A088679 -id:A088679 - cp's OEIS frontend

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

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

Offset: 0

Reinhard Zumkeller, Feb 12 2002

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

			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 + ...

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

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.

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

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 *)

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

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

A347289 Number of independent sets in the binomial tree of order n.

Original entry on oeis.org

2, 3, 8, 60, 3456, 11612160, 132090377011200, 17175244766164688547348480000, 291347192866832125410134687322211469174161539072000000000, 84034354923469245337680441503007090893711465882978424632224243601869256327175152475648504794972160000000000000000

Offset: 0

Kevin Ryde, Sep 26 2021

nonn

The binomial tree of order 0 is a single root vertex and order n>=1 is an order n-1 with another order n-1 joined as a subtree of the root.
Going by induction with this construction shows the number of sets where the root is in the set is with(n) = A052129(n), and the number where the root is not in the set is without(n) = A088679(n+1).
Among the total a(n), the respective proportions are without(n)/a(n) = (n+1)/(n+2) and with(n)/a(n) = 1/(n+2).
Also, a(n-1) is the number of maximum independent sets in binomial tree order n.
Tree n can be constructed from tree n-1 by adding a new child under each vertex.  Each independent set in tree n-1 corresponds one-to-one with a maximum independent set in tree n by putting each new child in or out of the set opposite to its parent.

			For n=5, the product formula is a(5) = 7 * 5 * 4^2 * 3^4 * 2^8 = 11612160.

Kevin Ryde, Table of n, a(n) for n = 0..12

Cf. A052129, A088679.

a(n) = my(P=1); for(k=2,n,P=sqr(P)*k); (n+2)*P;

a(n) = (n+2) * Product_{k=2..n} k^(2^(n-k)).
a(n) = A052129(n) + A088679(n+1).
a(n) = a(n-1)^2 - A052129(n-1)^2.

cp's OEIS Frontend

A180261 a(n) = A088679(n) / 48.

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

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A347289 Number of independent sets in the binomial tree of order n.

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