A001147(n+1)^2-1)/8, where A001147(n+1) = 3*5*...*(2n+1).

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A166750 a(n) = (A001147(n))^3 = 2^(3*n)*GAMMA(n+1/2)^3/Pi^(3/2).

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A248652 Union of the factorial numbers (A000142) and the double factorials of odd numbers (A001147).

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Mathematica

Extensions

A249349 (A001147(n+1)-1)/2, equals the index of A249348(n) within the triangular numbers A000217.

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Programs

PARI

PARI

Formula

A263801 Partial sums of odd double factorials (A001147) with alternating signs.

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Programs

Mathematica

PARI

Formula

A350464 Table read by rows. Interpolating the swinging factorial (A056040) and the double factorial (A001147).

Views

Author

Keywords

Examples

Crossrefs

Programs

Mathematica

Formula

A358987 Omit the trailing 5 from double factorial of odd numbers (A001147(n)).

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Author

Keywords

Comments

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Crossrefs

Programs

Mathematica

Formula

A000779 a(n) = 2*(2n-1)!!-(n-1)!*2^(n-1), where (2n-1)!! is A001147(n).

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Magma

Maple

A249348 a(n) = (A001147(n+1)^2-1)/8, where A001147(n+1) = 35...*(2n+1).

A166750 a(n) = (A001147(n))^3 = 2^(3n)GAMMA(n+1/2)^3/Pi^(3/2).

A000779 a(n) = 2(2n-1)!!-(n-1)!2^(n-1), where (2n-1)!! is A001147(n).

A034405 Let f(x) = (Pi - 2arctan(1/(sqrt(x)sqrt(x+2))))/(2sqrt(x)sqrt(x+2)), take (-1)^n*(n-th derivative from right at x=0) and multiply by A001147(n+1).