OFFSET
1,1
COMMENTS
Also, numbers with a record number of proper factorizations that form the base lengths of Pythagorean hyperrectangles.
Terms tend to be integer multiples of smaller terms; program can be modified to leverage this to find larger candidate terms more quickly.
Though total factorization counts can serve as a rough predictor of Pythagorean counts, this sequence has significant non-overlap with A033833 (record total proper factorizations).
EXAMPLE
a(1) = 12, 1 example: {3, 4} (3 * 4 = 12 and 3^2 + 4^2 = 5^2).
a(2) = 48, 2 examples: {2, 2, 2, 2, 3} (2 * 2 * 2 * 2 * 3 = 48 and 2^2 + 2^2 + 2^2 + 2^2 + 3^2 = 5^2), {6, 8}.
a(3) = 108, 3 examples: {3, 6, 6}, {9, 12}, {2, 6, 9}.
a(4) = 240, 4 examples: {2, 2, 2, 3, 10}, {2, 2, 6, 10}, {2, 4, 5, 6}, {10, 24}.
a(5) = 864, 7 examples: {3, 12, 24}, {3, 8, 36}, {2, 3, 6, 24}, {2, 12, 36}, {6, 12, 12}, {4, 12, 18}, {8, 9, 12}.
PROG
(PARI) r()={my(i=if(!#s2, 0, #s2[#s2])%#s1[1]+1); if(i==1, listput(s2, List())); for(i2=0, s1[#s1][i], listput(s2[#s2], i2); if(i<#s1[#s1], r(), p=prod(i3=1, #s2[#s2], v[1][i3]^s2[#s2][i3]); if(p>1 && (!#f || p>=f[#f]), listput(f, p); listput(s1, s1[#s1]); for(i3=1, #s2[#s2], s1[#s1][i3]-=s2[#s2][i3]); if(vecsum(s1[#s1])>0, r(), if(issquare(sum(i3=1, #f, f[i3]^2)), c++)); listpop(s1); listpop(f))); listpop(s2[#s2])); if(i==1, listpop(s2))}
cmax=0; for(t=2, 100000, v=Vec(factor(t)); s1=List([v[2]]); s2=List(); f=List(); c=-1; r(); if(c>cmax, cmax=c; print(t)))
CROSSREFS
KEYWORD
nonn
AUTHOR
Charles L. Hohn, Jan 06 2025
STATUS
approved