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%I #14 Aug 02 2017 10:42:31
%S 0,1,2,1,0,3,1,1,4,1,0,0,2,0,1,2,5,1,0,1,6,1,3,2,1,1,0,0,0,7,1,1,0,8,
%T 1,0,2,2,2,1,4,9,1,0,0,1,3,0,1,5,2,0,0,1,0,3,10,1,1,1,11,1,0,0,0,0,2,
%U 3,1,6,3,1,1,0,1,0,12,1,7,2,4,1,0,0,2,13
%N Irregular triangle T(n,k) = A112798(n,1) followed by first differences of A112798(n).
%C Irregular triangle T(n,k) = first differences of indices of prime divisors p of n.
%C Row lengths = (big) Omega(n) = A001222(n).
%C Row sums = A061395(n).
%C Row maxima = A286469(n).
%C We can concatenate the rows 1 <= n <= 28 as none of the values of k in this range exceed 9: {0, 1, 2, 10, 3, 11, 4, 100, 20, 12, 5, 101, 6, 13, 21, 1000, 7, 110, 8, 102, 22, 14, 9, 1001, 30, 15, 200, 103}; a(29) = {10}, which would require a digit greater than 9.
%C a(1) = 0 by convention.
%C a(0) is not defined (i.e., null set). a(n) is defined for positive nonzero n.
%C a(p) = A000720(p) for p prime.
%C a(p^e) = A000720(p) followed by (e - 1) zeros.
%C a(Product(p^e)) is the concatenation of the a(p^e) of the unitary prime power divisors p^e of n, sorted by the prime p (i.e. the function a(n) mapped across the terms of row n of A141809).
%C a(A002110(n)) = an array of n 1s.
%C T(n,k) could be used to furnish A054841(n). We read data in row n of T(n,k). If T(n,1) = 0, then write 0. If T(n,1) > 0, then increment the k-th place from the right. For k > 1, increment the k-th place to the right of the last-incremented place.
%C T(n,k) can be used to render n in decimal. If T(n,1) = 0, then write 1. If T(n,1) > 0, then multiply 1 by A000720(T(n,1)). For k > 1, multiply the previous product by pi(x) = A000720(x) of the running total of T(n,k) for each k.
%C Ignoring zeros in row n > 1 and decoding the remaining values of T(n,k) as immediately above yields the squarefree kernel of n = A007947(n).
%C Leading zeros of a(n) are trimmed, but as in decimal notation numbers that include leading zeros symbolize the same n as without them. Zeros that precede nonzero values merely multiply implicit 1 by itself until we encounter nonzero values. Thus, {0,0,2} = 1*1*pi(2) = 3, as {2} = pi(2) = 3. Because of this no row n > 1 has 0 for k = 1 of T(n,k).
%C Interpreting n written in binary as a row of a(n) yields A057335(n).
%H Michael De Vlieger, <a href="/A287352/b287352.txt">Table of n, a(n) for n = 1..15568</a> (rows 1 <= n <= 5000).
%F T(n,1) = A117798(n,1); T(n,k) = A117798(n,k) - A117798(n, k - 1) for 2 <= k <= A001222(n).
%e a(1) = {0} by convention.
%e a(2) = {pi(2)} = {1}.
%e a(4) = {pi(2), pi(2) - pi(2)}, = {1, 0} since 4 = 2 * 2.
%e a(6) = {pi(2), pi(3) - pi(2)} = {1, 1} since 6 = 2 * 3.
%e a(12) = {pi(2), pi(2) - pi(2), pi(3) - pi(2) - pi(2)} = {1, 0, 1}, since 12 = 2 * 2 * 3.
%e The triangle starts:
%e 1: 0;
%e 2: 1;
%e 3: 2;
%e 4: 1, 0;
%e 5: 3;
%e 6: 1, 1;
%e 7: 4;
%e 8: 1, 0, 0;
%e 9: 2, 0;
%e 10: 1, 2;
%e 11: 5;
%e 12: 1, 0, 1;
%e 13: 6;
%e 14: 1, 3;
%e 15: 2, 1;
%e 16: 1, 0, 0, 0;
%e 17: 7;
%e 18: 1, 1, 0;
%e 19: 8;
%e 20: 1, 0, 2;
%e ...
%t Table[Prepend[Differences@ #, First@ #] & Flatten[FactorInteger[n] /. {p_, e_} /; p > 0 :> ConstantArray[PrimePi@ p, e]], {n, 41}] // Flatten (* _Michael De Vlieger_, May 23 2017 *)
%Y Cf. A000720, A001222, A007947, A054841, A057335, A061395, A112798, A141809, A286469.
%K nonn,tabf,easy
%O 1,3
%A _Michael De Vlieger_, May 23 2017
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