0,2

VarScheme(k,n) = (n*k+1)*(VarScheme(k,n-1) + k^n), VarScheme(k,0) = 1. a(n) is the third row of this scheme, a(n) = VarScheme(2,n).

k | n -> the array A126062:

[0]..1,..1,...1,.....1,......1,.......1,.........1,..........1,............1

[1]..1,..4,..15,....64,....325,....1956,.....13699,.....109600,.......986409

[2]..1,..9,..65,...511,...4743,...52525,....683657,...10256775,....174369527

[3]..1,.16,.175,..2020,..27313,..440896,...8390875,..184647364,...4616348125

[4]..1,.25,.369,..5629,.100045,.2122449,..53163625,.1542220261,..50895431301

[5]..1,.36,.671,.12736,.280581,.7376356,.229151411,.8252263296,.338358810761

The second row counts the variations of n distinct objects A007526.

The second column is sequence A000290. The third column is sequence A005917.

P. Luschny, Variants of 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

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

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

Sequence in context: A036731 A020234 A154996 this_sequence A103459 A100311 A120286

Adjacent sequences: A128192 A128193 A128194 this_sequence A128196 A128197 A128198

easy,nonn

Peter Luschny (peter(AT)luschny.de), Feb 26 2007