

A343251


a(n) is the least k0 <= n such that v_5(n), the 5adic 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.


3



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
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OFFSET

1,2


COMMENTS

Conjecture: a(n) > 0 for all n >= 1.


LINKS

Dario T. de Castro, Table of n, a(n) for n = 1..1000


EXAMPLE

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 padic 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.


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 *)


PROG

(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


CROSSREFS

Cf. A112765, A343249, A343250.
Sequence in context: A294236 A280694 A270313 * A327464 A318308 A003324
Adjacent sequences: A343248 A343249 A343250 * A343254 A343255 A343256


KEYWORD

nonn


AUTHOR

Dario T. de Castro, Apr 09 2021


STATUS

approved



