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If rows of Pascal's triangle (OEIS's A007318) after their content concatenation {1-1, 1-2-1, 1-3-3-1, 1-4-6-4-1, 1-5-10-10-5-1, 1-6-15-20-15-6-1, 1-7-21-35-35-21-7-1 and so on } be considered as palindromes, presented in different bases, then central (largest) numbers (terms) in the "n"-s row of Pascal triangle (see A007318's table presentation of this sequence) increased by 1 (which is OEIS's A051920(n)) could be considered as the base radix for those palindromes, which finally look like {11, 121, 1331, 14641, 15AA51, 16FKF61, 17LZZL71, ...}. Then conversion of such palindromes from their bases to decimal yields series of consecutive, incrementally increased by 1 powers from 0 to 7 (for initial numbers shown below) and further on (those powers are Pascal's triangle row enumerations counting from 0): 3,16,125,4096,248832,113379904,94931877133, ... (that is 3^1, 4^2, 5^3, 8^4, 12^5, 22^6, 37^7, ...).

PS Robert Israel, in his answer below, gave the formula, explaining this result - thanks.

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This is a consequence of the binomial expansion of $11^n=(10+1)^n$. Until you carry, you will have a palindrome in any base if you read it as $11_b^n=(10_b+1)^b$

I don't know where the $12^5, 22^6, 37^7$ come from.

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$15AA51_{11}=(11_{10}+1)^5=12_{10}^5$ follows the pattern. The next term would be $16GJG61_{21}=22^6$. You get forced to a given base (or higher) by making sure the central term doesn't carry. – Ross Millikan Feb 7 '13 at 18:44

I'm confused. If you use the base b = A051920(n), $\sum_{i=0}^n {n \choose i} b^i = (1+b)^n$. So you should get
$3^1,4^2,5^3,8^4,12^5,22^6,37^7, \ldots$. Why do you have those $11$'s?

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