A329369 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.
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, 7, 115, 15, 31, 1, 63, 31, 77, 15, 145, 37, 81, 7, 245, 69, 155, 17, 261, 37, 77, 3, 391, 115, 261, 31, 445, 69, 145, 7, 675, 115, 245, 15, 391, 31, 63, 1, 127, 63
Offset: 0
Examples
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.
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.
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.
Links
- Mikhail Kurkov, Table of n, a(n) for n = 0..8191
- Max Alekseyev, Recursion for the sum with Stirling numbers of both kinds, answer to question on MathOverflow (2024).
- R. Ehrenborg and E. Steingrimsson, The excedance set of a permutation, Advances in Appl. Math., 24, 284-299, 2000.
- Peter J. Taylor, Correctness of the algorithm for the A329369, A347205 and related sequences, answer to question on MathOverflow (2024).
Crossrefs
Cf. A000120, A000523, A001511, A007814, A008292, A023416, A036987, A053645, A062253, A062254, A062255, A063250, A080791, A091344, A091347, A091348, A110501, A136126, A152884 (lexicographic order), A173018, A329718, A344902, A344947, A357990, A358612, A373183, A379817.
Similar recurrences: A006014, A124758, A217924, A243499, A284005, A290615, A341392, A347204, A347205.
Cf. A360288.
Programs
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Maple
g:= proc(n) option remember; 2^padic[ordp](n, 2) end: a:= proc(n) option remember; `if`(n=0, 1, (h-> a(h)+ `if`(n::odd, 0, (t-> a(h-t)+a(n-t))(g(h))))(iquo(n, 2))) end: seq(a(n), n=0..100); # Alois P. Heinz, Jan 30 2023 -
Mathematica
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]]]; a /@ Range[0, 65] (* Jean-François Alcover, Feb 13 2020 *)
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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
Formula
a(2n+1) = a(n) for n >= 0.
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).
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).
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).
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)).
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))).
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 1: this formula is equivalent to inverse modulo 2 binomial transform of A284005.
Sum_{k=0..2^n-1} a(k) = (n+1)! for n >= 0.
a((4^n-1)/3) = A110501(n+1) for n >= 0.
a(2^2*(2^n-1)) = A091344(n+1),
a(2^3*(2^n-1)) = A091347(n+1),
a(2^4*(2^n-1)) = A091348(n+1).
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).
Conjecture 2: 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
From Mikhail Kurkov, Jan 23 2023: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 3: a(n) = A357990(n, 1) for n >= 0.
Conjecture 4: a(2^m*(2k+1)) = Sum_{i=1..wt(k) + 2} i!*i^m*A358612(k, i)*(-1)^(wt(k) - i) for m >= 0, k >= 0 where wt(n) = A000120(n).
Conjecture 5: 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)
Conjecture 6: 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
From Mikhail Kurkov, Jul 10 2024: (Start)
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.
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)).
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)].
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).
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).
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.
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).
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).
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).
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.
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).
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)
From Mikhail Kurkov, Jan 27 2025: (Start)
The formulas below are not related to R. Ehrenborg's and E. Steingrimsson's work.
Conjecture 8: a(2^m*(2^n*(2k+1)-1)) = Sum_{i=0..m} Sum_{j=0..m-i} Sum_{q=0..i} binomial(m-i,j)*(m-j+1)^n*a(2^(q+1)*k)*L(m,i,q)*(-1)^j for m >= 0, n > 0, k >= 0 where L(n,k,m) = W(n-m,k-m,m+1) for n > 0, 0 <= k < n, 0 <= m <= k and where W(n,k,m) = (k+m)*W(n-1,k,m) + (n-k)*W(n-1,k-1,m) + [m > 1]*W(n,k,m-1) for 0 <= k < n, m > 0 with W(0,0,m) = 1, W(n,k,m) = 0 for n < 0 or k < 0.
In particular, W(n, k, 1) = A173018(n, k), W(n, k, 2) = A062253(n, k), W(n, k, 3) = A062254(n, k) and W(n, k, 4) = A062255(n, k).
Conjecture 9: a(n) = b(n,wt(n)) for n >= 0 where b(2n+1,k) = b(n,k) + (wt(n)-k+2)*b(n,k-1), b(2n,k) = (wt(n)-k+1)*b(2n+1,k) for n > 0, k > 0 with b(n,0) = A341392(n) for n >= 0, b(0,k) = 0 for k > 0 and where wt(n) = A000120(n) (see A379817).
More generally, a(2^m*(2k+1)) = ((m+1)!)^2*b(k,wt(k)-m) - Sum_{j=1..m} Stirling1(m+2,j+1)*a(2^(j-1)*(2k+1)) for m >= 0, k >= 0. Here we also consider that b(n,k) = 0 for k < 0. (End)
Conjecture 10: if we change b(n,0) = A341392(n) given above to b(n,0) = A341392(n)*x^n, then nonzero terms of the resulting polynomials for b(n,wt(n)) form c(n,k) such that a(n) = Sum_{k=0..A080791(n)} c(n,k) for n >= 0 where c(n,k) = (Product_{i=0..k-1} (1 + 1/A000120(floor(n/2^(A000523(n)-i))))) * Sum_{j=max{0,k-A080791(n)+A080791(A053645(n))}..A080791(A053645(n))} c(A053645(n),j) for n > 0, k >= 0 with c(0,0) = 1, c(0,k) = 0 for k > 0. - Mikhail Kurkov, Jun 19 2025
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