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a(n) = (2*n+4)!!/4!!, related to A000165 (even double factorials).
+20
13
1, 6, 48, 480, 5760, 80640, 1290240, 23224320, 464486400, 10218700800, 245248819200, 6376469299200, 178541140377600, 5356234211328000, 171399494762496000, 5827582821924864000, 209792981589295104000, 7972133300393213952000, 318885332015728558080000
OFFSET
0,2
COMMENTS
Row m=4 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.
LINKS
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
FORMULA
a(n) = (2*n+4)!!/4!!.
E.g.f.: 1/(1-2*x)^3.
a(n) ~ 2^(-1/2)*Pi^(1/2)*n^(5/2)*2^n*e^-n*n^n*{1 + 37/12*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Nov 23 2001
a(n) = (n+2)!*2^(n-1). - Zerinvary Lajos, Sep 23 2006. [corrected by Gary Detlefs, Apr 29 2019]
a(n) = 2^n*A001710(n+2). - R. J. Mathar, Feb 22 2008
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 6)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 6*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 6*x/(1 - 2*x/(1 - 8*x/(1 - 4*x/(1 - 10*x/(1 - 6*x/(1 - ... - (2*n + 4)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 6*x/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 8*sqrt(e) - 12.
Sum_{n>=0} (-1)^n/a(n) = 8/sqrt(e) - 4. (End)
a(n) = A052587(n+2) for n > 0. - M. F. Hasler, Nov 10 2024
MAPLE
a:= proc(n) option remember; `if`(n=0, 1, 2*(n+2)*a(n-1)) end:
seq(a(n), n=0..20); # Alois P. Heinz, Apr 29 2019
seq(2^(n-1)*(n+2)!, n=0..20); # G. C. Greubel, Nov 11 2019
MATHEMATICA
Table[2^(n-1)(n+2)!, {n, 0, 20}] (* Jean-François Alcover, Oct 05 2019 *)
Table[(2n+4)!!/8, {n, 0, 20}] (* Harvey P. Dale, Apr 06 2023 *)
PROG
(PARI) vector(21, n, 2^(n-2)*(n+1)! ) \\ G. C. Greubel, Nov 11 2019
(PARI) apply( {A051578(n)=(n+2)!<<(n-1)}, [0..18]) \\ M. F. Hasler, Nov 10 2024
(Magma) [2^(n-1)*Factorial(n+2): n in [0..20]]; // G. C. Greubel, Nov 11 2019
(Sage) [2^(n-1)*factorial(n+2) for n in (0..20)] # G. C. Greubel, Nov 11 2019
(GAP) List([0..20], n-> 2^(n-1)*Factorial(n+2) ); # G. C. Greubel, Nov 11 2019
CROSSREFS
Cf. A000165, A001147(n+1), A002866(n+1), A051577 (rows m=0..3), A051579, A051580, A051581, A051582, A051583.
Cf. A052587 (essentially the same).
KEYWORD
easy,nonn
STATUS
approved
a(n) = (2*n+6)!!/6!!, related to A000165 (even double factorials).
+20
11
1, 8, 80, 960, 13440, 215040, 3870720, 77414400, 1703116800, 40874803200, 1062744883200, 29756856729600, 892705701888000, 28566582460416000, 971263803654144000, 34965496931549184000, 1328688883398868992000
OFFSET
0,2
COMMENTS
Row m=6 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.
LINKS
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
FORMULA
a(n) = (2*n+6)!!/6!!.
E.g.f.: 1/(1-2*x)^4.
a(n) = n!*2^(n-4)/3, n>=3. - Zerinvary Lajos, Sep 23 2006
G.f.: G(0)/2, where G(k)= 1 + 1/(1 - x/(x + 1/(2*k+8)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 02 2013
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 8)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 8*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 8*x/(1 - 2*x/(1 - 10*x/(1 - 4*x/(1 - 12*x/(1 - 6*x/(1 - ... - (2*n + 6)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 8*x/(1 - 10*x/(1 - 2*x/(1 - 12*x/(1 - 4*x/(1 - 14*x/(1 - 6*x/(1 - ... - (2*n + 8)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 48*sqrt(e) - 78.
Sum_{n>=0} (-1)^n/a(n) = 30 - 48/sqrt(e). (End)
MAPLE
seq( mul(2*j+6, j=1..n), n=0..20); # G. C. Greubel, Nov 11 2019
MATHEMATICA
Table[2^n*Pochhammer[4, n], {n, 0, 20}] (* G. C. Greubel, Nov 11 2019 *)
Table[(2n+6)!!/6!!, {n, 0, 20}] (* Harvey P. Dale, Mar 03 2022 *)
PROG
(PARI) vector(20, n, prod(j=1, n-1, 2*j+6) ) \\ G. C. Greubel, Nov 11 2019
(Magma) [1] cat [(&*[2*j+6: j in [1..n]]): n in [1..20]]; // G. C. Greubel, Nov 11 2019
(Sage) [product( (2*j+6) for j in (1..n)) for n in (0..20)] # G. C. Greubel, Nov 11 2019
(GAP) List([0..20], n-> Product([1..n], j-> 2*j+6) ); # G. C. Greubel, Nov 11 2019
CROSSREFS
Cf. A000165, A001147(n+1), A002866(n+1), A051577, A051578, A051579 (rows m=0..5), A051581, A051582, A051583.
KEYWORD
easy,nonn
STATUS
approved
a(n) = (2*n+8)!!/8!!, related to A000165 (even double factorials).
+20
10
1, 10, 120, 1680, 26880, 483840, 9676800, 212889600, 5109350400, 132843110400, 3719607091200, 111588212736000, 3570822807552000, 121407975456768000, 4370687116443648000, 166086110424858624000, 6643444416994344960000
OFFSET
0,2
COMMENTS
Row m=8 of the array A(3; m,n) := (2*n+m)!!/m!!, m >= 0, n >= 0.
LINKS
A. N. Stokes, Continued fraction solutions of the Riccati equation, Bull. Austral. Math. Soc. Vol. 25 (1982), 207-214.
FORMULA
a(n) = (2*n+8)!!/8!!.
E.g.f.: 1/(1-2*x)^5.
a(n) = (n+4)!*2^(n-1)/12. - Zerinvary Lajos, Sep 23 2006
From Peter Bala, May 26 2017: (Start)
a(n+1) = (2*n + 10)*a(n) with a(0) = 1.
O.g.f. satisfies the Riccati differential equation 2*x^2*A(x)' = (1 - 10*x)*A(x) - 1 with A(0) = 1.
G.f. as an S-fraction: A(x) = 1/(1 - 10*x/(1 - 2*x/(1 - 12*x/(1 - 4*x/(1 - 14*x/(1 - 6*x/(1 - ... - (2*n + 8)*x/(1 - 2*n*x/(1 - ...))))))))) (by Stokes 1982).
Reciprocal as an S-fraction: 1/A(x) = 1/(1 + 10*x/(1 - 12*x/(1 - 2*x/(1 - 14*x/(1 - 4*x/(1 - 16*x/(1 - 6*x/(1 - ... - (2*n + 10)*x/(1 - 2*n*x/(1 - ...)))))))))). (End)
From Amiram Eldar, Dec 11 2022: (Start)
Sum_{n>=0} 1/a(n) = 384*sqrt(e) - 632.
Sum_{n>=0} (-1)^n/a(n) = 384/sqrt(e) - 232. (End)
MAPLE
seq(2^n*pochhammer(5, n), n=0..20); # G. C. Greubel, Nov 12 2019
MATHEMATICA
(2Range[0, 20]+8)!!/8!! (* Harvey P. Dale, Feb 03 2013 *)
Table[2^n*Pochhammer[5, n], {n, 0, 20}] (* G. C. Greubel, Nov 12 2019 *)
PROG
(PARI) vector(20, n, n--; (n+4)!*2^(n-1)/12) \\ Michel Marcus, Feb 09 2015
(Magma) F:=Factorial; [2^n*F(n+4)/F(4): n in [0..20]]; // G. C. Greubel, Nov 12 2019
(Sage) f=factorial; [2^n*f(n+4)/f(4) for n in (0..20)] # G. C. Greubel, Nov 12 2019
(GAP) F:=Factorial;; List([0..20], n-> 2^n*F(n+4)/F(4) ); # G. C. Greubel, Nov 12 2019
CROSSREFS
Cf. A000165, A001147(n+1), A002866(n+1).
Cf. A051577, A051578, A051579, A051580, A051581 (rows m=0..7), A051583.
KEYWORD
easy,nonn
STATUS
approved
Decimal expansion of the number whose Pierce expansion has the sequence of double factorial numbers (A000165) as coefficients.
+20
5
3, 5, 2, 8, 0, 6, 4, 3, 8, 1, 0, 6, 6, 5, 0, 0, 3, 6, 4, 6, 2, 1, 2, 3, 6, 0, 5, 3, 1, 0, 7, 3, 0, 0, 8, 6, 3, 1, 1, 1, 4, 5, 9, 6, 9, 4, 4, 4, 9, 9, 0, 1, 7, 4, 0, 2, 7, 4, 9, 4, 6, 3, 1, 0, 7, 1, 8, 6, 4, 7, 0, 1, 5, 3, 3, 6, 5, 6, 5, 4, 4, 1, 4, 5, 6, 9, 0, 9, 1, 8, 9, 6, 0, 9, 4, 8, 3, 3, 9
OFFSET
0,1
LINKS
Eric W. Weisstein, Pierce Expansion.
Eric W. Weisstein, Engel Expansion.
MAPLE
P:=proc(n) local a, i, j, k, w; a:=0; w:=1; for i from 0 by 1 to n do k:=i; j:=i-2; while j>0 do k:=k*j; j:=j-2; od; if (i=0 or i=1) then k:=1; fi; if i=2 then k:=2; fi; w:=w*k; a:=a+(-1)^i/w; print(evalf(a, 100)); od; end: P(100);
MATHEMATICA
RealDigits[N[(Sum[(-1)^n*Product[1/((k + 1)!!), {k, 1, n}], {n, 1, 250}]), 100]][[1]] (* G. C. Greubel, Jan 01 2017 *)
CROSSREFS
KEYWORD
easy,nonn,cons,changed
AUTHOR
STATUS
approved
Decimal expansion of the inverse of the number whose Engel expansion has the sequence of double factorial numbers (A000165) as coefficients.
+20
5
3, 7, 1, 8, 9, 6, 7, 8, 6, 2, 4, 4, 2, 5, 5, 8, 4, 7, 8, 3, 9, 5, 5, 1, 5, 3, 1, 1, 0, 6, 8, 3, 4, 0, 0, 3, 3, 4, 4, 1, 4, 2, 1, 6, 5, 0, 6, 7, 9, 1, 3, 0, 0, 2, 2, 8, 1, 1, 2, 5, 3, 9, 1, 1, 3, 8, 9, 3, 4, 8, 3, 0, 4, 4, 4, 1, 7, 6, 7, 7, 6, 4, 3, 0, 9, 3, 0, 2, 6, 3, 3, 1, 0, 7, 2, 5, 3, 6, 5
OFFSET
0,1
LINKS
Eric W. Weisstein, Pierce Expansion.
Eric W. Weisstein, Engel Expansion.
MAPLE
P:=proc(n) local a, i, j, k, w; a:=0; w:=1; for i from 0 by 1 to n do k:=i; j:=i-2; while j>0 do k:=k*j; j:=j-2; od; if (i=0 or i=1) then k:=1; fi; if i=2 then k:=2; fi; w:=w*k; a:=a+1/w; print(evalf(1/a, 100)); od; end: P(100);
MATHEMATICA
RealDigits[N[(1/Sum[Product[1/((k - 1)!!), {k, 1, n}], {n, 1, 250}]), 100]][[1]] (* G. C. Greubel, Jan 01 2016 *)
CROSSREFS
KEYWORD
easy,nonn,cons,changed
AUTHOR
STATUS
approved
A coefficient tree from the list partition transform relating A000129, A000142, A000165, A110327, and A110330.
+20
3
1, 2, 6, 2, 24, 24, 120, 240, 24, 720, 2400, 720, 5040, 25200, 15120, 720, 40320, 282240, 282240, 40320, 362880, 3386880, 5080320, 1451520, 40320, 3628800, 43545600, 91445760, 43545600, 3628800, 39916800, 598752000, 1676505600, 1197504000, 199584000, 3628800
OFFSET
0,2
COMMENTS
Construct the infinite array of polynomials
a(0,t) = 1
a(1,t) = 2
a(2,t) = 6 + 2 t
a(3,t) = 24 + 24 t
a(4,t) = 120 + 240 t + 24 t^2
a(5,t) = 720 + 2400 t + 720 t^2
a(6,t) = 5040 + 25200 t + 15120 t^2 + 720 t^3
This array is the reciprocal array of the following array b(n,t) under the list partition transform and its associated operations described in A133314.
b(0,t) = 1, b(1,t) = -2, b(2,t) = -2*(t-1), b(n,t) = 0 for n>2.
Then A000165(n) = a(n,1).
Lower triangular matrix A110327 = binomial(n,k)*a(n-k,2).
n! * A000129(n+1) = a(n,2) = A110327(n,0).
A110330 = matrix inverse of binomial(n,k)*a(n-k,2) = binomial(n,k)*b(n-k,2).
A000142(n+1) = a(n,0).
From Peter Bala, Sep 09 2013: (Start)
Let {P(n,x)}n>=0 be a polynomial sequence. Koutras has defined generalized Eulerian numbers associated with the sequence P(n,x) as the coefficients A(n,k) in the expansion of P(n,x) in a series of factorials of degree n, namely P(n,x) = Sum_{k=0..n} A(n,k)* binomial(x+n-k,n). The choice P(n,x) = x^n produces the classical Eulerian numbers of A008292. Let now P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial polynomial. Then the present table is the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x). See A228955 for the generalized Eulerian numbers associated with the polynomial sequence P(n,2*x + 1). (End)
LINKS
M. V. Koutras, Eulerian numbers associated with sequences of polynomials, The Fibonacci Quarterly, 32 (1994), 44-57.
FORMULA
E.g.f. for the polynomials b(.,t), introduced above, is 1 - 2x - (t-1) * x^2; therefore e.g.f. for the polynomials a(.,t), which are the row polynomials of this array, is 1 / ( 1 - 2x - (t-1) * x^2 ) = (t-1) / ( t - ( 1 + x*(t-1) )^2 ).
Also, a(n,t) = (1 - t*u^2)^(n+1) (D_u)^n [ 1 / (1 - t*u^2) ] with eval. at u = 1/t. Compare A076743.
a(n,t) = n!*Sum_{k>=0} binomial(n+1,2k+1) * t^k = n!*Sum_{k>=0} A034867(n,k) * t^k.
Additional relations are given by formulas in A133314.
From Peter Bala, Sep 09 2013: (Start)
Recurrence equation: T(n+1,k) = (n+2 +2*k)T(n,k) + (n +2 -2*k)T(n,k-1).
Let P(n,x) = x*(x + 1)*...*(x + n - 1) denote the n-th rising factorial.
T(n,k) = Sum_{j=0..k+1} (-1)^(k+1-j)*binomial(n+1,k+1-j)*P(n,2*j) for n >= 1.
The row polynomial a(n,t) satisfies t*a(n,t)/(1 - t)^(n+1) = Sum_{j>=1} P(n,2*j)*t^j. For example, for n = 3 we have t*(24 + 24*t)/(1 - t)^4 = 2*3*4*t + (4*5*6)*t^2 + (6*7*8)*t^3 + ..., while for n = 4 we have t*(120 + 240*t + 24*t^2)/(1 - t)^5 = (2*3*4*5)*t + (4*5*6*7)*t^2 + (6*7*8*9)*t^3 + .... (End)
EXAMPLE
Triangle begins as:
1;
2;
6, 2;
24, 24;
120, 240, 24;
720, 2400, 720;
5040, 25200, 15120, 720;
40320, 282240, 282240, 40320;
362880, 3386880, 5080320, 1451520, 40320;
3628800, 43545600, 91445760, 43545600, 3628800;
MAPLE
for n from 0 to 10 do
seq( n!*binomial(n+1, 2*k+1), k = 0..floor(n/2) )
end do; # Peter Bala, Sep 09 2013
MATHEMATICA
Table[n!*Binomial[n+1, 2*k+1], {n, 0, 10}, {k, 0, Floor[n/2]}]//Flatten (* G. C. Greubel, Dec 30 2019 *)
PROG
(PARI) T(n, k) = n!*binomial(n+1, 2*k+1);
for(n=0, 10, for(k=0, n\2, print1(T(n, k), ", "))) \\ G. C. Greubel, Dec 30 2019
(Magma) [Factorial(n)*Binomial(n+1, 2*k+1): k in [0..Floor(n/2)], n in [0..10]]; // G. C. Greubel, Dec 30 2019
(Sage) [[factorial(n)*binomial(n+1, 2*k+1) for k in (0..floor(n/2))] for n in (0..10)] # G. C. Greubel, Dec 30 2019
(GAP) Flat(List([0..10], n-> List([0..Int(n/2)], k-> Factorial(n)*Binomial(n+1, 2*k+1) ))); # G. C. Greubel, Dec 30 2019
CROSSREFS
Cf. A228955.
KEYWORD
easy,nonn,tabf
AUTHOR
Tom Copeland, Oct 30 2007, Nov 29 2007, Nov 30 2007
EXTENSIONS
Removed erroneous and duplicate statements. - Tom Copeland, Dec 03 2013
STATUS
approved
Triangle by rows, generated from the odd integers and related to A000165.
+20
3
1, 1, 1, 4, 3, 1, 24, 18, 5, 1, 192, 144, 40, 7, 1, 1920, 1440, 400, 70, 9, 1, 23040, 17280, 4800, 840, 108, 11, 1, 322560, 241920, 67200, 11760, 1512, 154, 13, 1, 5160960, 3870720, 1075200, 188160, 24192, 2464, 208, 15, 1
OFFSET
0,4
COMMENTS
Row sums = A000165, the double factorial numbers: (1, 2, 8, 48, 384,...).
Left border = A002866 and the eigensequence of the odd integers prefaced with a 1.
LINKS
FORMULA
Eigentriangle of triangle A158405 (odd integers in every row: (1, 3, 5,...)); the inverse of:
1;
-1, 1;
-1, -3, 1;
-1, -3, -5, 1;
-1, -3, -5, -7, 1;
...
EXAMPLE
First few rows of the triangle:
1;
1, 1;
4, 3, 1;
24, 18, 5, 1;
192, 144, 40, 7, 1;
1920, 1440, 400, 70, 9, 1;
23040, 17280, 4800, 840, 108, 11, 1;
322560, 241920, 67200, 11760, 1512, 154, 13, 1;
...
MAPLE
T:= proc(n) option remember; local M;
M:= (Matrix(n+1, (i, j)-> `if`(i=j, 1, `if`(i>j, -2*j+1, 0)))^(-1));
seq(M[n+1, k], k=1..n+1)
end:
seq(T(n), n=0..10); # Alois P. Heinz, Feb 27 2012
MATHEMATICA
T[n_] := T[n] = Module[{M}, M = Table[If[i == j, 1, If[i>j, -2*j+1, 0]], {i, 1, n+1 }, {j, 1, n+1}] // Inverse; M[[n+1]]]; Table[T[n], {n, 0, 10}] // Flatten (* Jean-François Alcover, Mar 09 2015, after Alois P. Heinz *)
CROSSREFS
KEYWORD
nonn,tabl,changed
AUTHOR
Gary W. Adamson, Feb 22 2012
EXTENSIONS
Typo in term 17 corrected by Alois P. Heinz, Dec 06 2012
STATUS
approved
Hankel transform of double factorial numbers n!*2^n=A000165(n).
+20
2
1, 4, 256, 589824, 86973087744, 1282470362637926400, 2723154477021188283432960000, 1133321924829207204666583887642624000000, 120746421332702772771144114237731253721340313600000000
OFFSET
0,2
COMMENTS
By the properties of the Hankel transform, a(n)=2^(n(n+1))*A055209(n).
Also Hankel transform of A000354, A010844, A082032. - Philippe Deléham, Jan 23 2008
LINKS
Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.
FORMULA
a(n) = Product_{k=1..n} (2k)^(2(n-k+1)).
a(n) ~ 2^((n+1)^2) * Pi^(n+1) * n^(n^2 + 2*n + 5/6) / (A^2 * exp(3*n^2/2 + 2*n - 1/6)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Feb 24 2019
MATHEMATICA
Table[Product[(2k)^(2(n-k+1)), {k, n}], {n, 0, 10}] (* Harvey P. Dale, Apr 11 2013 *)
PROG
(PARI) for(n=0, 10, print1(prod(k=1, n, (2*k)^(2*(n-k+1))), ", ")) \\ G. C. Greubel, Oct 14 2018
(Magma) [1] cat [(&*[(2*k)^(2*(n-k+1)): k in [1..n]]): n in [1..10]]; // G. C. Greubel, Oct 14 2018
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Jan 23 2008
STATUS
approved
Numbers n such that n!*2^n + n + 1 = A000165(n) + n + 1 is prime.
+20
1
0, 2, 4, 12, 18, 36
OFFSET
1,2
COMMENTS
For comparison, A000165(n)+1 is prime for n = 0, 1, 259, 16708, 18655,... (A256594).
a(6) > 25000, if it exists. - Giovanni Resta, Jun 01 2018
EXAMPLE
n=4, 2^4*4! + 4 + 1 = 389 is prime.
MATHEMATICA
Select[Range[100], PrimeQ[#! 2^# + # + 1] &] (* Giovanni Resta, May 31 2018 *)
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Labos Elemer, Feb 08 2000
EXTENSIONS
a(1)=0 inserted by Sean A. Irvine, Nov 03 2021
STATUS
approved
Prime values of n!*2^n+n+1 = A000165(n)+n+1.
+20
1
2, 11, 389, 1961990553613, 1678343852714360832019, 25563186766285862273530264901662157745369907200000037
OFFSET
1,1
COMMENTS
For a(6), n = 36. There are no additional terms up to n=1000. - Harvey P. Dale, Aug 06 2012
Primes for n = 0, 2, 4, 12, 18, 36, and no others for n < 5001. - Robert G. Wilson v, Aug 07 2012
EXAMPLE
If n=4, 2^4*4!+4+1 = 389 is prime, so 389 is a term.
MATHEMATICA
Select[Table[n! 2^n+n+1, {n, 0, 1000}], PrimeQ] (* Harvey P. Dale, Aug 06 2012 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Labos Elemer, Feb 08 2000
STATUS
approved

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