A139526 Triangle A061356 read right to left.
1, 1, 2, 1, 6, 9, 1, 12, 48, 64, 1, 20, 150, 500, 625, 1, 30, 360, 2160, 6480, 7776, 1, 42, 735, 6860, 36015, 100842, 117649, 1, 56, 1344, 17920, 143360, 688128, 1835008, 2097152, 1, 72, 2268, 40824, 459270, 3306744, 14880348, 38263752, 43046721, 1, 90, 3600, 84000, 1260000, 12600000, 84000000, 360000000, 900000000, 1000000000
Offset: 2
Examples
(1) times (1) = (1) (1 1) * (1 2) = (1 2) (1 2 1 ) * (1 3 9) = (1 6 9) (1 3 3 1) * (1 4 16 64) = (1 12 48 64) etc.
References
- R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA. Second ed. 1994.
- Peter D. Schumer (2004), Mathematical Journeys, page 168, Proposition 16.1 (c)
Links
- P. Bala, Fractional iteration of a series inversion operator
- Wikipedia, Lambert W function
Crossrefs
Programs
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Maple
A061356 := proc(n,k) binomial(n-2,k-1)*(n-1)^(n-k-1); end: A139526 := proc(n,k) A061356(n,n-k-1) ; end: for n from 2 to 14 do for k from 0 to n-2 do printf("%d,",A139526(n,k)) ; od: od: # R. J. Mathar, May 22 2008
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Mathematica
T[n_, k_] := (n - 1)^k*Binomial[n - 2, n - k - 2]; Table[T[n, k], {n, 2, 11}, {k, 0, n - 2}] // Flatten (* Jean-François Alcover, Jun 13 2023 *)
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PARI
for(n=2,12,forstep(k=n-1,1,-1,print1(binomial(n-2, k-1)*(n-1)^(n-k-1)","))) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 10 2008
Formula
E.g.f. (with offset 1) Sum_{n >= 1} (1 + n*t)^(n-1)*x^n/n! = x + (1 + 2*t)*x^2/2! + (1 + 6*t + 9*t^2)*x^3/3! + .... For properties of this function see Graham et al., equations 5.60, 5.61 and 7.71. The e.g.f. is the series reversion with respect to x of the function log(1 + x)/(1 + x)^t, which is the e.g.f. for a signed version of A028421. - Peter Bala, Jul 18 2013
From Peter Bala, Nov 16 2015: (Start)
E.g.f. with offset 0 and constant term 1: A(x,t) = ( Sum_{n >= 0} (n + 1)^(n-1)*t^n*x^n/n! )^(1/t). This is the generalized exponential series E_t(x) in the terminology of Graham et al., Section 5.4.
A(x,t)^m = 1 + Sum_{n >= 1} m*(m + n*t)^(n-1)*x^n/n!.
log(A(x,t)) = Sum_{n >= 1} (n*t)^(n-1)*x^n/n! = 1/t*T(t*x), where T(z) is Euler's tree function. See A000169.
A(x,t) = ( 1/x* Revert( x*exp(-x*t)) )^(1/t), where Revert is the series reversion operator with respect to x.
Extensions
More terms from R. J. Mathar and Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2008
Comments