Identify the misprint in a sequence

Question

Given the sequence, 1,3,11,31,69,113,223,351,521,739. Identify the misprint.

I tried reasoning out but couldn't. Do I have to do some numerical analysis like forward differences? Have no clue. Kindly help.

Answer 1

If you take the forward differences, you get

$$2,8,20,38,44,110,128,170,218\;.$$

And if you take the second differences, you get

$$6,12,18,6,66,18,42,48\;.$$

It would be nice if we could replace $6,66,18$ by $24,30,36$, as the sequence would then be arithmetic. That would require changing $44$ to $62$ and $110$ to $92$, which in turn would accomplished if we changed $113$ to $131$.

Answer 2

$1, 3, 11, 31, 69, 131, 223, 351, 521, 739, \ldots$ is OEIS sequence A071568. I found it by doing a search for 521,739.

EDIT: In general, suppose you have a finite sequence $a_0, \ldots, a_n$ and you suspect that there is a polynomial $f(x)$ of degree $< n$ such that $f(j) = a_j$ for all but one of $j = 0, \ldots, n$. The polynomial interpolating $f(j) = a_j$ for $j \in S$ is $$ P_S(x) = \sum_{j \in S} a_j \prod_{i \in S \backslash \{j\}} \dfrac{x - i}{j-i}$$ The coefficient of $x^{|S|-1}$ in this polynomial is $$c_S = \sum_{j \in S} \dfrac{a_j}{\prod_{i \in S \backslash \{j\}} (j-i) }$$ which you want to be $0$.

In your example $[a_0, \ldots, a_{9}] = [1,3,11,31,69,113,223,351,521,739]$, the coefficient $c_{\{0\ldots,9\} \backslash j} = 0$ only for $j = 5$, and we have $P_{\{0,1,2,3,4,6,7,8,9\}}(x) = x^3 + x + 1$.

In the other example you mentioned to Brian Scott, $[a_0, \ldots, a_7] = [25,21,18,18,27,45,76,123]$, $c_{\{0,\ldots,7\}} = 0$ only for $j = 3$, and $P_{\{0,1,2,4,5,6,7\}}(x) = x^3/2 - x^2 - 7 x/2 + 25$. This matches the data except for $P(3) = 19$ instead of $18$.

Answer 3

take 3 orders of backward differences to get 1,2,3,1,11,3,7,8 and note that if you change 113 to 131 the differences become 1,2,3...8

Answer 4

Technically speaking a sequence can be anything. The assumption is that there is some simple rule. But if we don't know the rule we have to guess.

As a rule of thumb I start by looking at the differences to see if that gives me a hint. The differences are 2,8,20,38,44,110,128,170,218 the 44 looks out of place. The differences of those are 6,12,18,6,66,18,42,48, I'm seeing sums of 6.

If the differences were 6,12,18,24,30, 36,42,48 things would be ....better. That would make the previous difference be 2,8,20,38,62,92,128,170,218 which would make the first series 1,3,11,31,69,131,223,351,521, 739.

The 131 was mistyped as 113.

===

BTW to get the sequence:

6,12,18,24,30, 36,42,48 => $c_n = 6*n$

2,8,20,38,62,92,128,170,218 => $b_n = 2 + \sum 6i = 2 + 3(n-1)n$

1,3,11,31,69,131,223,351,521, 739 => $a_n = 1 + \sum something_to_do_with_square_powers = something_to_do_with_cube_powers$.

(I can never remember nor derive the damned sum equations and even if I did I just know I'm going to make a careless arithmetic error. It'd be impossible for me not to...)

So $a_n = an^3 + bn^2 + cn + d$. $a_0 = 1$, $a_1 = 3$, $a_2 = 11$ and $a_3=31$ is enough for me to conclude the sequence is $a_n = n^3 + n + 1$.