A052129 a(0) = 1; thereafter a(n) = n*a(n-1)^2.

1, 1, 2, 12, 576, 1658880, 16511297126400, 1908360529573854283038720000, 29134719286683212541013468732221146917416153907200000000

Offset: 0

Author

Reinhard Zumkeller, Feb 12 2002

Keywords

• nonn
• nice

Comments

Somos's quadratic recurrence sequence.

Iff n is prime (n>2), the n-adic valuation of a(2n) is 3*A001045(n) (three times the values at the prime indices of Jacobsthal numbers), which is 2^n+1. For example: the 11-adic valuation at a(22) = 2049 = 3*A001045(11)= 683. 3*683 = 2^11+1 = 2049. True because: When n is prime, n-adic valuation is 1 at A052129(n), then doubles as n-increases to 2n, at which point 1 is added; thus A052129(2n) = 2^n+1. Since 3*A001045(n) = 2^n+1, n-adic valuation of A052129(2n) = 3*A001045(n) when n is prime. - Bob Selcoe, Mar 06 2014

Unreduced denominators of: f(1) = 1, f(n) = f(n-1) + f(n-1)/(n-1). - Daniel Suteu, Jul 29 2016

Examples

			a(3) = 3*a(2)^2 = 3*(2*a(1)^2)^2 = 3*(2*(1*a(0)^2)^2)^2 = 3*(2*(1*1^2)^2)^2 = 3*(2*1)^2 = 3*4 = 12.
G.f. = 1 + x + 2*x^2 + 12*x^3 + 576*x^4 + 1658880*x^5 + 16511297126400*x^6 + ...
		

References

• S. R. Finch, Mathematical Constants, Cambridge University Press, Cambridge, 2003, p. 446.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..12
• Sung-Hyuk Cha, On the k-ary Tree Combinatorics.
• Chao-Ping Chen, Sharp inequalities and asymptotic series related to Somos' quadratic recurrence constant, Journal of Number Theory, 172 (2017), 145-159.
• Chao-Ping Chen and X. F. Han, On Somos' quadratic recurrence constant, Journal of Number Theory, Volume 166, September 2016, Pages 31-40.
• Olivier Golinelli, Remote control system of a binary tree of switches - II. balancing for a perfect binary tree, arXiv:2405.16968 [cs.DM], 2024. See p. 16.
• Jesús Guillera and Jonathan Sondow, Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent, arXiv:math/0506319 [math.NT], 2005-2006.
• Jesús Guillera and Jonathan Sondow, Double integrals and infinite products for some classical constants via analytic continuations of Lerch's transcendent, Ramanujan J. 16 (2008), 247-270.
• Dawei Lu and Zexi Song, Some new continued fraction estimates of the Somos' quadratic recurrence constant, Journal of Number Theory, 155 (2015), 36-45.
• Dawei Lu, Xiaoguang Wang, and Ruiqing Xu, Some New Exponential-Function Estimates of the Somos' Quadratic Recurrence Constant, Results in Mathematics 74(1) (2019), Article 6.
• Gergo Nemes, On the coefficients of an asymptotic expansion related to Somos' quadratic recurrence constant, Applicable Analysis and Discrete Mathematics, 5(1) (2011), 60-66.
• Jörg Neunhäuserer, On the universality of Somos' constant, arXiv:2006.02882 [math.DS], 2020.
• Jonathan Sondow and Petros Hadjicostas, The generalized-Euler-constant function gamma(z) and a generalization of Somos's quadratic recurrence constant, arXiv:math/0610499 [math.CA], 2006.
• Jonathan Sondow and Petros Hadjicostas, The generalized-Euler-constant function gamma(z) and a generalization of Somos's quadratic recurrence constant, J. Math. Anal. Appl. 332 (2007), 292-314.
• Xu You and Di-Rong Chen, Improved continued fraction sequence convergent to the Somos' quadratic recurrence constant, Mathematical Analysis and Applications, 436(1) (2016), 513-520.
• Eric Weisstein's World of Mathematics, Somos's Quadratic Recurrence Constant.

Crossrefs

Cf. A000142, A001045, A030450, A088679, A112302, A116603, A123851, A123852, A123853, A123854, A238462 (2-adic valuation).

Programs

Mathematica

Join[{1},RecurrenceTable[{a[1]==1,a[n]==n a[n-1]^2},a,{n,10}]]  (* Harvey P. Dale, Apr 26 2011 *)
a[ n_] := If[ n < 1, Boole[n == 0], Product[ (n - k)^2^k, {k, 0, n - 1}]]; (* Michael Somos, May 24 2013 *)
a[n_] := Product[ k^(2^(n - k)), {k,1,n}] (* Jonathan Sondow, Mar 17 2014 *)
NestList[{#[[1]]+1,#[[1]]*#[[2]]^2}&,{1,1},10][[All,2]] (* Harvey P. Dale, Jul 30 2018 *)

PARI

{a(n) = if( n<1, n==0, prod(k=0, n-1, (n - k)^2^k))}; /* Michael Somos, May 24 2013 */

Formula

a(n) ~ s^(2^n) / (n + 2 - 1/n + 4/n^2 - 21/n^3 + 138/n^4 - 1091/n^5 + ...) where s = 1.661687949633... (see A112302) and A116603. - Michael Somos, Apr 02 2006

a(n) = n * A030450(n - 1) if n>0. - Michael Somos, Oct 22 2006

a(n) = (a(n-1) + a(n-2)^2) * (a(n-1) / a(n-2))^2. - Michael Somos, Mar 20 2012

a(n) = product_{k=1..n} k^(2^(n-k)). - Jonathan Sondow, Mar 17 2014

A088679(n+1)/a(n) = n+1. -Daniel Suteu, Jul 29 2016