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A375971
Positions of records in A375970.
2
1, 7, 12, 24, 121, 337, 512, 722, 840, 4704, 4900, 23762, 28560, 29767, 166464, 235224, 647149, 970224, 1940449, 4920547, 14070000, 20346212, 32959080, 42508287, 79346175, 307546368, 319311720, 450982499, 491756160, 921166587
OFFSET
1,2
COMMENTS
Numbers k such that the k-th square pyramidal number A000330(k) is divisible by a square larger than any square dividing A000330(i) for 1 <= i < k.
From David A. Corneth, Sep 13 2024: (Start)
To ease the search one could split the positive integers into classes mod 6. For example numbers of the form m = 6*k + 1 have m*(m+1)*(2*m+1)/6 = (6*k + 1)*(6*k + 2)*(12*k + 3)/6 = (6*k + 1)*(3*k + 1)*(4*k + 1). Using such factorizations prevents factorizing larger numbers as m, m+1 and 2*m+1 are pairwise coprime.
An additional optimization would be to stop checking if 1 or 2 factors are checked knowing the term cannot produce a record and so at least skipping the third. (End)
EXAMPLE
a(3) = 12 because A000330(12) = 650 = 2 * 5^2 * 13 is divisible by 5^2, which is greater than any square dividing A000330(i) for 1 <= i < 12.
From David A. Corneth, Sep 13 2024: (Start)
24 is in the sequence as A000330(24) = 24 * 25 * 49 / 6 = 4 * 25 * 49. The largest square dividing 4 is 4, the largest square dividing 25 is 25 and the largest square dividing 49 is 49.
So the largest k such that k^2 divides 4 * 25 * 49 is sqrt(4)*sqrt(25)*sqrt(49) = 2*5*7 = 70, a record found at position 24 in A375970. (End)
MAPLE
g:= proc(n) local t, s, F; t:= n*(n+1)*(2*n+1)/6;
F:= ifactors(t)[2];
mul(s[1]^floor(s[2]/2), s=F)
end proc:
R:= NULL: m:= 0: count:= 0:
for k from 1 while count < 20 do
v:= g(k);
if v > m then m:= v; R:= R, k; count:= count+1; fi
od:
R;
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Robert Israel, Sep 04 2024
EXTENSIONS
a(25) from Michael S. Branicky, Sep 06 2024
a(26)-a(31) from David A. Corneth, Sep 08 2024
STATUS
approved