Pascal's triangle curiosity

Question

I noticed the following pattern in the rows of Pascal's triangle:

$$ 1 = 11^0\\ 11 = 11^1\\ 121= 11^2\\ 1331= 11^3\\ 14641=11^4 $$

at this point I thought maybe this pattern would follow indefinitely, but the next row is $15101051=7\times2157293$. The next two rows aren't powers of 11 either, they are $1615201561=43\times 37562827$ and $172135352171=29\times5935701799$. Are the first 5 rows the only rows in the Pascal triangle that are powers of 11? The following row 18285670562881 is a prime (at least according to Linux factor function). How often primes appear in these rows? This might be fun to show or investigate. I am just curious.

I just wrote a Python script to generate a few more rows of the Pascal triangle. Each row was also factored up to row n=19:

from scipy.special import binom
from sympy import factorint

for n in range(5, 20):
  row = ''
  print('n=%d' % n)
  for k in range(n+1):
    binom_nk = int(binom(n, k))
    row = row+str(binom_nk)
  print(row)
  row_number = int(row)
  factors=factorint(row_number)
  print(factors)

n=5 15101051 {7: 1, 2157293: 1}

n=6 1615201561 {43: 1, 37562827: 1}

n=7 172135352171 {29: 1, 5935701799: 1}

n=8 18285670562881 {18285670562881: 1}

n=9 193684126126843691 {5647: 1, 34298587945253: 1}

n=10 1104512021025221012045101 {13: 1, 197: 1, 4649: 1, 92768668286052709: 1}

n=11 1115516533046246233016555111 {11: 1, 101410593913295112092414101: 1}

n=12 1126622049579292479249522066121 {523: 1, 4637: 1, 464557485113006356820471: 1}

n=13 11378286715128717161716128771528678131 {369268429: 1, 180642383: 1, 170574866715037030033: 1}

n=14 11491364100120023003343230032002100136491141 {41: 1, 86399681: 1, 3910184543: 1, 829618322366629399154147: 1}

n=15 11510545513653003500564356435500530031365455105151 {17: 1, 113: 1, 149: 1, 337: 1, 5362678549: 1, 3533441: 1, 2118359: 1, 2972851397279413777: 1}

n=16 116120560182043688008114401287011440800843681820560120161 {7: 1, 13: 1, 10567: 1, 835035230611: 1, 144614294599341081090704192088907914583: 1}

n=17 1171366802380618812376194482431024310194481237661882380680136171 {71: 1, 89: 1, 1583: 1, 11969: 1, 82361: 1, 8401361: 1, 6505783: 1, 2173376383488959250244793817754469: 1}

n=18 118153816306085681856431824437584862043758318241856485683060816153181 {31: 1, 59: 1, 4421: 1, 28979: 1, 248303053: 1, 2030710802068204317879252665132300005649014383107: 1}

n=19 1191719693876116282713250388755829237892378755825038827132116283876969171191 {4506133: 1, 37569563: 1, 70854794496642568411: 1, 99349280033387580503673432191635187608339: 1}

Answer 1

This happens because:

$$11^{n} = (10 + 1)^{n} = 10^{n} + \binom{n}{1}10^{n-1}+\binom{n}{2}10^{n-2} + \dots + \binom{n}{n-1}10^{1} + 10^{0}$$

By binomial expansion. The pattern first breaks down at $n=5$ because it is the first time one of the binomial coefficients is at least $10$. If you used a higher base (say, hexadecimal), the pattern would continue for longer.

You have associated it with Pascal's Triangle because each row of Pascal's Triangle contains the respective binomial coefficients.

Answer 2

I found this Magic 11's

This even works until the 7th row

Answer 3

$11^5 = 161051$ but you can get this by using the 1 from the 10s as a carry figure on your 15101051 and the pattern will still work