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A343253
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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.
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5
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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
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OFFSET
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1,2
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COMMENTS
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Conjecture: a(n) is the greatest power of a prime different from 11 that divides n.
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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