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%I #21 Mar 30 2021 01:19:21
%S 1,3,3,15,15,224,224,4095,76544,512000,9511424,20916224,410572799,
%T 672358400,2985984000,1004293914624,1004293914624,78942076928000,
%U 610877575397375,83179139563520000,490473044848410624,6878928869130239999,185974097225789210624,1708887984313466880000,68817755280574852890624
%N a(n) = least k such that 1+2+3+...+k (k-th triangular number) is a multiple of n!; a(n) = least k such that A232096(k) >= n.
%C a(n) = least k such that A232096(k) >= n.
%C Each A000217(a(n)) is divisible by A118381(n).
%C Each a(n) or a(n)+1 is divisible by 2*A060818(n) = A086117(n+1).
%C Each a(n) or a(n)+1 is divisible by A060828(n), and similarly for all the higher bases.
%C If we were instead searching for the first occurrence where A232096 gets a new distinct value, then we would have another sequence, b, which would start as: 1, 3, 4, 15, 32, 224, 575, 4095, ... as those distinct values do not appear in monotone order, being for n>=1, A232096(b(n)) = 1, 3, 2, 5, 4, 7, 6, 8, 9, 10, ...
%e a(5) = 15 as binomial(15 + 1, 2) = 120 is the smallest binomial that is divisible by 5! = 120. - _David A. Corneth_, Mar 29 2021
%o (Scheme)
%o (define (A232097 n) (let ((increment (* 2 (A060818 n)))) (let loop ((k increment)) (cond ((>= (A232096 (- k 1)) n) (- k 1)) ((>= (A232096 k) n) k) (else (loop (+ k increment)))))))
%o ;; Alternative, very naive and slow version:
%o (define (A232097v2 n) (let loop ((k 1)) (if (>= (A232096 k) n) k (loop (+ 1 k)))))
%o (PARI) a(n) = { my(p = 2*n!, f = factor(p), res = oo); for(i = 2^(#f~-1), 2^#f~-1, b = binary(i); pr = prod(j = 1, #f~, f[j,1]^(b[j]*f[j, 2])); ipr = p/pr; for(j = -1, 0, c = lift(chinese(Mod(-1-j, ipr), Mod(j, pr))); if(c > 0, res = min(res, c)))) res } \\ _David A. Corneth_, Mar 29 2021
%Y Cf. A000217, A232096. A232101 gives the ratio A000217(a(n)) / n!
%K nonn
%O 1,2
%A _Antti Karttunen_, Nov 18 2013
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