# Modification of the Fibonacci Sequences

A chart of Fibonacci sequences gives:

Fibonacci number :$$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...$$

tribonacci number :$$1, 1, 2, 4, 7, 13, 24, 44, 81, ...$$

tetranacci number :$$1, 1, 2, 4, 8, 15, 29, 56, 108, ...$$

pentanacci number :$$1, 1, 2, 4, 8, 16, 31, 61, 120, ...$$

hexanacci number :$$1, 1, 2, 4, 8, 16, 32, 63, 125, ...$$

heptanacci number :$$1, 1, 2, 4, 8, 16, 32, 64, 127, ...$$

However, I am wondering if there is a modification of the fibonacci sequence such that the output is the sum of ALL previous terms. Some of the later sequences, such as the hexanacci and heptanacci seem to follow the desired pattern, but fail at $$63$$ (hexanacci) and $$127$$ (heptanacci). Is there such a sequence that can be written in a summation form where each following term is the summation of all previous terms? I cannot seem to find any formal definition for such a sequence, just the ones listed above.

## 1 Answer

So what you want is that,

$$a_n=\sum_{k=1}^{n-1} a_{k}$$

$$a_1=1$$

$$a_2=1$$

Then

$$a_3=2$$

$$a_4=4$$

$$a_5=8$$

Seems like for $n \geq 2$.

$$a_n=2^{n-2}$$

Proof:

This works for $n=2$ and $2^{k-2}=(2-1)2^{k-2}=2^{k-1}-2^{k-2}$. So that by telescoping,

$$\sum_{k=1}^{n-1} a_{k}$$

$$=1+\sum_{k=2}^{n-1} (2^{k-1}-2^{k-2})$$

$$=1+(2^{n-2}-2^0)$$

$$=2^{n-2}$$

• In general if $a_0=x$ and $a_1=y$ then $a_2=x+y,\; a_3=2(x+y), \;a_4=4(x+y),\;a_5=8(x+y),\;$ etc. Feb 26, 2017 at 5:09