Ramanujan's Tau Function and Euler's Pentagonal Number Theorem

Ramanujan found that when the infinite series,

x(1 – x)²⁴(1 – x²) ²⁴(1 – x³)²⁴(1 – x⁴)²⁴ …  is multiplied out we obtain

1x – 24x² + 252x³ – 1472x⁴ + 4830x⁵ – 6048x⁶ – 16744x⁷ + 84480x⁸ – … 

This is now known as the Ramanujan tau function.

Now if we divide by x then

(1 – x)²⁴(1 – x²)²⁴(1 – x³)²⁴(1 – x⁴)²⁴ …  

= {(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …}²⁴  

 = 1 – 24x + 252x² – 1472x³ + 4830x⁴ – 6048x⁵ – 16744x⁶ + 84480x⁷ – …  

So the tau values for each term thereby remain unchanged.

Interestingly, Euler earlier discovered that

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …

= x⁰ – x¹ – x²  + x⁵ + x⁷ – x¹² – x¹⁵ + x²² + x²⁶ – …

So the powers of the x terms in this infinite series comprise the generalised pentagonal numbers

i.e. 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, 40, …

This thereby means that we can show a fascinating connection as between these generalised pentagonal numbers and the corresponding Ramanujan tau values.

Thus {x⁰ – x¹ – x²  + x⁵ + x⁷ – x¹² – x¹⁵ + x²² + x²⁶ – …}²⁴

= 1 – 24x + 252x² – 1472x³ + 4830x⁴ – 6048x⁵ – 16744x⁶ + 84480x⁷ – …  

Then a related Euler formula shows that

(1 – x)(1 – x²)(1 – x³)(1 – x⁴) …      =  1 – {x/(1 – x)} + x³/{(1 – x)(1 – x²)}

– x⁶/{(1 – x)(1 – x²)( 1 – x³)} + …

So this latter expression involves (in the sole x term of each numerator) the triangular numbers 1, 3, 6, 10, 15, …

Therefore we can equally show a fascinating connection as between the Ramanujan tau values and the triangular numbers.