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1, 3, 16, 125, 4096, 248832, 113379904, 94931877133, 722204136308736, 9223372036854775808, 1117730665547154976408576, 214633637635011206805784100864, 397495155639882245867698528490622976, 1135797931555041090259334993227408493600768

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a(n) = (C(n, floor(n/2)) + 2)^n for n >= 0.

1, 3, 16, 125, 4096, 248832, 113379904, 94931877133, 2148541983608805523456, 500609261666673206194904576, 223991529323659151853921232282624, 783074711359940122081340283858644854784722204136308736, 9223372036854775808, 1117730665547154976408576, 214633637635011206805784100864, 397495155639882245867698528490622976

G. C. Greubel, <a href="/A193242/b193242.txt">Table of n, a(n) for n = 0..59</a>

Table[(Binomial[n, Floor[n/2]] + 2)^n, {n, 0, 20}] (* G. C. Greubel, Feb 20 2017 *)

(PARI) for(n=0, 20, print1((binomial(n, floor(n/2)) + 2)^n, ", ")) \\ G. C. Greubel, Feb 20 2017

Corrected a(8) onward - G. C. Greubel, Feb 20 2017

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(C(n, [floor(n/2]) ) + 2)^n for n >= 0.

If the terms in each row of a Pascal's triangle (see the tabular presentation of A007318) 1, 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 be are concatenated (if necessary) and considered as palindromes, represented in different bases, then A051920(n) for n>=0 could be considered as the smallest base radix, for which those palindromes are comprised composed of single digits/letters. Those palindromes will look like: 1, 11, 121, 1331, 14641, 15AA51, 16FKF61, 17LZZL71, ... . Conversion of such palindromes from their above mentioned bases to decimal yields this sequence of the consecutive ascending powers. Such powers are enumerations of the rows in a Pascal's triangle, counting from 0, namely: 1, 3, 16, 125, 4096, 248832, 113379904, 94931877133, ... (that is 1^0, 3^1, 4^2, 5^3, 8^4, 12^5, 22^6, 37^7, ...). In general those powers could be described as (A051920(n)+1)^n for n >= 0. Another property of above discussed the palindromes discussed above is that the value of the sum of their digits/letters constitutes sum to 2^n.

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a(n) = (A051920(n) + 1)^n.

a(n) = (A001405(n) + 2)^n.

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