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%I #61 Jul 12 2022 18:31:29
%S 1,2,3,4,1,3,7,8,9,2,11,4,13,7,3,16,17,9,19,4,7,11,23,8,1,13,27,7,29,
%T 3,31,32,11,17,7,9,37,19,13,8,41,7,43,11,9,23,47,16,49,2,17,13,53,27,
%U 11,8,19,29,59,4,61,31,9,64,13,11,67,17,23,7,71,9,73,37,3,19,11,13,79,16
%N a(n) is the least k0 <= n such that v_5(n), the 5-adic order of n, can be obtained by the formula: v_5(n) = log_5(n / L_5(k0, n)), where L_5(k0, n) is the lowest common denominator of the elements of the set S_5(k0, n) = {(1/n)*binomial(n, k), with 0 < k <= k0 such that k is not divisible by 5} or 0 if no such k0 exists.
%C Conjecture: a(n) is the greatest power of a prime different from 5 that divides n.
%H Dario T. de Castro, <a href="/A343251/b343251.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 = 12, a(12) = 4. To understand this result, consider the largest set S_5, which is the S_5(k0=12, 12). According to the definition, S_5(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 5. The elements of S_5(12, 12) are {1, 11/2, 55/3, 165/4, 0, 77, 66, 165/4, 55/3, 0, 1, 1/12}, where the zeros were inserted pedagogically to identify the skipped terms, i.e., when k is divisible by 5. At this point we verify which of the nested subsets {1}, {1, 11/2}, {1, 11/2, 55/3}, {1, 11/2, 55/3, 165/4}, ... will match for the first time the p-adic order's formula. If k varies from 1 to 4 (instead of 12) we see that the lowest common denominator of the set S_5(4, 12) will be 12. So, L_5(4, 12) = 12 and the equation v_5(12) = log_5(12/12) yields a True result. Then we may say that a(12) = 4 specifically because 4 was the least k0.
%t j = 3;
%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}];
%t (* alternate code *)
%t a[n_] := Module[{k = 1, v = IntegerExponent[n, 5]}, While[Log[5, n/LCM @@ Denominator[Binomial[n, Select[Range[k], ! Divisible[#, 5] &]]/n]] != v, k++]; k]; Array[a, 100] (* _Amiram Eldar_, Apr 23 2021 *)
%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=5) = {my(k=1, v=valuation(n, p)); for (k=1, n, if (isok(k, n, v, p), return(k)););} \\ _Michel Marcus_, Apr 23 2021
%Y Cf. A112765, A343249, A343250, A343252, A343253, A345531.
%K nonn
%O 1,2
%A _Dario T. de Castro_, Apr 09 2021
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