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%I #19 Dec 07 2019 08:19:39
%S 1,2,28,284,2849,28500,285008,2850083,28500834,285008350,2850083504,
%T 28500835049,285008350498,2850083504986,28500835049863,
%U 285008350498633,2850083504986335,28500835049863359,285008350498633597,2850083504986335973
%N Number of digits of the (10^n)-th tetranacci number (A000078(10^n)).
%C a(n)/10^n converges to 0.28500835...
%H Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007, <a href="/A097353/b097353.txt">Table of n, a(n) for n = 0..25</a>
%H Tito Piezas III, <a href="http://www.oocities.org/titus_piezas/Fibonacci.htm">On Fibonacci Numbers and Their Friends</a>.
%F a(n) = floor(log_10(r) + (10^n-2)*log_10(x)) + 1 for n >= 1, where x is the positive real root of the tetranacci limit equation x^4 - x^3 - x^2 - x - 1 = 0, x = 1.92756... and r is the positive real root of the tetranacci auxiliary equation 563r^4 - 20r^2 - 5r - 1 = 0, r = 0.293813... - Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007
%e Let t(n) = A000078(n). Then we have t(1) = 0, t(10) = 56, t(100) = 2505471397838180985096739296, with respectively 1, 2, 28 and 284 digits.
%p # This Maple code will at least get the first few terms correctly!
%p f:=proc(n) option remember; if n <= 2 then RETURN(0); fi; if n = 3 then RETURN(1); fi; f(n-1) + f(n-2) + f(n-3) +f(n-4); end; for n from 0 to 4 do lprint(f(10^n), length(f(10^n))); od;
%t a = b = c = 0; d = i = 1; Do[e = a + b + c + d; a = b; b = c; c = d; d = e; If[n == 10^i, Print[Length[IntegerDigits[e]]]; i++ ], {n, 4, 10^6}] (* _Ryan Propper_, Jul 22 2005 *)
%o (PARI) \p 100 x=solve(x=1.9274,1.9276,x^4-x^3-x^2-x-1); r=solve(x=0.2937,0.2939,563*x^4-20*x^2-5*x-1); for(k=1,25,n=10^k;print1(floor( (log(r)+(n-2)*log(x))/log(10) )+1",")) \\ Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007
%K nonn,base
%O 0,2
%A _Michael Taktikos_, Sep 17 2004
%E 2 more terms from _Ryan Propper_, Jul 22 2005
%E More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007
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