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Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.
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%I #192 Aug 11 2024 14:44:03

%S 1,1,3,1,7,3,7,1,15,7,17,3,31,7,15,1,31,15,37,7,69,17,37,3,115,31,69,

%T 7,115,15,31,1,63,31,77,15,145,37,81,7,245,69,155,17,261,37,77,3,391,

%U 115,261,31,445,69,145,7,675,115,245,15,391,31,63,1,127,63

%N Number of permutations of {1,2,...,m} with excedance set constructed by taking m-i (0 < i < m) if b(i-1) = 1 where b(k)b(k-1)...b(1)b(0) (0 <= k < m-1) is the binary expansion of n.

%C Another version of A152884.

%C The excedance set of a permutation p of {1,2,...,m} is the set of indices i such that p(i) > i; it is a subset of {1,2,...,m-1}.

%C Great work on this subject was done by R. Ehrenborg and E. Steingrimsson, so most of the formulas given below are just their results translated into the language of the sequences which are related to the binary expansion of n.

%C Conjecture: equivalently, number of open tours by a biased rook on a specific f(n) X 1 board, which ends on a white cell, where f(n) = A070941(n) = floor(log_2(2n)) + 1 and cells are colored white or black according to the binary representation of 2n. A cell is colored white if the binary digit is 0 and a cell is colored black if the binary digit is 1. A biased rook on a white cell moves only to the left and otherwise moves only to the right. - _Mikhail Kurkov_, May 18 2021

%C Conjecture: this sequence is an inverse modulo 2 binomial transform of A284005. - _Mikhail Kurkov_, Dec 15 2021

%H Mikhail Kurkov, <a href="/A329369/b329369.txt">Table of n, a(n) for n = 0..8191</a>

%H Max Alekseyev, <a href="https://mathoverflow.net/a/474668/231922">Recursion for the sum with Stirling numbers of both kinds</a>, answer to question on MathOverflow (2024).

%H R. Ehrenborg and E. Steingrimsson, <a href="http://dx.doi.org/10.1006/aama.1999.0671">The excedance set of a permutation</a>, Advances in Appl. Math., 24, 284-299, 2000.

%H Peter J. Taylor, <a href="https://mathoverflow.net/a/470752/231922">Correctness of the algorithm for the A329369, A347205 and related sequences</a>, answer to question on MathOverflow (2024).

%F a(2n+1) = a(n) for n >= 0.

%F a(2n) = a(n) + a(n - 2^f(n)) + a(2n - 2^f(n)) for n > 0 with a(0) = 1 where f(n) = A007814(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).

%F a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) for m > 0, n >= 0 (equivalent to proposition 2.5 at the page 287, see R. Ehrenborg and E. Steingrimsson link).

%F a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n)) for n > 0 with a(0) = 1 where g(n) = A053645(n), h(n) = A063250(n) (equivalent to proposition 2.1 at the page 286, see R. Ehrenborg and E. Steingrimsson link).

%F a(2n) = 2*a(n + g(n)) + a(2*g(n)) for n > 0, floor(n/3) < 2^(floor(log_2(n))-1) (in other words, for 2^m + k where 0 <= k < 2^(m-1), m > 0) with a(0) = 1 (just a special case of the previous formula, because for 2^m + k where 0 <= k < 2^(m-1), m > 0 we have 2^h(n) = n - g(n)).

%F a(2n) = a(f(n,-1)) + a(f(n,0)) + a(f(n,1)) for n > 0 with a(0) = 1 where f(n,k) = 2*(f(floor(n/2),k) + n mod 2) + k*A036987(n) for n > 1 with f(1,k) = abs(k) (equivalent to a(2n) = a(2*g(n)) + a(2n - 2^h(n)) + a(2*g(n) + 2^h(n))).

%F a(n) = Sum_{j=0..2^wt(n) - 1} (-1)^(wt(n) - wt(j)) Product_{k=0..wt(n) - 1} (1 + wt(floor(j/2^k)))^T(n,k) for n > 0 with a(0) = 1 where wt(n) = A000120(n), T(n,k) = T(floor(n/2), k - n mod 2) for k > 0 with T(n,0) = A001511(n) (equivalent to theorem 6.3 at page 296, see R. Ehrenborg and E. Steingrimsson link). Here T(n, k) - 1 for k > 0 is the length of the run of zeros between k-th pair of ones from the right side in the binary expansion of n. Conjecture: this formula is equivalent to inverse modulo 2 binomial transform of A284005.

%F Sum_{k=0..2^n-1} a(k) = (n+1)! for n >= 0.

%F a((4^n-1)/3) = A110501(n+1) for n >= 0.

%F a(2^2*(2^n-1)) = A091344(n+1),

%F a(2^3*(2^n-1)) = A091347(n+1),

%F a(2^4*(2^n-1)) = A091348(n+1).

%F More generally, a(2^m*(2^n-1)) = a(2^n*(2^m-1)) = S(n+1,m) for n >= 0, m >= 0 where S(n,m) = Sum_{k=1..n} k!*k^m*Stirling2(n,k)*(-1)^(n-k) (equivalent to proposition 6.5 at the page 297, see R. Ehrenborg and E. Steingrimsson link).

%F Conjecture: a(n) = (1 + A023416(n))*a(g(n)) + Sum_{k=0..floor(log_2(n))-1} (1-R(n,k))*a(g(n) + 2^k*(1 - R(n,k))) for n > 1 with a(0) = 1, a(1) = 1, where g(n) = A053645(n) and where R(n,k) = floor(n/2^k) mod 2 (at this moment this is the only formula here, which is not related to R. Ehrenborg's and E. Steingrimsson's work and arises from another definition given above, exactly conjectured definition with a biased rook). Here R(n,k) is the (k+1)-th bit from the right side in the binary expansion of n. - _Mikhail Kurkov_, Jun 23 2021

%F Conjectures from _Mikhail Kurkov_, Jan 23 2023: (Start)

%F The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.

%F a(n) = A357990(n, 1) for n >= 0.

%F a(2^m*(2k+1)) = Sum_{i=1..wt(k) + 2} i!*i^m*A358612(k, i)*(-1)^(wt(k) - i + 2) for m >= 0, k >= 0 where wt(n) = A000120(n).

%F a(2^m*(2^n - 2^p - 1)) = Sum_{i=1..n} i!*i^m*(-1)^(n - i)*((i - p + 1)*Stirling2(n, i) - Stirling2(n - p, i - p) + Sum_{j=0..p-2} (p - j - 1)*Stirling2(n - p, i - j)/j! Sum_{k=0..j} (i - k)^p*binomial(j, k)*(-1)^k) for n > 2, m >= 0, 0 < p < n - 1. Here we consider that Stirling2(n, k) = 0 for n >= 0, k < 0. (End)

%F Conjecture: a(2^m*n + q) = Sum_{i=A001511(n+1)..A000120(n)+1} A373183(n, i)*a(2^m*(2^(i-1)-1) + q) for n >= 0, m >= 0, q >= 0. Note that this formula is recursive for n != 2^k - 1. Also, it is not related to R. Ehrenborg's and E. Steingrimsson's work. - _Mikhail Kurkov_, Jun 05 2024

%F From _Mikhail Kurkov_, Jul 10 2024: (Start)

%F Conjecture: a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*(-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) for m >= 0, n >= 0, k >= 0 with a(0) = 1.

%F Proof: start with a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n) given above and rewrite it as a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) a(2^i*(2^(n-1)*(2k+1) - 1)).

%F Then conjecture that a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=1..m+1} a(2^i*k)*f(n, m, i). From that it is obvious that f(0, m, i) = [i = (m+1)].

%F After that use a(2^m*(2^n*(2k+1) - 1)) = Sum_{i=0..m} binomial(m+1, i) Sum_{j=1..i+1} a(2^j*k)*f(n-1, i, j) = Sum_{i=1..m+1} a(2^i*k) Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i). From that it is obvious that f(n, m, i) = Sum_{j=i-1..m} binomial(m+1, j)*f(n-1, j, i).

%F Finally, all we need is to show that basic conditions and recurrence for f(n, m, i) gives f(n, m, i) = (-1)^(m-i+1)*Sum_{j=i..m+1} j^n*Stirling1(j, i)*Stirling2(m+1, j) (see _Max Alekseyev_ link).

%F Conjecture: a(2^m*(2k+1)) = a(2^(m-1)*k) + (m+1)*a(2^m*k) + Sum_{i=1..m-1} a(2^m*k + 2^i) for m > 0, k >= 0.

%F Proof: start with a(2^(m+1)*(2k+1)) = a(2^m*k) + (m+2)*a(2^(m+1)*k) + Sum_{i=1..m} a(2^(m+1)*k + 2^i).

%F Then use a(2^m*(4k+1)) = a(2^m*k) + (m+1)*a(2^(m+1)*k) + Sum_{i=1..m-1} a(2^(m+1)*k + 2^i).

%F From that we get a(2^(m+1)*(2k+1)) - a(2^m*k) - (m+2)*a(2^(m+1)*k) - a(2^(m+1)*k + 2^m) = a(2^m*(4k+1)) - a(2^m*k) - (m+1)*a(2^(m+1)*k).

%F Finally, a(2^(m+1)*(2k+1)) = a(2^(m+1)*k) + a(2^m*(2*k+1)) + a(2^m*(4k+1)) which agrees with the a(2^m*(2n+1)) = a(2^m*n) + a(2^(m-1)*(2n+1)) + a(2^(m-1)*(4n+1)) given above.

%F This formula can be considered as an alternative to a(2^m*(2n+1)) = Sum_{k=0..m} binomial(m+1,k) a(2^k*n). There are algorithms for both these formulas that allow you to calculate them without recursion. However, even though it is necessary to calculate binomial coefficients in the mentioned formula, the triple-looped algorithm for it still works faster (see _Peter J. Taylor_ link).

%F It looks like you can also change v2 in the mentioned algorithm to vector with elements a(2^m*(2^(i+A007814(n+1)-1)-1) + q) to get a(2^m*n + q) instead of a(n). This may have common causes with formula that uses A373183 given above. (End)

%e a(1) = 1 because the 1st excedance set is {m-1} and the permutations of {1,2,...,m} with such excedance set are 21, 132, 1243, 12354 and so on, i.e., for a given m we always have 1 permutation.

%e a(2) = 3 because the 2nd excedance set is {m-2} and the permutations of {1,2,...,m} with such excedance set are 213, 312, 321, 1324, 1423, 1432, 12435, 12534, 12543 and so on, i.e., for a given m we always have 3 permutations.

%e a(3) = 1 because the 3rd excedance set is {m-2, m-1} and the permutations of {1,2,...,m} with such excedance set are 231, 1342, 12453 and so on, i.e., for a given m we always have 1 permutation.

%p g:= proc(n) option remember; 2^padic[ordp](n, 2) end:

%p a:= proc(n) option remember; `if`(n=0, 1, (h-> a(h)+

%p `if`(n::odd, 0, (t-> a(h-t)+a(n-t))(g(h))))(iquo(n, 2)))

%p end:

%p seq(a(n), n=0..100); # _Alois P. Heinz_, Jan 30 2023

%t a[n_] := a[n] = Which[n == 0, 1, OddQ[n], a[(n-1)/2], True, a[n/2] + a[n/2 - 2^IntegerExponent[n/2, 2]] + a[n - 2^IntegerExponent[n/2, 2]]];

%t a /@ Range[0, 65] (* _Jean-François Alcover_, Feb 13 2020 *)

%o (PARI) upto(n) = my(A, v1); v1 = vector(n+1, i, 0); v1[1] = 1; for(i=1, n, v1[i+1] = v1[i\2+1] + if(i%2, 0, A = 1 << valuation(i/2, 2); v1[i/2-A+1] + v1[i-A+1])); v1 \\ _Mikhail Kurkov_, Jun 06 2024

%Y Cf. A000120, A001511, A004754, A007814, A023416, A036987, A053645, A062383, A063250, A089309, A091344, A091347, A091348, A110501, A110962, A136126, A152884 (lexicographic order), A329718, A340375, A344902, A344947, A357990, A358612, A373183.

%Y Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A341392, A347204, A347205.

%K nonn

%O 0,3

%A _Mikhail Kurkov_, Nov 12 2019