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Irregular triangle T(n,k) starting with 2^n followed by p_k^e_k = p_k^floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.
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%I #12 Feb 11 2022 17:31:02

%S 1,2,4,3,8,3,16,9,5,32,27,25,7,64,27,25,7,128,81,25,7,256,243,125,49,

%T 11,512,243,125,49,11,1024,729,625,343,121,13,2048,729,625,343,121,13,

%U 4096,2187,625,343,121,13,8192,6561,3125,2401,1331,169,17

%N Irregular triangle T(n,k) starting with 2^n followed by p_k^e_k = p_k^floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.

%C T(0,1) = 1 by convention.

%C T(n,1) = 2^n. T(n,k) = p_k^e_k such that p_k^T(n,k) is the largest 1 < p_k^e_k < p_(k-1)^e_(k-1).

%H Michael De Vlieger, <a href="/A347288/b347288.txt">Table of n, a(n) for n = 0..10367</a> (rows 0 <= n <= 300, flattened)

%F T(n,1) = 2^n; T(n,k) = p_k^floor(log_p_k(p_(k-1)^T(n,k-1))).

%F A347385(n,k) = p_k^T(n,k).

%F A089576(n) = row lengths.

%F A347284(n) = product of row n.

%e Row 0 contains {1} by convention.

%e Row 1 contains {2} since no nonzero exponent e exists such that 3^e < 2^1.

%e Row 2 contains {4,3} since 3^1 < 2^2 yet 3^2 > 2^2. (We assume hereinafter that the powers listed are the largest possible smaller than the immediately previous term.)

%e Row 3 contains {8,3} since 2^3 > 3^1.

%e Row 4 contains {16,9,5} since 2^4 > 3^2 > 5^1, etc.

%e Triangle begins:

%e 2 3 5 7 11 13 17 ...

%e --------------------------------------------------

%e 0: 1

%e 1: 2

%e 2: 4 3

%e 3: 8 3

%e 4: 16 9 5

%e 5: 32 27 25 7

%e 6: 64 27 25 7

%e 7: 128 81 25 7

%e 8: 256 243 125 49 11

%e 9: 512 243 125 49 11

%e 10: 1024 729 625 343 121 13

%e 11: 2048 729 625 343 121 13

%e 12: 4096 2187 625 343 121 13

%e 13: 8192 6561 3125 2401 1331 169 17

%e 14: 16384 6561 3125 2401 1331 169 17

%e ...

%t {{1}}~Join~Array[Most@ NestWhile[Block[{p = Prime[#2]}, Append[#1, p^Floor@ Log[p, #1[[-1]]]]] & @@ {#, Length@ # + 1} &, {2^#}, #[[-1]] > 1 &] &, 1 (* _Michael De Vlieger_, Aug 28 2021 *)

%Y Cf. A000217, A000961, A089576, A347284.

%K nonn,tabf,easy

%O 0,2

%A _Michael De Vlieger_, Aug 28 2021