A343251 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.

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, 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, 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

Offset: 1

Author

Dario T. de Castro, Apr 09 2021

Keywords

• nonn

Comments

Conjecture: a(n) is the greatest power of a prime different from 5 that divides n.

Examples

			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.
		

Links

• Dario T. de Castro, Table of n, a(n) for n = 1..1000
• Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.

Crossrefs

Cf. A112765, A343249, A343250, A343252, A343253, A345531.

Programs

Mathematica

j = 3;
Nmax = 250;
Array[val, Nmax];
Do[val[i] = 0, {i, 1, Nmax}];
Do[flag = 0;
  Do[If[(flag == 0 &&
      Prime[j]^IntegerExponent[n, Prime[j]] ==
       n/LCM[Table[
           If[Divisible[k, Prime[j]], 1,
            Denominator[(1/n) Binomial[n, k]]], {k, 1, k}] /.
          List -> Sequence]), val[n] = k; flag = 1;, Continue], {k, 1,
     n, 1}], {n, 1, Nmax}];
tabseq = Table[val[i], {i, 1, Nmax}];
(* alternate code *)
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 *)

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