1 1 4 0 20 ? what is the sixth number ?

Question

1 1 4 0 20 ? what is the sixth number ?

Answer 1

The cerebration of one's mind is unpredictable.. but I thought d  solution in zis  way! ! ! i followed my mentation!!!

Let

T[1] = 1

T[2] = 1 T[N] =(N-1) +  [T(1)+T(2)]^(2^(N-3))  FOR N= ODD NATURAL NUMBERS

T[N] =  [T(1)+T(2)]^(2^(N-3)) - (N)                 FOR N = EVEN NATURAL NUMBERS

SO T[6[ will b as it is even then go to formula # 2 & apply it that gives an answer of

so 2^8 - 6 = 250 . . . (i'm not takin 3 terms as initials)

and the simplest to think of d q is 1*4 = 4 1-1 = 0 4*5 = 20 4-0 = 4 is d answer!(this is d sol given by other person not me  my answer to zis is 250)

ALTERNATE:

let the numbers be t1 t2 etc

t[n] = 20*(t[n-3]-t[n-2])^t[n-1] for n>4

so initial terms are t[1]=1 t[2]=1 and t[3]=4

now t[4] = 20*0^4=0

t[5] = 20*(-3)^0=20

and hence t[6] = 20*(4^20)

BUT D RECOMMENDED ANSWER IS THE FIRST SOLUTION I GAVE U THAT'S 250 . . .

Answer 2

8 (it's a phone number)