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A076620 Coefficient of x^a(n) in (x+1)*(x+2)*...*(x+n) is the largest one.

%I A076620 #27 Mar 07 2021 17:47:57
%S A076620 0,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,
%T A076620 3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,
%U A076620 3,3,3,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4,4
%N A076620 Coefficient of x^a(n) in (x+1)*(x+2)*...*(x+n) is the largest one.
%H A076620 David A. Corneth, <a href="/A076620/b076620.txt">Table of n, a(n) for n = 1..10000</a>
%F A076620 Is a(n) - floor(log(n)) bounded?
%e A076620 In (x+1)(x+2)(x+3) = x^3 + 6*x^2 + 11*x + 6, the largest coefficient (11) appears at x^1, hence a(3)=1.
%o A076620 (PARI) a(n) = my(p=prod(j=1, n, x+j), m=vecmax(Vec(p))); for (i=0, poldegree(p), if (polcoef(p, i)==m, return(i))); \\ _Michel Marcus_, Feb 19 2021
%o A076620 (PARI) first(n) = {res = vector(n); my(r = 1); v = [1]; for(i = 1, n, v = concat(0, v) + concat(v, 0)*i; for(j = r + 1, #v, if(v[j] > v[j - 1], r++ , next ); ); res[i] = r-1 ); res } \\ _David A. Corneth_, Feb 21 2021
%o A076620 (Python)
%o A076620 from sympy import prod, Poly
%o A076620 from sympy.abc import x
%o A076620 def A076620(n):
%o A076620     y = Poly(prod(x+i for i in range(1,n+1))).all_coeffs()[::-1]
%o A076620     return y.index(max(y)) # _Chai Wah Wu_, Mar 07 2021
%Y A076620 Cf. A065048, A130534, A309237.
%K A076620 nonn
%O A076620 1,7
%A A076620 _Benoit Cloitre_, Nov 10 2002