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A347285
Irregular triangle T(n,k) starting with n followed by e_k = floor(log_p_k(p_(k-1)^e_(k-1))) such that e_k > 0.
6
0, 1, 2, 1, 3, 1, 4, 2, 1, 5, 3, 2, 1, 6, 3, 2, 1, 7, 4, 2, 1, 8, 5, 3, 2, 1, 9, 5, 3, 2, 1, 10, 6, 4, 3, 2, 1, 11, 6, 4, 3, 2, 1, 12, 7, 4, 3, 2, 1, 13, 8, 5, 4, 3, 2, 1, 14, 8, 5, 4, 3, 2, 1, 15, 9, 6, 4, 3, 2, 1, 16, 10, 6, 4, 3, 2, 1, 17, 10, 6, 4, 3, 2, 1
OFFSET
0,3
COMMENTS
Irregular triangle T(n,k) starting with n followed by e_k corresponding to the largest 1 < p_k^e_k < p_(k-1)^e_(k-1).
T(0,1) = 0 by convention; 0 is not allowed for n > 0.
T(n,k) > T(n,k+1). The least first difference among row n is 1.
Conjecture: let S be the sum of the absolute values of the first differences of terms in row n. For all n > 0, n - S = 1. - Michael De Vlieger, Aug 27 2021
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..10367 (rows 0 <= n <= 300, flattened)
FORMULA
T(n,1) = n; T(n,k) = floor(log_p_k(p_(k-1)^T(n,k-1))).
A000217(A089576(n)) <= sum of terms in row n.
EXAMPLE
Row 0 contains {0} by convention.
Row 1 contains {1} since we can find no nonzero exponent e such that 3^e < 2^1.
Row 2 contains {2,1} 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.)
Row 3 contains {3,1} since 2^3 > 3^1.
Row 4 contains {4,2,1} since 2^4 > 3^2 > 5^1, etc.
Triangle begins:
0
1
2 1
3 1
4 2 1
5 3 2 1
6 3 2 1
7 4 2 1
8 5 3 2 1
9 5 3 2 1
10 6 4 3 2 1
11 6 4 3 2 1
12 7 4 3 2 1
13 8 5 4 3 2 1
14 8 5 4 3 2 1
15 9 6 4 3 2 1
16 10 6 4 3 2 1
...
MATHEMATICA
Array[NestWhile[Block[{p = Prime[#2]}, Append[#1, {p^#, #} &@ Floor@ Log[p, #1[[-1, 1]]]]] & @@ {#, Length@ # + 1} &, {{2^#, #}}, #[[-1, -1]] > 1 &][[All, -1]] &, 18, 0] // Flatten
CROSSREFS
Cf. A000217, A000961, A089576 (row lengths).
Sequence in context: A120385 A216477 A195836 * A132460 A238800 A375266
KEYWORD
nonn,tabf,easy
AUTHOR
Michael De Vlieger, Aug 26 2021
STATUS
approved