A343253 a(n) is the least k0 <= n such that v_11(n), the 11-adic order of n, can be obtained by the formula: v_11(n) = log_11(n / L_11(k0, n)), where L_11(k0, n) is the lowest common denominator of the elements of the set S_11(k0, n) = {(1/n)*binomial(n, k), with 0 < k <= k0 such that k is not divisible by 11} or 0 if no such k0 exists.

1, 2, 3, 4, 5, 3, 7, 8, 9, 5, 1, 4, 13, 7, 5, 16, 17, 9, 19, 5, 7, 2, 23, 8, 25, 13, 27, 7, 29, 5, 31, 32, 3, 17, 7, 9, 37, 19, 13, 8, 41, 7, 43, 4, 9, 23, 47, 16, 49, 25, 17, 13, 53, 27, 5, 8, 19, 29, 59, 5, 61, 31, 9, 64, 13, 3, 67, 17, 23, 7, 71, 9, 73, 37, 25, 19, 7, 13, 79, 16

Offset: 1

Author

Dario T. de Castro, May 31 2021

Keywords

• nonn

Comments

Conjecture: a(n) is the greatest power of a prime different from 11 that divides n.

Examples

			For n = 12, a(12) = 4. To understand this result, consider the largest set S_11, which is the S_11(k0=12, 12). According to the definition, S_11(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 11. The elements of S_11(12, 12) are {1, 11/2, 55/3, 165/4, 66, 77, 66, 165/4, 55/3, 11/2, 0, 1/12}, where the zero was inserted pedagogically to identify the skipped term, i.e., when k is divisible by 11. 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_11(4, 12) will be 12. So, L_11(4, 12) = 12 and the equation v_11(12) = log_11(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, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.

Crossrefs

Cf. A343249, A343250, A343251, A343252.

Programs

Mathematica

j = 5;
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, 11]}, While[Log[11, n/LCM @@ Denominator[Binomial[n, Select[Range[k], ! Divisible[#, 11] &]]/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=11) = {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