This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A343250 #67 Jul 12 2022 18:33:56
%S A343250 1,2,1,4,5,2,7,8,1,5,11,4,13,7,5,16,17,2,19,5,7,11,23,8,25,13,1,7,29,
%T A343250 5,31,32,11,17,7,4,37,19,13,8,41,7,43,11,5,23,47,16,49,25,17,13,53,2,
%U A343250 11,8,19,29,59,5,61,31,7,64,13,11,67,17,23,7,71,8,73,37,25,19,11,13,79,16
%N A343250 a(n) is the least k0 <= n such that v_3(n), the 3-adic order of n, can be obtained by the formula: v_3(n) = log_3(n / L_3(k0, n)), where L_3(k0, n) is the lowest common denominator of the elements of the set S_3(k0, n) = {(1/n)*binomial(n, k), with 0 < k <= k0 such that k is not divisible by 3} or 0 if no such k0 exists.
%C A343250 Conjecture: a(n) is the greatest power of a prime different from 3 that divides n.
%H A343250 Dario T. de Castro, <a href="/A343250/b343250.txt">Table of n, a(n) for n = 1..1000</a>
%H A343250 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 A343250 For n = 10, a(10) = 5. To understand this result, consider the largest set S_3, which is the S_3(k0=10, 10). According to the definition, S_3(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 3. The elements of S_3(10, 10) are: {1, 9/2, 0, 21, 126/5, 0, 12, 9/2, 0, 1/10}, where the zeros were put pedagogically to identify the skipped terms, i.e., when k is divisible by 3. At this point we verify which of the nested subsets {1}, {1, 9/2}, {1, 9/2, 0}, {1, 9/2, 0, 21}, {1, 9/2, 0, 21, 126/5},... will match for the first time the p-adic order’s formula. If k vary from 1 to 5 (instead of 10) we see that the lowest common denominator of the set S_3(5, 10) will be 10. So, L_3(5, 10) = 10 and the equation v_3(10) = log_3(10/10) yields a True result. Then we may say that a(10) = 5 specifically because 5 was the least k0.
%t A343250 j = 2;
%t A343250 Nmax = 250;
%t A343250 Array[val, Nmax];
%t A343250 Do[val[i] = 0, {i, 1, Nmax}];
%t A343250 Do[flag = 0;
%t A343250 Do[If[(flag == 0 &&
%t A343250 Prime[j]^IntegerExponent[n, Prime[j]] ==
%t A343250 n/LCM[Table[
%t A343250 If[Divisible[k, Prime[j]], 1,
%t A343250 Denominator[(1/n) Binomial[n, k]]], {k, 1, k}] /.
%t A343250 List -> Sequence]), val[n] = k; flag = 1;, Continue], {k, 1,
%t A343250 n, 1}], {n, 1, Nmax}];
%t A343250 tabseq = Table[val[i], {i, 1, Nmax}];
%t A343250 (* alternate code *)
%t A343250 a[n_] := Module[{k = 1, v = IntegerExponent[n, 3]}, While[Log[3, n/LCM @@ Denominator[Binomial[n, Select[Range[k], ! Divisible[#, 3] &]]/n]] != v, k++]; k]; Array[a, 100] (* _Amiram Eldar_, Apr 23 2021 *)
%o A343250 (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 A343250 isok(k, n, v, p) = p^v == n/Lp(k, n, p);
%o A343250 a(n, p=3) = {my(k=1, v=valuation(n, p)); for (k=1, n, if (isok(k, n, v, p), return(k)););} \\ _Michel Marcus_, Apr 22 2021
%Y A343250 Cf. A007949, A343249, A343251, A343252, A343253, A345531.
%K A343250 nonn
%O A343250 1,2
%A A343250 _Dario T. de Castro_, Apr 09 2021