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 A060351 If the binary expansion of n has k bits, let S be the subset of [k-1] such that i is in S if the i-th bit of n is a 1 (with the first bit being the least significant bit); a(n) is the number of permutations of [k] with descent set S. 19
 1, 1, 1, 1, 1, 2, 2, 1, 1, 3, 5, 3, 3, 5, 3, 1, 1, 4, 9, 6, 9, 16, 11, 4, 4, 11, 16, 9, 6, 9, 4, 1, 1, 5, 14, 10, 19, 35, 26, 10, 14, 40, 61, 35, 26, 40, 19, 5, 5, 19, 40, 26, 35, 61, 40, 14, 10, 26, 35, 19, 10, 14, 5, 1, 1 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,6 COMMENTS a(n) is the number of permutations in the symmetric group S_k such that n = 2^(k-1) + the sum of 2^(i-1), where i is a descent of the permutation and k = number of digits in the binary expansion of n. If n=4m then a(n)-a(n+1)+a(n+2)-a(n+3) = 0. This follows from Theorem 10 of my paper arXiv:0801.0072v1. E.g., a(20)-a(21)+a(22)-a(23) = 9-16+11-4 = 0. - Vladimir Shevelev, Jan 07 2008 Denote by {n,k} the number of permutations of {0,1,...n} such that the binary expansion of k with n-1 digits (the expansion is allowed to begin with 0's) indicates a fixed distribution of "up"(1) and "down"(0) points. The numbers {n,k} are called "up-down coefficients" of permutations, since they have many similar properties to binomial coefficients C(n,k) (see Shevelev et al. references). The sequence lists the rows numbers {n,k} as a triangle read by rows (see example). - Vladimir Shevelev, Feb 13 2014 REFERENCES I. Niven, A combinatorial problem of finite sequences, Nieuw Arch. Wisk. (3) 16 (1968), 116-123. LINKS Alois P. Heinz, Rows n = 0..14, flattened (rows n=1..12 from Wouter Meeussen) N. G. de Bruijn, Permutations with given ups and downs, Nieuw Arch. 18 (1970), 61-65. Vladimir Shevelev, On the Basis Polynomials in the Theory of Permutations with Prescribed Up-Down Structure, arXiv:0801.0072 [math.CO], 2007-2010. See remarks following Theorem 22. Vladimir Shevelev, The number of permutations with prescribed up-down structure as a function of two variables, INTEGERS, 12 (2012), #A1. V. Shevelev and J. Spilker, Up-down coefficients for permutations, Elem. Math. 68 (2013), 115-127. FORMULA {n+1,2*k} + {n+1,2*k+1) = (n+1)*{n,k}, {n+2,4*k} + {n+2,4*k+2} = {n+2, 4*k+1} + {n+2,4*k+1} + {n+2,4*k+3} = (n+2)*(n+1)/2 * {n,k}, etc. Sum_{i=0..2^r-1} {n,i} = n*(n-1)*...*(n-r+1). For n >= 1, 0 <= k < 2^(n-1), {n,k} <= {n,r_n}, where r_n=(2^n-2)/3, if n is odd, r_n=(2^n-1)/3, if n is even. Equality holds iff k=r_n or 2^(n-1)-r_n-1, which corresponds the case of alternating permutations. De Bruijn mentioned that Niven knew the latter result, but he never published this statement. A proof can be found in the Shevelev and Spilker reference (Section 5). Many other equalities, recursions and unequalities can be found in Shevelev and Shevelev-Spilker references. - Vladimir Shevelev, Feb 13 2014 EXAMPLE Interpreted as a triangle: 1; 1; 1, 1; 1, 2, 2, 1; 1, 3, 5, 3, 3, 5, 3, 1; 1, 4, 9, 6, 9, 16, 11, 4, 4, 11, 16, 9, 6, 9, 4, 1; 1, 5, 14, 10, 19, 35, 26, 10, 14, 40, 61, 35, 26, 40, 19, 5, 5, 19, 40, 26, 35, 61, 40, 14, 10, 26, 35, 19, 10, 14, 5, 1; ... From Vladimir Shevelev, Feb 13 2014: (Start) Consider {4,2} (see comments). k=010 (4-1 binary digits). So {4,2} is the number of down-up-down permutations of {1,2,3,4}. We have 5 such permutations (2,1,4,3),(3,1,4,2),(3,2,4,1),(4,1,3,2) and (4,2,3,1). Thus {4,2}=5. Over rows, the sequence has the form: {0,0} {1,0} {2,0} {2,1} {3,0} {3,1} {3,2} {3,3} {4,0} {4,1} {4,2} {4,3} {4,4} {4,5} {4,6} {4,7} ... such that the i-th row contains ceiling(2^(i-1)) entries with row sum i!, i>=0. (End) The binary expansion of n=11 is 1011, which has k=4 digits. Of the first k-1=3 bits, starting from the least significant bit on the right, the first and second are 1, so S={1,2}. The a(11)=3 permutations of [k]={1,2,3,4} with descent set S={1,2} are {3,2,1,4}, {4,2,1,3}, and {4,3,1,2}. - Danny Rorabaugh, Apr 02 2015 MAPLE ct := proc(k) option remember; local i, out, n; if k=0 then RETURN(1); fi; n := floor(evalf(log[2](k)))+1; if k=2^n or k=2^(n+1)-1 then RETURN(1); fi; out := 0; for i from 1 to n do if irem(iquo(k, 2^(i-1)), 2) = 1 and irem(iquo(2*k, 2^(i-1)), 2) =0 then out := out+(n-1)!/(i-1)!/(n-i)!* ct(floor(irem(k, 2^(i-1))+2^(i-2)))*ct(iquo(k, 2^i)); fi; od; out; end: seq(ct(i), i=0..64); # second Maple program: b:= proc(u, o, t) option remember; expand(`if`(u+o=0, 1, add(b(u-j, o+j-1, t+1)*x^floor(2^(t-1)), j=1..u)+ add(b(u+j-1, o-j, t+1), j=1..o))) end: T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n, 0\$2)): seq(T(n), n=0..7); # Alois P. Heinz, Sep 08 2020 # third Maple program: b:= proc(u, o, t) option remember; `if`(u+o=0, `if`(t=0, 1, 0), `if`(irem(t, 2)=0, add(b(u-j, o+j-1, iquo(t, 2)), j=1..u), add(b(u+j-1, o-j, iquo(t, 2)), j=1..o))) end: T:= (n, k)-> b(n, 0, 2*k): seq(seq(T(n, k), k=0..ceil(2^(n-1))-1), n=0..7); # Alois P. Heinz, Sep 12 2020 MATHEMATICA <

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