A097353 Number of digits of the (10^n)-th tetranacci number (A000078(10^n)).
1, 2, 28, 284, 2849, 28500, 285008, 2850083, 28500834, 285008350, 2850083504, 28500835049, 285008350498, 2850083504986, 28500835049863, 285008350498633, 2850083504986335, 28500835049863359, 285008350498633597, 2850083504986335973
Offset: 0
Examples
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.
Links
- Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007, Table of n, a(n) for n = 0..25
- Tito Piezas III, On Fibonacci Numbers and Their Friends.
Programs
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Maple
# This Maple code will at least get the first few terms correctly! 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;
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Mathematica
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 *) -
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
Formula
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
Extensions
2 more terms from Ryan Propper, Jul 22 2005
More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), May 01 2007
Comments