A286090 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.
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, 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, 1, 41, 9, 3, 3, 0, 0, 0, 0, 3, 3, 4, 2, 39, 11, 6, 4, 2, 1
Offset: 1
Examples
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. 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. The first block of m of 10 by 10 is: [3 7 8 15 15 21 19 27 29 31] [3 2 3 2 6 3 8 7 7 9] [2 3 2 1 2 2 2 4 3 3] [3 2 1 3 1 2 2 3 3 4] [2 4 2 1 2 0 2 0 2 1] [2 3 2 2 0 4 0 1 3 1] [1 5 2 2 2 0 3 0 0 2] [2 3 4 3 0 1 0 4 0 0] [2 3 3 3 2 3 0 0 3 0] [1 4 3 4 1 1 2 0 0 5]
Links
- David A. Corneth, Table of n, a(n) for n = 1..10011 (first 142 rows flattened.)
Programs
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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} \\ Diagonals m[i, j] where i + j <= n. 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}
Formula
for n > 1, m[n, n] = A014085(n).
Comments