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

Combinations 1

 
 
Solution :

 
Combinations 2

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

Combinations 3

which satisfy the equation.

 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged , . Bookmark the permalink.

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.

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