A375971 - OEIS

%I #24 Sep 14 2024 06:47:25

%S 1,7,12,24,121,337,512,722,840,4704,4900,23762,28560,29767,166464,

%T 235224,647149,970224,1940449,4920547,14070000,20346212,32959080,

%U 42508287,79346175,307546368,319311720,450982499,491756160,921166587

%N Positions of records in A375970.

%C 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.

%C From _David A. Corneth_, Sep 13 2024: (Start)

%C 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.

%C 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)

%e 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.

%e From _David A. Corneth_, Sep 13 2024: (Start)

%e 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.

%e 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)

%p g:= proc(n) local t,s,F; t:= n*(n+1)*(2*n+1)/6;

%p   F:= ifactors(t)[2];

%p   mul(s[1]^floor(s[2]/2), s=F)

%p end proc:

%p R:= NULL: m:= 0: count:= 0:

%p for k from 1 while count < 20 do

%p   v:= g(k);

%p   if v > m then m:= v; R:= R,k; count:= count+1; fi

%p od:

%p R;

%Y Cf. A000330, A375970, A375973.

%K nonn,more

%O 1,2

%A _Robert Israel_, Sep 04 2024

%E a(25) from _Michael S. Branicky_, Sep 06 2024

%E a(26)-a(31) from _David A. Corneth_, Sep 08 2024