A333625 - OEIS

A333625

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

%I #11 Jul 19 2020 11:53:11

%S 1,8,8,27,27,216,512,216,512,648,648,686,12096,46656,262144,46656,

%T 262144,12096,686,192000,139968,192000,139968,1866240,179712,74088,

%U 91125,74088,91125,179712,1866240,343000,1000000,5832000,4251528,5832000,80621568,13824000,1073741824

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

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

%C 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.

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

%C 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)).

%H Michael De Vlieger, <a href="/A333625/b333625.txt">Table of n, a(n) for n = 1..9999</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%e Relationship of this sequence to A334556 and A333624:

%e n A334556(n) a(n) Row n of A333624

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

%e 1 1 1 0

%e 2 11 8 3

%e 3 13 8 3

%e 4 39 27 0, 3

%e 5 57 27 0, 3

%e 6 83 216 3, 3

%e 7 91 512 9

%e 8 101 216 3, 3

%e 9 109 512 9

%e 10 151 648 3, 4

%e 11 233 648 3, 4

%e 12 543 686 1, 0, 0, 3

%e 13 599 12096 6, 3, 0, 1

%e 14 659 46656 6, 6

%e 15 731 262144 18

%e 16 805 46656 6, 6

%e ...

%e 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).

%e 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.

%e 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.

%e 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.

%t 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]]]]

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

%K nonn

%O 1,2

%A _Michael De Vlieger_, May 13 2020