## 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

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

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

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

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

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

math grad - Interest: Number theory
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### 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.

• benvitalis says:

$binomial(binomial(n+2,2), \,2)$
$= \; 1/8 \; (n^2+3 \, n)^2 \, + \, 1/4 \, (n^2+3 \, n)$
$= \; 1/8 \; n (n+1) \, (n+2) \, (n+3)$