A333625 Read terms e = T(n,k) in A333624 as Product(prime(k)^e) for n in A334556.

1, 8, 8, 27, 27, 216, 512, 216, 512, 648, 648, 686, 12096, 46656, 262144, 46656, 262144, 12096, 686, 192000, 139968, 192000, 139968, 1866240, 179712, 74088, 91125, 74088, 91125, 179712, 1866240, 343000, 1000000, 5832000, 4251528, 5832000, 80621568, 13824000, 1073741824

Offset: 1

Author

Michael De Vlieger, May 13 2020

Keywords

• nonn

Comments

Row a(n) of A067255 = row A334556(n) of A333624.

An XOR-triangle t(n) is an inverted 0-1 triangle formed by choosing a top row the binary rendition of n and having each entry in subsequent rows be the XOR of the two values above it, i.e., A038554(n) applied recursively until we reach a single bit.

Let T(n,k) address the terms in the k-th position of row n in A333624.

This sequence encodes T(n,k) via A067255 to succinctly express the number of zero-triangles in A334556(n). To decode a(n) => A333624(A334556(n)), we use A067255(a(n)).

Examples

			Relationship of this sequence to A334556 and A333624:
       n A334556(n) a(n)  Row n of A333624
       -----------------------------------
       1     1        1   0
       2    11        8   3
       3    13        8   3
       4    39       27   0, 3
       5    57       27   0, 3
       6    83      216   3, 3
       7    91      512   9
       8   101      216   3, 3
       9   109      512   9
      10   151      648   3, 4
      11   233      648   3, 4
      12   543      686   1, 0, 0, 3
      13   599    12096   6, 3, 0, 1
      14   659    46656   6, 6
      15   731   262144   18
      16   805    46656   6, 6
      ...
Let b(n) = n written in binary and let L(n) = ceiling(log_2(n)) = A070939(n). Let => be a single iteration of XOR across pairs of bits in b(n). Let t(n) be the XOR triangle initiated by b(n).
a(1) = 0, since b(1) = 1 and row 1 of A333624 is {0}. Since the XOR triangle t(1) that results from a single 1-bit merely consists of that bit and since there are no zeros in the triangle t(1), we write the single term zero in row n of A333624. thus a(n) = prime(1)^0 = 2^0 = 1.
a(2) = 8 because row A334556(2) of A333624 (i.e., the 11th row) has {3}. b(11) = 1011 => 110 => 01 => 1 (a rotationally symmetrical t(11)). We have 3 isolated zeros thus row 11 of A333624 = {3}, therefore a(2) = prime(1)^3 = 2^3 = 8.
a(4) = 27 because row A334556(4) of A333624 (i.e., the 39th row) has {0, 3}. b(39) = 100111 => 10100 => 1110 => 001 => 01 => 1 (a rotationally symmetrical t(39)). We have 3 isolated triangles of zeros with edge length 2, thus row 39 of A333624 = {0, 3}, therefore a(4) = prime(1)^0 * prime(2)^3 = 2^0 * 3^3 = 27.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..9999
• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A038554, A067255, A070939, A333624, A334591, A334556.

Programs

Mathematica

With[{s = Rest[Import["https://oeis.org/A334556/b334556.txt", "Data"][[All, -1]] ]}, Map[With[{w = NestWhileList[Map[BitXor @@ # &, Partition[#, 2, 1]] &, IntegerDigits[#, 2], Length@ # > 1 &]}, If[Length@ # == 0, 1, Times @@ Flatten@ MapIndexed[Prime[#2]^#1 &, #] &@ ReplacePart[ConstantArray[0, Max@ #[[All, 1]]], Map[#1 -> #2 & @@ # &, #]]] &@ Tally@ Flatten@ Array[If[# == 1, Map[If[First@ # == 1, Nothing, Length@ #] &, Split@ w[[#]] ], Map[If[First@ # == -1, Length@ #, Nothing] &, Split[w[[#]] - Most@ w[[# - 1]] ] ]] &, Length@ w]] /. -Infinity -> 0 &, s[[1 ;; 36]]]]