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.

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

Author

Michael De Vlieger, Aug 26 2021

Keywords

• nonn
• tabf
• easy

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

Examples

			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
  ...
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 0..10367 (rows 0 <= n <= 300, flattened)

Crossrefs

Cf. A000217, A000961, A089576 (row lengths).

Programs

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

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.