Triangular numbers which are the sum of 3 pentagonal numbers

 
 
Pentagonal numbers:   n \, (3 \,n - 1)/2
 
 
Let   T_n   represent the n-th triangular number and   P_n   the n-th pentagonal number

For any integer   k \geq  2,   if   n \; = \; k^2,   then the sum

P_{n-k} \; + \; P_n \; + \; P_{n+k}    equals    T_{3 \,n}
 

P_{n-k} \; + \; P_n + \; P_{n+k}

= \; (3 \,(n-k)^2 \; - \; (n-k))/2 \; + \; (3 \,n^2 - n)/2 \; + \; (3 \,(n+k)^2 \; - \; (n+k))/2

= \; 3/2 \; (2 \, k^2 \; + \; 3 \, n^2 \; - \; n)

since   n \; = \; k^2

= \; 3/2 \; (3 \, k^4 \; + \; k^2)

= \; 3/2 \; k^2 \; (3 \, k^2 \; + \; 1)

= \; T_{3 \,n}

 
 
And,

T_{3 \,n + 2} \; = \; 3 \, P_{n+1}

T_{12 \,n + 5} \; = \; 3 \, P_{4 \,n + 2}

T_{6 \,n + 2} \; = \; 3 \, P_{2 \,n + 1}

 

Any others?

 
 
 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

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