A343249 - OEIS

%I #80 Jul 12 2022 12:22:45

%S 1,1,3,1,5,3,7,1,9,5,11,3,13,7,5,1,17,9,19,5,7,11,23,3,25,13,27,7,29,

%T 5,31,1,11,17,7,9,37,19,13,5,41,7,43,11,9,23,47,3,49,25,17,13,53,27,

%U 11,7,19,29,59,5,61,31,9,1,13,11,67,17,23,7,71,9,73,37,25,19,11,13,79,5

%N a(n) is the least k0 <= n such that v_2(n), the 2-adic order of n, can be obtained by the formula: v_2(n) = log_2(n / L_2(k0, n)), where L_2(k0, n) is the lowest common denominator of the elements of the set S_2(k0, n) = {(1/n)*binomial(n, k), with 0 < k <= k0 such that k is not divisible by 2} or 0 if no such k0 exists.

%C Conjecture: a(n) is the greatest power of a prime different from 2 that divides n.

%H Dario T. de Castro, <a href="/A343249/b343249.txt">Table of n, a(n) for n = 1..1000</a>

%H Dario T. de Castro, <a href="http://math.colgate.edu/~integers/w61/w61.pdf">P-adic Order of Positive Integers via Binomial Coefficients</a>, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.

%e For n = 15, a(15) = 5. To understand this result, consider the largest set S_2, which is the S_2(k0=15, 15). According to the definition, S_2(n, n) is the set of elements of the form (1/n)*binomial(n, k), where k goes from 1 to n, skipping the multiples of 2. The elements of S_2(15, 15) are: {1, 0, 91/3, 0, 1001/5, 0, 429, 0, 1001/3, 0, 91, 0, 7, 0, 1/15}, where the zeros were put pedagogically to identify the skipped terms, i.e., when k is divisible by 2. At this point we verify which of the nested subsets {1}, {1, 0}, {1, 0, 91/3}, {1, 0, 91/3, 0}, {1, 0, 91/3, 0, 1001/5},... will match for the first time the p-adic order’s formula. If k vary from 1 to 5 (instead of 15) we see that the lowest common denominator of the set S_2(5, 15) will be 15. So, L_2(5, 15) = 15 and the equation v_2(15) = log_2(15/15) yields a True result. Then we may say that a(15) = 5 specifically because 5 was the least k0.

%t j = 1;

%t Nmax = 250;

%t Array[val, Nmax];

%t Do[val[i] = 0, {i, 1, Nmax}];

%t Do[flag = 0;

%t   Do[If[(flag == 0 &&

%t       Prime[j]^IntegerExponent[n, Prime[j]] ==

%t        n/LCM[Table[

%t            If[Divisible[k, Prime[j]], 1,

%t             Denominator[(1/n) Binomial[n, k]]], {k, 1, k}] /.

%t           List -> Sequence]), val[n] = k; flag = 1;, Continue], {k, 1,

%t      n, 1}], {n, 1, Nmax}];

%t tabseq = Table[val[i], {i, 1, Nmax}];

%o (PARI) Lp(k, n, p) = {my(list = List()); for (i=1, k, if (i%p, listput(list, binomial(n,i)/n));); lcm(apply(denominator, Vec(list)));}

%o isok(k, n, v, p) = p^v == n/Lp(k, n, p);

%o a(n, p=2) = {my(k=1, v=valuation(n, p)); for (k=1, n, if (isok(k, n, v, p), return(k));); n;} \\ _Michel Marcus_, Apr 22 2021

%Y Cf. A007814, A343250, A343251, A343252, A343253, A345531.

%K nonn

%O 1,3

%A _Dario T. de Castro_, Apr 09 2021