A375973 -id:A375973 - cp's OEIS frontend

A375970 a(n) is the largest number k such that k^2 divides the square pyramidal number A000330(n).

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 5, 3, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 70, 5, 3, 3, 1, 1, 1, 4, 4, 1, 1, 1, 1, 5, 1, 2, 6, 1, 1, 1, 1, 1, 1, 2, 14, 35, 5, 1, 1, 3, 3, 2, 2, 1, 1, 1, 11, 1, 5, 4, 4, 1, 1, 3, 1, 1, 1, 2, 2, 7, 5, 5, 1, 1, 1, 2, 6, 3, 1, 1, 13, 1, 1, 10, 2, 1, 1, 1, 1, 1, 3, 4, 4, 7

Offset: 1

Robert Israel, Sep 04 2024

nonn

a(n)^2 is the largest square that divides n*(n+1)*(2*n+1)/6.

			a(12) = 5 because A000330(12) = 650 = 2 * 5^2 = 13 and 5^2 is the largest square dividing 650.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000188, A000330, A119356, A375971, A375973.

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:
map(g, [$1..100]);

a(n) = my(m=n*(n+1)*(2*n+1)/6); sqrtint(m/core(m)); \\ Michel Marcus, Sep 06 2024

a(n) = A000188(A000330(n)).

A375971 Positions of records in A375970.

Original entry on oeis.org

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

Robert Israel, Sep 04 2024

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)

			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)

Cf. A000330, A375970, A375973.

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;

a(25) from Michael S. Branicky, Sep 06 2024

a(26)-a(31) from David A. Corneth, Sep 08 2024

cp's OEIS Frontend

A375970 a(n) is the largest number k such that k^2 divides the square pyramidal number A000330(n).

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