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# (a(n),b(n))-Pascal triangle

The (a(n),b(n))-Pascal triangle, or (an,bn)-Pascal triangle, is a generalization of the (a,b)-Pascal triangle where ana(n) and bnb(n) are integer sequences, keeping the original Pascal triangle recurrence rule unchanged for the interior cells of the triangle.

 n = 0 b0 1 a1 b1 2 a2 ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {1-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {1-i}{0}}b_{i}\,}$ b2 3 a3 ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {2-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {2-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {2-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {2-i}{0}}b_{i}\,}$ b3 4 a4 ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {3-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {3-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {3-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {3-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {3-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {3-i}{0}}b_{i}\,}$ b4 5 a5 ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {4-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {4-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {4-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {4-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {4-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {4-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {4-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {4-i}{0}}b_{i}\,}$ b5 6 a6 ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {5-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {5-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {5-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {5-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {5-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {5-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {5-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {5-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {5-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {5-i}{0}}b_{i}\,}$ b6 7 a7 ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {6-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {6-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {6-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {6-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {6-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {6-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {6-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {6-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {6-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {6-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {6-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {6-i}{0}}b_{i}\,}$ b7 8 a8 ${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {7-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {7-i}{6}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {7-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {7-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {7-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {7-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {7-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {7-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {7-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {7-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {7-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {7-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {7-i}{6}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {7-i}{0}}b_{i}\,}$ b8 9 a9 ${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {8-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {8-i}{7}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {8-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {8-i}{6}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {8-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {8-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {8-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {8-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {8-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {8-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {8-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {8-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {8-i}{6}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {8-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {8-i}{7}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {8-i}{0}}b_{i}\,}$ b9 10 a10 ${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {9-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {9-i}{8}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {9-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {9-i}{7}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {9-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {9-i}{6}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {9-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {9-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {9-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {9-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {9-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {9-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {9-i}{6}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {9-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {9-i}{7}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {9-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {9-i}{8}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {9-i}{0}}b_{i}\,}$ b10 11 a11 ${\displaystyle \scriptstyle \sum _{i=1}^{10}{\binom {10-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {10-i}{9}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {10-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {10-i}{8}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {10-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {10-i}{7}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {10-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {10-i}{6}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {10-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {10-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {10-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {10-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {10-i}{6}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {10-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {10-i}{7}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {10-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {10-i}{8}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {10-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {10-i}{9}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{10}{\binom {10-i}{0}}b_{i}\,}$ b11 12 a12 ${\displaystyle \scriptstyle \sum _{i=1}^{11}{\binom {11-i}{0}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {11-i}{10}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{10}{\binom {11-i}{1}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {11-i}{9}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {11-i}{2}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {11-i}{8}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {11-i}{3}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {11-i}{7}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {11-i}{4}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {11-i}{6}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {11-i}{5}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{6}{\binom {11-i}{5}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{5}{\binom {11-i}{6}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{7}{\binom {11-i}{4}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{4}{\binom {11-i}{7}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{8}{\binom {11-i}{3}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{3}{\binom {11-i}{8}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{9}{\binom {11-i}{2}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{2}{\binom {11-i}{9}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{10}{\binom {11-i}{1}}b_{i}\,}$ ${\displaystyle \scriptstyle \sum _{i=1}^{1}{\binom {11-i}{10}}a_{i}+\,}$${\displaystyle \scriptstyle \sum _{i=1}^{11}{\binom {11-i}{0}}b_{i}\,}$ b12 j = 0 1 2 3 4 5 6 7 8 9 10 11 12

## Recursion rule

The (an,bn)-Pascal triangle recursion rule is:

${\displaystyle T_{(a_{n},b_{n})}(n,n)=b_{n},\ n\geq 0,\,}$
${\displaystyle T_{(a_{n},b_{n})}(n,0)=a_{n},\ n\geq 1,\,}$
${\displaystyle T_{(a_{n},b_{n})}(n,j)=T_{(a_{n},b_{n})}(n-1,j-1)+T_{(a_{n},b_{n})}(n-1,j),\ 0

## Formulae

${\displaystyle T_{(a_{n},b_{n})}(n,n)=b_{n},\ n\geq 0,\,}$
${\displaystyle T_{(a_{n},b_{n})}(n,0)=a_{n},\ n\geq 1,\,}$
${\displaystyle T_{(a_{n},b_{n})}(n,j)=\sum _{i=1}^{n-j}{\binom {n-1-i}{j-1}}a_{i}+\sum _{i=1}^{j}{\binom {n-1-i}{n-1-j}}b_{i},\ 0

## Multiplicative (p(2n),p(2n+1))-Pascal triangle giving a Gödel encoding of coefficients

 n = 0 ${\displaystyle 2\,}$ 1 ${\displaystyle 3\,}$ ${\displaystyle 5\,}$ 2 ${\displaystyle 7\,}$ ${\displaystyle \scriptstyle 3\cdot 5\,}$ ${\displaystyle 11\,}$ 3 ${\displaystyle 13\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 11\,}$ ${\displaystyle 17\,}$ 4 ${\displaystyle 19\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\,}$ ${\displaystyle \scriptstyle 3^{2}\cdot 5^{2}\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 11\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 11\cdot \,}$ ${\displaystyle \scriptstyle 17\,}$ ${\displaystyle 23\,}$ 5 ${\displaystyle 29\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\,}$ ${\displaystyle \scriptstyle 3^{3}\cdot 5^{3}\cdot 7^{2}\cdot \,}$ ${\displaystyle \scriptstyle 11\cdot 13\,}$ ${\displaystyle \scriptstyle 3^{3}\cdot 5^{3}\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 11^{2}\cdot 17\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 11\cdot \,}$ ${\displaystyle \scriptstyle 17\cdot 23\,}$ ${\displaystyle 31\,}$ 6 ${\displaystyle 37\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 41\,}$ 7 ${\displaystyle 43\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 47\,}$ 8 ${\displaystyle 53\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\cdot 43\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 59\,}$ 9 ${\displaystyle 61\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\cdot 43\cdot 53\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 67\,}$ 10 ${\displaystyle 71\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\cdot 43\cdot 53\cdot \,}$ ${\displaystyle \scriptstyle 61\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 73\,}$ 11 ${\displaystyle 79\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\cdot 43\cdot 53\cdot \,}$ ${\displaystyle \scriptstyle 61\cdot 71\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 83\,}$ 12 ${\displaystyle 89\,}$ ${\displaystyle \scriptstyle 3\cdot 5\cdot 7\cdot \,}$ ${\displaystyle \scriptstyle 13\cdot 19\cdot 29\cdot \,}$ ${\displaystyle \scriptstyle 37\cdot 43\cdot 53\cdot \,}$ ${\displaystyle \scriptstyle 61\cdot 71\cdot 79\,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle \scriptstyle \,}$ ${\displaystyle 97\,}$ j = 0 1 2 3 4 5 6 7 8 9 10 11 12

The multiplicative (p2n,p2n+1)-Pascal triangle gives a Gödel encoding of the (an,bn)-Pascal triangle, i.e.:

${\displaystyle M_{(p_{2n},p_{2n+1})}(n,j)=\prod _{i=1}^{2n+1}{p_{i}}^{\alpha _{i}}\,}$

where the exponents ${\displaystyle \scriptstyle \alpha _{i}\,}$ are the coefficients in

${\displaystyle T_{(a_{n},b_{n})}(n,j)=\sum _{i=1}^{2n+1}{\alpha _{i}}{c_{i}},\,}$

where

${\displaystyle c_{2k+1}=b_{k},\ k\geq 0,\,}$
${\displaystyle c_{2k}=a_{k},\ k\geq 1.\,}$

For example:

${\displaystyle M_{(p_{2n},p_{2n+1})}(5,2)=3^{3}\cdot 5^{3}\cdot 7^{2}\cdot 11\cdot 13\,}$

corresponds to

${\displaystyle T_{(a_{n},b_{n})}(5,2)=3a_{1}+3b_{1}+2a_{2}+b_{2}+a_{3}\,}$

### Multiplicative recursion rule

The multiplicative (p2n,p2n+1)-Pascal triangle recursion rule is:

${\displaystyle M_{(p_{2n},p_{2n+1})}(n,n)=p_{2n+1},\ n\geq 0,\,}$
${\displaystyle M_{(p_{2n},p_{2n+1})}(n,0)=p_{2n},\ n\geq 1,\,}$
${\displaystyle M_{(p_{2n},p_{2n+1})}(n,j)=M_{(p_{2n},p_{2n+1})}(n-1,j-1)\cdot M_{(p_{2n},p_{2n+1})}(n-1,j),\ 0

where ${\displaystyle \scriptstyle p_{i},\ i\geq 1\,}$ is the ith prime.