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A343249
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a(n) is the least k0 <= n such that v_2(n), the 2-adic order of n, can be obtained by the formula: v_2(n) = log_2(n / L_2(k0, n)), where L_2(k0, n) is the lowest common denominator of the elements of the set S_2(k0, n) = {(1/n)*binomial(n, k), with 0 < k <= k0 such that k is not divisible by 2} or 0 if no such k0 exists.
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6
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1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 5, 1, 17, 9, 19, 5, 7, 11, 23, 3, 25, 13, 27, 7, 29, 5, 31, 1, 11, 17, 7, 9, 37, 19, 13, 5, 41, 7, 43, 11, 9, 23, 47, 3, 49, 25, 17, 13, 53, 27, 11, 7, 19, 29, 59, 5, 61, 31, 9, 1, 13, 11, 67, 17, 23, 7, 71, 9, 73, 37, 25, 19, 11, 13, 79, 5
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OFFSET
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1,3
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COMMENTS
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Conjecture: a(n) is the greatest power of a prime different from 2 that divides n.
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LINKS
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EXAMPLE
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For n = 15, a(15) = 5. To understand this result, consider the largest set S_2, which is the S_2(k0=15, 15). According to the definition, S_2(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 2. The elements of S_2(15, 15) are: {1, 0, 91/3, 0, 1001/5, 0, 429, 0, 1001/3, 0, 91, 0, 7, 0, 1/15}, where the zeros were put pedagogically to identify the skipped terms, i.e., when k is divisible by 2. At this point we verify which of the nested subsets {1}, {1, 0}, {1, 0, 91/3}, {1, 0, 91/3, 0}, {1, 0, 91/3, 0, 1001/5},... will match for the first time the p-adic order’s formula. If k vary from 1 to 5 (instead of 15) we see that the lowest common denominator of the set S_2(5, 15) will be 15. So, L_2(5, 15) = 15 and the equation v_2(15) = log_2(15/15) yields a True result. Then we may say that a(15) = 5 specifically because 5 was the least k0.
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MATHEMATICA
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j = 1;
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}];
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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=2) = {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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