A286090 - OEIS

%I #27 Jan 06 2018 18:46:49

%S 3,7,3,8,2,2,15,3,3,3,15,2,2,2,2,21,6,1,1,4,2,19,3,2,3,2,3,1,27,8,2,1,

%T 1,2,5,2,29,7,2,2,2,2,2,3,2,31,7,4,2,0,0,2,4,3,1,32,9,3,3,2,4,2,3,3,4,

%U 1,41,9,3,3,0,0,0,0,3,3,4,2,39,11,6,4,2,1

%N Square matrix m read by antidiagonals. For j <= i, m[i, j] is how often sqrtint(gpf(c)) = j where i^2 <= c < (i+1)^2; for j >= i, m[i, j] is how often sqrtint(p) = i where p is a prime factor of c counted with multiplicity and i^2 <= c < (i+1)^2.

%C The greatest prime factor (see A006530) of 1 is 1 by convention. The definitions for m[i, j] where j >= i and j <= i give no contradiction for j = i.

%H David A. Corneth, <a href="/A286090/b286090.txt">Table of n, a(n) for n = 1..10011 (first 142 rows flattened.)</a>

%F for n > 1, m[n, n] = A014085(n).

%e Let i = 2 and sqrtint(n) = the square root of n rounded down to an integer. The integers c such that i^2 = 2^2 <= c < 3^2 = (i+1)^2 are 4, 5, 6, 7 and 8. The greatest prime factors h of these terms are 2, 5, 3, 7 and 2 respectively. sqrtint(h) are 1, 2, 1, 2 and 1 respectively. 1 occurs thrice here, 2 occurs twice, giving m[2, 1] = 3 and m[2, 2] = 2.

%e m[1, 2] = 7 because the prime factorizations of 4, 5, 6, 7 and 8 are 2^2, 5, 2*3, 7, 2^3. The prime 2 occurs 6 times and sqrtint(2) = 1. The prime factor 3 occurs once and sqrtint(3) = 1. Therefore m[1, 2] = 6 + 1 = 7. p = 5 and p = 7 each occur once and each have sqrtint(p) = 2. Therefore m[2, 2] = 2 as found earlier.

%e The first block of m of 10 by 10 is:

%e [3 7 8 15 15 21 19 27 29 31]

%e [3 2 3  2  6  3  8  7  7  9]

%e [2 3 2  1  2  2  2  4  3  3]

%e [3 2 1  3  1  2  2  3  3  4]

%e [2 4 2  1  2  0  2  0  2  1]

%e [2 3 2  2  0  4  0  1  3  1]

%e [1 5 2  2  2  0  3  0  0  2]

%e [2 3 4  3  0  1  0  4  0  0]

%e [2 3 3  3  2  3  0  0  3  0]

%e [1 4 3  4  1  1  2  0  0  5]

%o (PARI) squaremat(n) = {my(m=matrix(n,n)); m[1,1] = 3; for(i=2, n, for(c = i^2, (i+1)^2-1, f=factor(c); for(j=1, matsize(f)[1], m[sqrtint(f[j,1]), i] += f[j, 2]); m[i, sqrtint(f[matsize(f)[1], 1])]++));for(i=2, n, m[i,i] = m[i,i] >> 1); m}

%o \\ Diagonals m[i, j] where i + j <= n.

%o upto(n) = {my(v=vector(binomial(n+1, 2)), m = squaremat(n), t = 0); for(i=1, n, for(j=0, i-1, t++; v[t] = m[j+1, i-j])); v}

%Y Cf. A006530 (greatest prime factor, gpf), A014085.

%K nonn,tabl

%O 1,1

%A _David A. Corneth_, May 02 2017 and May 17 2017