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a(n+1) is smallest number with a(n+1)^n > a(n)^(n+1).
3

%I #21 Aug 11 2019 13:07:23

%S 1,2,3,5,8,13,20,31,48,74,114,176,271,417,642,988,1521,2341,3603,5545,

%T 8533,13131,20207,31096,47853,73639,113320,174383,268350,412951,

%U 635471,977896,1504837,2315721,3563551,5483776,8438716,12985930,19983416

%N a(n+1) is smallest number with a(n+1)^n > a(n)^(n+1).

%C A kind of discrete exponential, since the series of n-th differences resembles the original sequence. The principle of construction is F(a(n+1)) > G(a(n)) as in A059842 but slightly modified to F(n,a(n+1)) > F(n+1,a(n)) with F(n,x) = x^n. This seems to be a fruitful construction principle!

%H T. D. Noe, <a href="/A059923/b059923.txt">Table of n, a(n) for n = 1..300</a>

%F a(n+1) = floor( a(n)^(1+1/n) ) + 1; compare A080870. - _Paul D. Hanna_, Feb 21 2003

%F a(n) = floor(x^n), where x=1.53885131519173... - _Paul D. Hanna_, Feb 21 2003

%e We have a(5)=8 and therefore a(6) = 13 because 13^5 > 8^6.

%p a := proc(n) option remember: if n=1 then RETURN(1) fi: if n=2 then RETURN(2) fi: ceil(a(n-1)^((n)/(n-1))): end: Digits := 20: for n from 1 to 250 do printf(`%d,`,a(n)) od:

%t a[n_] := a[n] = Floor[a[n-1]^(1+1/(n-1))]+1; a[1] = 1; Table[a[n], {n, 1, 50}] (* _Jean-François Alcover_, Feb 03 2014, after _Paul D. Hanna_ *)

%t nxt[{n_,a_}]:={n+1,Floor[a^(1+1/n)]+1}; NestList[nxt,{1,1},40][[All,2]] (* _Harvey P. Dale_, Aug 11 2019 *)

%o (PARI) { default(realprecision, 200); a=1; for (n=1, 300, write("b059923.txt", n, " ", a); a=floor(a^(1 + 1/n)) + 1; ) } \\ _Harry J. Smith_, Jun 30 2009

%Y Cf. A059842, A080869, A080870.

%K easy,nice,nonn

%O 1,2

%A _Rainer Rosenthal_, Mar 03 2001

%E More terms from _James A. Sellers_, Mar 15 2001