Tritriangular Number: T_{t_n}

 
 
Tritriangular Number
http://mathworld.wolfram.com/TritriangularNumber.html

 

T_{t_n} \; = \; \dbinom{ \; \dbinom{n+2}{2} \; }{ \; 2 \; } \; = \; 1/8 \; n \,(n+1) \,(n+2) \,(n+3)

 

Determine

T_{t_{n+1}} \; - \; T_{t_n}

 
 
binomial(binomial(n, \, 2), \, 2) \; = \; n(n + 1)(n - 1)(n - 2) \,/ \,8

binomial(binomial(n+2,\, 2),\, 2) \; = \; n (n+1) (n+2) (n+3) \,/ \,8

 
(n+1) (n+2) (n+3) \,/ \,2
   = \; binomial(binomial(n+3,2),2) \; - \; binomial(binomial(n+2,2),2)

n (n+1) (n+2) \,/ \,2
   = \; binomial(binomial(n+2,2),2) \; - \; binomial(binomial(n+1,2),2)

 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

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

2 Responses to Tritriangular Number: T_{t_n}

  1. paul says:

    I might be reading it wrong but a triangular number with a triangular index is different to what you get, I seem to get 1/8 n (n + 1)(n^2 + n + 2)

    Expanding mine gives
    n/4 + (3 n^2)/8 + n^3/4 + n^4/8
    and expanding yours gives
    (3 n)/4 + (11 n^2)/8 + (3 n^3)/4 + n^4/8

    Anyway here’s what I get forT (tn+1)

    1/8 (2 + n + n^2) (4 + n + n^2)

    and one minus the other is

    1/8 (2 + n + n^2) (4 + n + n^2) – 1/8 n (n + 1)(n^2 + n + 2) = 1/2(n^2 + n + 2)

    Answer = 1/2(n^2 + n + 2)

    First 16 Tn’s are
    {1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136}

    for n= {2, 3, 4, 5}
    the T tn+1 is
    {10, 28, 66, 136}
    the T tn is
    {6, 21, 55, 120}
    difference is
    {4, 7, 11, 16}

    Paul.

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