 Chukchi
Topic Author
Posts: 915
Joined: December 15th, 2001, 3:43 am

### Pascal's Powers

Blaise Pascal (19 June 1623 ? 19 August 1662)https://en.wikipedia.org/wiki/Blaise_PascalIt is known that a sum of all elements in p high rows of the Pascal's Triangle each raised to the (p-1) power is divisible by p for all prime p.A181990 = Sum_{0 <= k <= m < p} (binomial(m, k)^(p - 1))/p, where p is the n-th prime.Prove that for p = 3 and 7 (and their powers like 3,9,27,... and 7,49,...) the sums of all elements in n = p^k high rows of the Pascal's Triangle each raised to the (n-1) = (p^k-1) power are divisible by n^2 = p^(2k) for all k>0. Chukchi
Topic Author
Posts: 915
Joined: December 15th, 2001, 3:43 am

### Pascal's Powers

NB The rows in Pascal's Triangle are counted from n = 0.The first p rows have indices from 0 to (p-1).http://mathworld.wolfram.com/PascalsTriangle.html
Last edited by Chukchi on July 16th, 2014, 10:00 pm, edited 1 time in total. ExSan
Posts: 4552
Joined: April 12th, 2003, 10:40 am

### Pascal's Powers

QuoteOriginally posted by: ChukchiNB The rows in Pascal's Triangle are counted from n = 0.The first p rows have indices from 0 to (p-1).http://mathworld.wolfram.com/PascalsTriangle.html how is this done?The plot above shows the binary representations for the first terms of a flattened Pascal's triangle. ??? ExSan
Posts: 4552
Joined: April 12th, 2003, 10:40 am

### Pascal's Powers

QuoteOriginally posted by: outrunQuoteOriginally posted by: ExSanQuoteOriginally posted by: ChukchiNB The rows in Pascal's Triangle are counted from n = 0.The first p rows have indices from 0 to (p-1).http://mathworld.wolfram.com/PascalsTriangle.html how is this done?The plot above shows the binary representations for the first terms of a flattened Pascal's triangle. ???1 2 10 1 01 0 1_________1 3 3 10 1 1 01 1 1 1_______1 4 6 4 1 0 1 1 1 00 0 1 0 01 0 0 0 1got it, thank you, I will work on the images later. cheers.  