$70 = 2 \times 5 \times 7$
Consider a Fibonacci-like sequence where $a(0) = 0$ and $a(1) = 1$ and the recurrence equation is $a(n) = 2a(n - 1) + a(n - 2)$ then the first twelve members of the sequence are:
$0, 1, 2, 5, 12, 29, 70, 169, 408, 985, 2378, 5741$
which includes today's number, $70$, at the sixth position. These are known as the Pell Numbers, see A000129.
All of these Fibonacci-like sequences can be generalised as
$a(0) = p, a(1) = q, a(n) = s.a(n - 1) + r.a(n - 2)$.
This is known as the Horodam Sequence, see Wolfram Mathworld.
In the case of the Pell Numbers the four constants, $(p, q, r, s)$, are $(0, 1, 1, 2)$
Consider a triangle of numbers constructed so that the first two lines are the same as Pascal's Triangle. All subsequent lines start and end with the number one, with all intervening numbers being the sum of the triangle of three numbers above them, giving the following:
1
1.... 1
1.... 3.... 1
1.... 5.... 5.... 1
1.... 7....13.... 7.... 1
1.... 9....25....25.... 9.... 1
1....11....41....63....41....11.... 1
If each line is summed then we get:
1 = 1
1 + 1 = 2
1 + 3 + 1 = 5
1 + 5 + 5 + 1 = 12
1 + 7 + 13 + 7 + 1 = 29
1 + 9 + 25 + 25 + 9 + 1 = 70
1 + 11 + 41 + 63 + 41 + 11 + 1 = 169
Inspection shows that the results of these sums are the same as the Pell Numbers (search for Patrick Costello in A000129).
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