A128196 -id:A128196 - cp's OEIS frontend

A126063 Triangle read by rows: see A128196 for definition.

1, 1, 2, 3, 6, 4, 15, 30, 20, 8, 105, 210, 140, 56, 16, 945, 1890, 1260, 504, 144, 32, 10395, 20790, 13860, 5544, 1584, 352, 64, 135135, 270270, 180180, 72072, 20592, 4576, 832, 128, 2027025, 4054050, 2702700, 1081080, 308880, 68640, 12480, 1920, 256

Offset: 0

N. J. A. Sloane, Feb 28 2007

			Triangle begins:
       1
       1,       2
       3,       6,       4
      15,      30,      20,       8
     105,     210,     140,      56,     16
     945,    1890,    1260,     504,    144,    32
   10395,   20790,   13860,    5544,   1584,   352,    64
  135135,  270270,  180180,   72072,  20592,  4576,   832,  128

Ivan Neretin, Rows n = 0..100, flattened
P. Luschny, Variants of Variations.

First column is A001147, second column is A097801.

The diagonal is A000079, the subdiagonal is A014480.

A126063 := (n,k) -> 2^k*doublefactorial(2*n-1)/ doublefactorial(2*k-1); seq(print(seq(A126063(n,k),k=0..n)),n=0..7); # Peter Luschny, Dec 20 2012

Flatten[Table[2^k (2n - 1)!!/(2k - 1)!!, {n, 0, 8}, {k, 0, n}]] (* Ivan Neretin, May 11 2015 *)

Let H be the diagonal matrix diag(1,2,4,8,...) and
let G be the matrix (n!! defined as A001147(n), -1!! = 1):
(-1)!!/(-1)!!
1!!/(-1)!! 1!!/1!!
3!!/(-1)!! 3!!/1!! 3!!/3!!
5!!/(-1)!! 5!!/1!! 5!!/3!! 5!!/5!!
...
Then T = G*H. [Gottfried Helms]
T(n,k) = 2^k*(2n - 1)!!/(2k - 1)!!. - Ivan Neretin, May 13 2015

A014480 Expansion of g.f. (1+2x)/(1-2x)^2.

Original entry on oeis.org

1, 6, 20, 56, 144, 352, 832, 1920, 4352, 9728, 21504, 47104, 102400, 221184, 475136, 1015808, 2162688, 4587520, 9699328, 20447232, 42991616, 90177536, 188743680, 394264576, 822083584, 1711276032, 3556769792, 7381975040, 15300820992, 31675383808, 65498251264

Offset: 0

N. J. A. Sloane

Number of binary trees of size n and height n-1, computed from size n=3 onward; i.e. A014480(n) = A073345(n+3,n+2). (For sizes n=0 through 2 there are no such trees.)
Also determinant of the n X n matrix M(i,j)=binomial(2i+2j,i+j). - Benoit Cloitre, Mar 27 2004
Subdiagonal in triangle displayed in A128196. - Peter Luschny, Feb 26 2007
From Jaume Oliver Lafont, Nov 08 2009: (Start)
From two BBP-type formulas by Knuth, (page 6 of the reference)
Sum_{n>=0} 1/a(n) = 2^(1/2)*log(1+2^(1/2))
Sum_{n>=0} (-1)^n/a(n) = 2^(1/2)*atan(1/2^(1/2))
(End)
Create a triangle with first column T(n,1)=1+4*n for n=0 1 2... The remaining terms T(r,c)=T(r,c-1)+T(r-1,c-1).  T(n,n+1)=a(n). - J. M. Bergot, Dec 18 2012

			(1 + 2*x)/(1-2*x)^2 = 1 + 6*x + 20*x^2 + 56*x^3 + 144*x^4 + 352*x^5 + 832*x^6 + ...

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
David Bailey, Peter Borwein, and Simon Plouffe, On the rapid computation of various polylogarithmic constants, in: L. Berggren, J. Borwein, and P. Borwein (eds.), Pi: A Source Book, Springer, New York, NY, 2000.
Index entries for linear recurrences with constant coefficients, signature (4,-4).

Cf. A054582, A118417, A128196, A132775, A134233.

Leftmost column of A167580 (shifted).

a014480 n = a014480_list !! n
a014480_list = 1 : 6 : map (* 4)
   (zipWith (-) (tail a014480_list) a014480_list)
-- Reinhard Zumkeller, Jan 22 2012

[2^n*(2*n + 1): n in [0..35]]; // Vincenzo Librandi, Oct 20 2014

a:=n-> sum(2^n*n^binomial(j,n)/2,j=1..n): seq(a(n),n=1..29); # Zerinvary Lajos, Apr 18 2009

CoefficientList[ Series[(1 + 2*x)/(1 - 2*x)^2, {x, 0, 28}], x]
LinearRecurrence[{4, -4}, {1, 6}, 29] (* Robert G. Wilson v, Dec 26 2012 *)
Table[2^n (2*n + 1), {n, 0, 28}] (* Fred Daniel Kline, Oct 20 2014 *)

Vec((1+2*x)/(1-2*x)^2+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

a(n) = (2n+1)*2^n = 4a(n-1)-4a(n-2) = 4*A052951(n-1) = a(n-1)+A052951(n) = a(n-1)*(2+4/(2n-1)) = A054582(n, n). - Henry Bottomley, May 16 2001
E.g.f.: x*cosh(sqrt(2)*x) = x + 6x^3/3! + 20x^5/5! + 56x^7/7! +... - Ralf Stephan, Mar 03 2005
From Reinhard Zumkeller, Apr 27 2006: (Start)
a(n) = A118416(n+1,n+1) = A118413(n+1,n+1);
A001511(a(n)) = A003602(a(n));
A117303(a(n)) = a(n). (End)
Row sums of triangle A132775 - Gary W. Adamson, Aug 29 2007
Row sums of triangle A134233 - Gary W. Adamson, Oct 14 2007
From Johannes W. Meijer, Nov 23 2009: (Start)
a(n) = 3*a(n-1) - 2^(n-1)*(2*n-5) with a(0) = 1.
a(n) = 3*a(n-1) - 2*a(n-2) + 2^n with a(0) = 1 and a(1) = 6.
(End)
G.f.: -G(0) where G(k) =  1 - (2*k+2)/(1 - x/(x - (k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 06 2012
E.g.f.: Q(0), where Q(k)= 1 + 4*x/( 1 - 1/(1 + 2*(k+1)/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 11 2013

A128195 a(n) = (2n + 1)(a(n - 1) + 2^n) for n >= 1, a(0) = 1.

Original entry on oeis.org

1, 9, 65, 511, 4743, 52525, 683657, 10256775, 174369527, 3313030741, 69573667065, 1600194389599, 40004859842375, 1080131215965309, 31323805263469097, 971037963168557815, 32044252784564570583, 1121548847459764557925, 41497307356011298342553, 1618394986884440655806799

Offset: 0

Peter Luschny, Feb 26 2007

P. Luschny, Variants of Variations.

Cf. A007526 (The number of variations), A128196 (A weighted sum of double factorials), A126062.

a := n -> `if`(n=0,1,(2*n+1)*(a(n-1)+2^n));

a[0] = 1; a[n_] := a[n] = (2*n+1)*(a[n-1] + 2^n); Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Jul 29 2013 *)

a(n) = A126062(2, n), double variations.
a(n) = (2n+1)!/(n! 2^n) Sum(k=0..n, 4^k*k!/(2k)!) [Gottfried Helms]
a(n) = 2^n (2n+1) Sum(k=0..n, Gamma(n+1/2)/Gamma(k+1/2))
a(n) = 2^(n+1) Gamma(n+3/2) Sum(k=0..n, 1/Gamma(k+1/2))
a(n) = A128196(n)*A005408(n)
a(n) = A128196(n+1)-A000079(n+1)
Recursive form:
a(n) = 2^(n+1)*v(n+1/2) with v(x) = if x <= 1 then x else x(v(x-1)+1).
a(n) = (2n+1)*(a(n-1)+2^n), a(0) = 1 [Wolfgang Thumser]
Note: The following constants will be used in the next formulas.
K = (1-exp(1)*Gamma(1/2,1))/Gamma(1/2)
M = sqrt(2)(1+exp(1)(Gamma(1/2)-Gamma(1/2,1)))
Generalized form: For x>0
a(x) = 2^(x+1)(x+1/2)(exp(1) Gamma(x+1/2,1) + K Gamma(x+1/2))
Asymptotic formula:
a(n) ~ 2^(n+5/2)*Gamma(n+3/2)
a(n) ~ (exp(1)+K)*2^(n+1)*(n+1/2)!
a(n) ~ M(2n+1)(2exp(-1)(n-1/(24*n+19/10*1/n)))^n

A128198 Array read by antidiagonals. A scheme of arrangements: ArrScheme(k,n) = VarScheme(k,n-1) + k^n; ArrScheme(k,0) = 1. VarScheme(k,n) = (nk+1)(VarScheme(k,n-1) + k^n); VarScheme(k,0) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 5, 1, 1, 4, 13, 16, 1, 1, 5, 25, 73, 65, 1, 1, 6, 41, 202, 527, 326, 1, 1, 7, 61, 433, 2101, 4775, 1957, 1

Offset: 0

Peter Luschny, Mar 02 2007

The second row (k=1) is sequence A000522 counting the arrangements of a set with n elements and the third row (k=2) is the sequence A128196. Cf. the scheme of variations A128195.

			Array begins:
[k=0] 1, 1, 1, 1, 1, 1, 1, 1
[k=1] 1, 2, 5, 16, 65, 326, 1957, 13700
[k=2] 1, 3, 13, 73, 527, 4775, 52589, 683785
[k=3] 1, 4, 25, 202, 2101, 27556, 441625, 8393062
[k=4] 1, 5, 41, 433, 5885, 101069, 2126545, 53180009
[k=5] 1, 6, 61, 796, 13361, 283706, 7391981, 229229536
[k=6] 1, 7, 85, 1321, 26395, 667651, 20743837, 767801905
[k=7] 1, 8, 113, 2038, 47237, 1386680, 50038129, 2152463090

P. Luschny, Variants of Variations.

Cf. A000522, A128195, A128196, A126062.

VarScheme := (k,n) -> `if`(n=0,1,(n*k+1)*(VarScheme(k,n-1)+k^n)); ArrScheme := (k,n) -> `if`(n=0,1, VarScheme(k,n-1)+k^n);

cp's OEIS Frontend

A126063 Triangle read by rows: see A128196 for definition.

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A014480 Expansion of g.f. (1+2*x)/(1-2*x)^2.

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A128195 a(n) = (2*n + 1)*(a(n - 1) + 2^n) for n >= 1, a(0) = 1.

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A128198 Array read by antidiagonals. A scheme of arrangements: ArrScheme(k,n) = VarScheme(k,n-1) + k^n; ArrScheme(k,0) = 1. VarScheme(k,n) = (n*k+1)*(VarScheme(k,n-1) + k^n); VarScheme(k,0) = 1.

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Maple

A014480 Expansion of g.f. (1+2x)/(1-2x)^2.

A128195 a(n) = (2n + 1)(a(n - 1) + 2^n) for n >= 1, a(0) = 1.

A128198 Array read by antidiagonals. A scheme of arrangements: ArrScheme(k,n) = VarScheme(k,n-1) + k^n; ArrScheme(k,0) = 1. VarScheme(k,n) = (nk+1)(VarScheme(k,n-1) + k^n); VarScheme(k,0) = 1.