## C(n,k-1) + C(n,k+1) = 2*C(n,k)

Determine all pairs (n, k) of integers such that   0 < k < n   and

Solution :

Here any integer value of   $c$   will yield nonnegative integer values of   $a$   and   $b$;
however, the condition   $0 < k < n$   requires that
$a = k + 1$   and   $b = n - k + 1$
are each at least 2.
Hence the values   $c \; = \; -2, \; -1, \; 0, \; 1, \; 2$   are excluded,
while every   $c \; \leq \; -3$   and   $c \; \geq \; 3$
will yield permissible values for

which satisfy the equation.

math grad - Interest: Number theory
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### 3 Responses to C(n,k-1) + C(n,k+1) = 2*C(n,k)

1. paul says:

There are two sets of recurrences for {n, k} both are defined with the same recurrence numbers but on different series, the recurrences are {3, -3, 1} and the generating numbers are:-
{7, 14, 23, 34…} and {2, 5, 9, 14..}. They are set up as follows

{7, 2}, {7, 5}
{14, 5},{14,9}
{23, 9}, {23,14}
{34, 14},{34,20}

The extended series are
n = {7,14,23,34,47,62,79,98,119,142,167,194,223,254,287,322,359,398,439,482}
k = {2,5,9,14,20,27,35,44,54,65,77,90,104,119,135,152,170,189,209,230}

The first few terms are

7 + 35 = 42
35 + 7 = 42
1001 + 3003 = 4004
3003 + 1001 = 4004
490314 + 1144066 = 1634380
1144066 + 490314 = 1634380
927983760 + 1855967520 = 2783951280
1855967520 + 927983760 = 2783951280
6973199770790 + 12551759587422 = 19524959358212
12551759587422 + 6973199770790 = 19524959358212

There are an infinite number.

Paul.