Vladimir Shevelev - cp's OEIS frontend

A294811 Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are triangular numbers (A000217). Then a(n) = b(n)/A010551(n).

1, 1, 1, 1, 2, 4, 6, 11, 16, 30, 48, 97, 157, 322, 524, 1077, 1777, 3684, 6157, 12876, 21684, 45520, 77212, 162533, 277608, 585993, 1006784, 2129433, 3677453, 7788711, 13514487, 28654668, 49933938, 105964856, 185377690, 393631445, 691101516, 1468137470

Offset: 0

Vladimir Shevelev and Peter J. C. Moses, Nov 09 2017

nonn

All terms are positive integers (for a proof, cf. comment in A293984). Note that a(1), a(2), a(3), a(4) remain the same if in the definition the triangular numbers are replaced by k-gonal numbers for k >= 5.

			Let n=3. For a permutation C={c_1,c_2,c_3}, set s = s(C) = c_1 - c_2 + c_3. We have the permutations:
1,2,3; s=2
1,3,2; s=0
2,1,3; s=4
2,3,1; s=0
3,1,2; s=4
3,2,1; s=2
Here there are 2 permutations for which {s} are triangular numbers (when s = 0). Further, since A010551(3) = 2, then a(3) = 1.
Let n=4. For a permutation C={c_1,c_2,c_3,c_4}, set s = s(C) = c_1 - c_2 + c_3 - c_4. We have the permutations:
1,2,3,4; s=-2
1,3,2,4; s=-4
2,1,3,4; s=0
2,3,1,4; s=-4
3,1,2,4; s=0
3,2,1,4; s=-2
1,2,4,3; s=0
1,3,4,2; s=0
2,1,4,3; s=2
2,3,4,1; s=2
3,1,4,2; s=4
3,2,4,1; s=4
1,4,2,3; s=-4
1,4,3,2; s=-2
2,4,1,3; s=-4
2,4,3,1; s=0
3,4,1,2; s=-2
3,4,2,1; s=0
4,1,2,3; s=2
4,1,3,2; s=4
4,2,1,3; s=0
4,2,3,1; s=4
4,3,1,2; s=0
4,3,2,1; s=2
Here there are 8 permutations for which {s} are triangular numbers (when s = 0). Further, since A010551(4) = 4, then a(4) = 8/4 = 2.

Alois P. Heinz, Table of n, a(n) for n = 0..250 (terms n=1..200 from Peter J. C. Moses)

Cf. A010551, A293857, A293984.

b:= proc(p, m, s) option remember; (n-> `if`(n=0, `if`(issqr(8*s+1), 1, 0),
      `if`(p>0, b(p-1, m, s+n), 0)+`if`(m>0, b(p, m-1, s-n), 0)))(p+m)
    end:
a:= n-> (t-> b(n-t, t, 0))(iquo(n, 2)):
seq(a(n), n=0..40);  # Alois P. Heinz, Sep 17 2020

polyQ[order_,n_]:=If[n==0,True,IntegerQ[(#-4+Sqrt[(#-4)^2+8 n (#-2)])/(2 (#-2))]&[order]];(*is a number polygonal?*)
Map[Total,Table[
possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2,Floor[(n+1)/2]^2];
filteredSums=Select[possibleSums,polyQ[3,#]&&#>-1&];
positions=Map[Flatten[{#,Position[possibleSums,#,1]-1}]&,filteredSums];
Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#[[2]]/2}]&,positions],{n,25}]] (* Peter J. C. Moses, Jan 02 2018 *)

a(0)=1 prepended by Alois P. Heinz, Sep 17 2020

A294812 Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are pentagonal numbers (A000326). Then a(n) = b(n)/A010551(n).

Original entry on oeis.org

1, 1, 1, 2, 4, 5, 6, 10, 23, 38, 70, 110, 196, 346, 759, 1250, 2313, 3982, 8433, 14520, 29437, 50466, 102830, 179587, 376439, 654374, 1343540, 2352149, 4916286, 8654120, 18065200, 31783592, 66233160, 117371504, 246610521, 436972949, 913862320, 1626523783

Offset: 1

Vladimir Shevelev and Peter J. C. Moses, Nov 09 2017

nonn

All terms are positive integers (for a proof, cf. comment in A293984).

Note that a(1), a(2), a(3), a(4) remain the same, if in the definition the pentagonal numbers are replaced by k-gonal numbers for k >= 3 other than k=4.

Peter J. C. Moses, Table of n, a(n) for n = 1..200

Cf. A010551, A293857, A293984, A294811.

polyQ[order_,n_]:=If[n==0,True,IntegerQ[(#-4+Sqrt[(#-4)^2+8 n (#-2)])/(2 (#-2))]&[order]];(*is a number polygonal?*)
Map[Total,Table[
possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2,Floor[(n+1)/2]^2];
filteredSums=Select[possibleSums,polyQ[5,#]&&#>-1&];
positions=Map[Flatten[{#,Position[possibleSums,#,1]-1}]&,filteredSums];
Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#[[2]]/2}]&,positions],{n,25}]] (* Peter J. C. Moses, Jan 02 2018 *)

A297003 a(n) = 2^(n-1), n=1,2,3; for n >= 4, a(n) is the number of the previous terms dividing n.

Original entry on oeis.org

1, 2, 4, 3, 1, 4, 2, 6, 3, 4, 2, 11, 2, 6, 4, 10, 2, 11, 2, 13, 4, 10, 2, 18, 2, 11, 4, 16, 2, 17, 2, 19, 7, 13, 3, 24, 2, 14, 7, 21, 2, 23, 2, 24, 5, 16, 2, 31, 4, 19, 6, 25, 2, 24, 6, 27, 7, 17, 2, 35, 2, 20, 9, 28, 5, 29, 2, 29, 6, 29, 2, 41, 2, 22, 8, 31

Offset: 1

Vladimir Shevelev, Dec 23 2017

			1-3) a(1)=1, a(2)=2 a(3)=4 by the definition;
4) Let n=4. From the previous terms {1,2,4} everyone divides 4, so a(4)=3;
5) Let n=5. From the previous terms {1,2,4,3} only 1 divides 5. So a(5)=1;
6) Let n=6. From the previous terms {1,2,4,3,1} exactly four divide 6. So a(6)=4; etc.

Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Cf. A088167.

first[n_] := Fold[Append[#1, Count[#1, k_ /; Divisible[#2, k]]] &,
  2^Range[0, Min[n - 1, 2]], Range[4, n]] (* Michael De Vlieger, Dec 23 2017 *)

first(n) = my(res = vector(n), c = 0); for(x = 1, min(n, 3), res[x] = 1<<(x-1)); for(x=4, n, for(k=1, x-1, if(x%res[k]==0, c++)); res[x] = c; c = 0); res \\ Iain Fox, Dec 23 2017

def A297003_list(leng):
    L = [1, 2, 4]
    if leng < 4: return L[0:leng]
    for n in (4..leng) :
        count = 0
        for l in L: count += int(l.divides(n))
        L.append(count)
    return L
print(A297003_list(76)) # Peter Luschny, Dec 24 2017

a(p) = 2, where p is prime, other than 3 and 5.

More terms from Peter J. C. Moses, Dec 23 2017

A295872 Decimal expansion of the first Ramanujan trigonometric constant (negated).

Original entry on oeis.org

7, 1, 7, 5, 1, 5, 0, 7, 9, 6, 4, 9, 9, 3, 9, 9, 3, 5, 1, 2, 0, 9, 5, 0, 5, 5, 9, 1, 7, 7, 9, 8, 6, 1, 1, 2, 1, 0, 8, 4, 5, 7, 6, 0, 1, 1, 5, 5, 2, 5, 0, 5, 7, 2, 1, 8, 3, 3, 0, 2, 8, 3, 0, 0, 2, 7, 9, 8, 1, 4, 6, 5, 0

Offset: 0

Vladimir Shevelev, Dec 09 2017

According to the famous Ramanujan identity, the constant r_1 has a representation: r_1 = Sum_{i = 1..3} (cos(2^i*Pi/7))^(1/3) (see formula). This identity was submitted in 1914 by Ramanujan as a problem (cf. [Berndt, Y. S. Choi, S. Y. Kang]). For proof, see first [V. Shevelev].

			r_1 =-0.7175150796499399351209505591779861121084576011552505721833028300279814650...

B. Bajorska-Harapinska, M. Pleszczynski, D. Slota and R. Witula, A few properties of Ramanujan cubic polynomials and Ramanujan cubic polynomials of the second kind, in book: Selected Problems on Experimental Mathematics, Gliwice 2017, pp. 181-200.
B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc. in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q524, JIMS VI, 1914).
S. Ramanujan, Notebooks (2 volumes), Tata Institute of Fundamental Research, Bombay, 1957.

B. C. Berndt, H. H. Chan, and L. C. Zhang, Radicals and units in Ramanujan's work, Acta Arith., 87 (1988), 145-158.
B. C. Berndt and S. Bhargava, Ramanujan - for Lowbrows, Amer. Math. Monthly, 100, no. 7, 1993, 644-656.
Vladimir Shevelev, Three Ramanujan's formulas, Kvant 6 (1988), 52-55 in Russian. English translation: Kvant Selecta 14 (1999), 139-144.
Vladimir Shevelev, On Ramanujan cubic polynomials, arXiv:0711.3420 [math.AC], 2007; South East Asian J. Math. & Math. Sci. 8 (2009), 113-122.

use RealDomain in solve(4*x^9 - 30*x^6 + 75*x^3 + 32 = 0) end use:
evalf(%, 79); # Peter Luschny, Dec 13 2017

RealDigits[(-(5 - 3*7^(1/3))/2)^(1/3), 10, 111][[1]] (* Robert G. Wilson v, Dec 13 2017 *)

-((3*7^(1/3)-5)/2)^(1/3) \\ Michel Marcus, Dec 10 2017

r_1 = ((5 - 3*7^(1/3))/2)^(1/3).

More terms from Michel Marcus, Dec 09 2017

A296448 Decimal expansion of the second Ramanujan trigonometric constant r_2.

Original entry on oeis.org

4, 9, 3, 4, 1, 4, 6, 2, 5, 9, 1, 8, 7, 8, 5, 6, 6, 4, 4, 2, 5, 6, 7, 2, 7, 5, 3, 3, 9, 3, 6, 7, 3, 4, 2, 6, 4, 3, 3, 7, 3, 7, 4, 7, 8, 3, 9, 9, 3, 7, 5, 0, 1, 8, 6, 3, 6, 6, 6, 4, 1, 7, 9, 5, 4, 9, 4, 7, 6, 7, 5, 8, 7, 8, 7, 8, 5, 9, 1, 8, 0, 5, 7, 4, 3, 2, 5, 1, 6, 9, 4, 1, 2, 9, 4, 5, 9, 7, 2, 4, 2, 8, 4, 0, 9

Offset: 0

Vladimir Shevelev, Dec 13 2017

According to the famous Ramanujan identity, the constant r_2 has a representation: r_2 = Sum_{i = 1..3} (cos(2^i*Pi/9))^(1/3) (see formula). This identity was submitted in 1914 by Ramanujan as a problem (cf. [Berndt, Y. S. Choi, S. Y. Kang]). For proof, see first [V. Shevelev].

			0.4934146259187856644256727533936734264337374783993750186366641795494767587...

B. Bajorska-Harapinska, M. Pleszczynski, D. Slota and R. Witula, A few properties of Ramanujan cubic polynomials and Ramanujan cubic polynomials of the second kind, in book: Selected Problems on Experimental Mathematics, Gliwice 2017, pp. 181-200.
S. Ramanujan, Notebooks (2 volumes), Tata Institute of Fundamental Research, Bombay, 1957.

B. C. Berndt, H. H. Chan, and L. C. Zhang, Radicals and units in Ramanujan's work, Acta Arith., 87 (1988), 145-158.
B. C. Berndt, Y. S. Choi, and S. Y. Kang, The problems submitted by Ramanujan to the Journal of Indian Math. Soc., in: Continued fractions, Contemporary Math., 236 (1999), 15-56 (see Q524, JIMS VI, 1914).
B. C. Berndt and S. Bhargava, Ramanujan - for Lowbrows, Amer. Math. Monthly, 100, no. 7, 1993, 644-656.
Vladimir Shevelev, Three Ramanujan's formulas, Kvant 6 (1988), 52-55 in Russian. English translation: Kvant Selecta 14 (1999), 139-144.
Vladimir Shevelev, On Ramanujan cubic polynomials, arXiv:0711.3420 [math.AC], 2007; South East Asian J. Math. & Math. Sci. 8 (2009), 113-122.

Cf. A295872.

use RealDomain in solve(8*x^9 + 72*x^6 + 216*x^3 - 27 = 0) end use:
evalf(%, 85); # Peter Luschny, Dec 13 2017

RealDigits[(3/2 (-2 + 3^(2/3)))^(1/3), 10, 111][[1]] (* Robert G. Wilson v, Dec 13 2017 *)

((3*9^(1/3) - 6)/2)^(1/3) \\ Michel Marcus, Dec 13 2017

r_2 = (3/2 (3^(2/3) -2))^(1/3)

More terms from Michel Marcus, Dec 13 2017

A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 2, 8, 0, 4, 0, 24, 0, 12, 0, 108, 0, 36, 576, 0, 720, 0, 144, 4608, 0, 4032, 0, 576, 0, 31680, 0, 31680, 0, 2880, 0, 288000, 0, 201600, 0, 14400, 2505600, 0, 2764800, 0, 1987200, 0, 86400, 30067200, 0, 28512000, 0, 14515200, 0, 518400

Offset: 1

Peter J. C. Moses and Vladimir Shevelev, Oct 18 2017

From Shevelev's comment in A008794 it follows that the last entries of rows, corresponding to the maximal possible i^2, form sequence A010551, n >= 1. Also note that the last entry of each row is the gcd of all its entries (for a proof see comment in A293857). - Vladimir Shevelev, Oct 26 2017

			Triangle begins
         0,      1;
         0,      1;
         2,      0,        2;
         8,      0,        4;
         0,     24,        0,     12;
         0,    108,        0,     36;
       576,      0,      720,      0,      144;
      4608,      0,     4032,      0,      576;
         0,  31680,        0,  31680,        0,  2880;
         0, 288000,        0, 201600,        0, 14400;
   2505600,      0,  2764800,      0,  1987200,     0,  86400;
  30067200,      0, 28512000,      0, 14515200,     0, 518400;
The compressed triangle resulting from the division of each entry by the last entry of its row begins as follows. If i is the index of the row, starting with i = 1 then this last entry is floor(i/2)! * (i - floor(i/2))!.
    0,   1;
    0,   1;
    1,   0,   1;
    2,   0,   1;
    0,   2,   0,   1;
    0,   3,   0,   1;
    4,   0,   5,   0,   1;
    8,   0,   7,   0,   1;
    0,  11,   0,  11,   0,   1;
    0,  20,   0,  14,   0,   1;
   29,   0,  32,   0,  23,   0,   1;
   58,   0,  55,   0,  28,   0,   1;
    0,  88,   0,  94,   0,  46,   0,   1;
    0, 169,   0, 146,   0,  53,   0,   1;
  263,   0, 282,   0, 283,   0,  86,   0,   1;
  526,   0, 515,   0, 383,   0,  97,   0,   1;
For the sense of the entries of this triangle see the [Shevelev] link (with a continuation there). Let B(n,i) be the set of permutations C of 1..n for which c_1 - c_2 + ... + (-1)^(n-1)*c_n = i^2, i >= 0. Then |B(n,i)| is the entry in the n-th row and i-th column of the first triangle. Let us call two permutations C_1 and C_2 equivalent if one of them is obtained from another by a permutation of its elements with odd indices and/or separately with even indices. Let b(n,i) be the entry in the n-th row and i-th column of the second triangle. Then b(n,i) is the maximal possible number of pairwise non-equivalent permutations which could be chosen in B(n,i). On the other hand, it is the smallest number of non-equivalent permutations in B(n,i) such that every other permutation in B(n,i) is equivalent to one of them. So in some sense b(n,i) is the dimension of B(n,i). In particular, b(n,i) = 0 corresponds to empty B(n,i). - _Vladimir Shevelev_, Nov 13 2017

Peter J. C. Moses, Table of the first 100 rows
Vladimir Shevelev, Basis in subsets of permutations described in A293783, SeqFan post Oct 27 2017.

Cf. A008794, A010551, A293857, A293984.

a293783=Flatten[Table[PadLeft[Riffle[#,Table[0,{Floor[(n-1)/4]}]/.{}->0],1+Floor[(1+n)/2]](Floor[n/2]!*(n-Floor[n/2])!)&[Reverse[Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#}]&,Map[2#(Floor[(n+1)/2] - #)&,Range[0,Floor[(n+1)/4]]]]]],{n,20}]] (* Peter J. C. Moses, Nov 01 2017 *)

Row sums of triangle give A293857.

If C = {c_1..c_n} is a permutation of {1..n}, then c_1 - c_2 + ... has the same parity as 1 + 2 + ... + n = n*(n+1)/2. So adjacent rows in the triangle for odd and even n have the same positions of 0's. These positions follow through one, beginning from the first position for n == 1,2 (mod 4) and from the second position for n == 3,0 (mod 4). - David A. Corneth and Vladimir Shevelev, Oct 19 2017

A293857 a(n) is the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are squares.

Original entry on oeis.org

1, 1, 1, 4, 12, 36, 144, 1440, 9216, 66240, 504000, 7344000, 73612800, 830995200, 9373190400, 181875456000, 2474319052800, 38246274662400, 572552876851200, 13783143886848000, 237527801118720000, 4658378696294400000, 86818505051013120000, 2488457229932298240000

Offset: 0

Vladimir Shevelev, Oct 17 2017

nonn

For a permutation C = {c_1..c_n} of {1..n}, set s(C) = c_1 - c_2 + ... + (-1)^(n-1)*c_n. Then max s(C) is square that is (ceil(n/2))^2 or A008794(n+1).
a(n)/n! is slowly and non-monotonically decreasing: 1, 1/2, 2/3, 1/2, 3/10, 1/5, 2/7, 8/35, 23/126, 5/36, 85/462, 71/462, ... .
Positions for which a(n) divisible by all primes <= n: 1, 4, 10, ... .
The smallest primes <= n not dividing a(n) or 0 if there is no such primes: 0, 2, 3, 0, 5, 5, 7, 5, 7, 0, 7, 7, ... .
Let k = floor(n / 2). Then a(n) = divisible by k! * (n-k)!. - David A. Corneth, Oct 18 2017. (For a proof, cf. comment in A293984. - Vladimir Shevelev, Nov 06 2017)

			Let n=3. For a permutation C={c_1,c_2,c_3}, set s = s(C) = c_1 - c_2 + c_3. We have the permutations:
1,2,3; s=2
1,3,2; s=0
2,1,3; s=4
2,3,1; s=0
3,1,2; s=4
3,2,1; s=2
Here there are 4 permutations for which {s} are squares. So a(3)=4.

Peter J. C. Moses, Table of n, a(n) for n = 0..200

Cf. A000290, A008794, A293783, A293984.

b:= proc(p, m, s) option remember; (n-> `if`(n=0, `if`(issqr(s), 1, 0),
      `if`(p>0, b(p-1, m, s+n), 0)+`if`(m>0, b(p, m-1, s-n), 0)))(p+m)
    end:
a:= n-> (t-> b(n-t, t, 0)*t!*(n-t)!)(iquo(n, 2)):
seq(a(n), n=0..28);  # Alois P. Heinz, Sep 17 2020

a293857=Table[Total[(Floor[n/2]!*(n-Floor[n/2])!)(Reverse[Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#}]&,Map[2#(Floor[(n+1)/2] - #)&,Range[0,Floor[(n+1)/4]]]]]
)],{n,25}] (* Peter J. C. Moses, Nov 01 2017 *)

From author's comment in A008794 it follows that a(n) >= A010551(n).

a(5)-a(12) from Peter J. C. Moses, Oct 17 2017
a(13)-a(23) from David A. Corneth, Oct 17 2017
a(0)=1 prepended by Alois P. Heinz, Sep 17 2020

A293984 a(n) = A293857(n)/A010551(n).

Original entry on oeis.org

1, 1, 1, 2, 3, 3, 4, 10, 16, 23, 35, 85, 142, 229, 369, 895, 1522, 2614, 4348, 10467, 18038, 32160, 54488, 130148, 226594, 414130, 710880, 1685496, 2958666, 5503780, 9544629, 22476690, 39724867, 74884360, 130949625, 306457174, 544777361, 1037587152, 1827129712

Offset: 0

Vladimir Shevelev, Oct 21 2017

nonn

Or row sums of the compressed triangle in A293783.
Conjecture: all terms are positive integers.
From David A. Corneth (with participation of Vladimir Shevelev), Oct 24 2017: (Start)
Conjecture is true. Proof.
1) Let C={c_1..c_n} be a permutation of {1..n}, d(C) be alternating sum c_1 - c_2 + ... +(-1)^(n-1)*c_n. Then max_{C in S_n}d(C) = A008794(n+1). Indeed, if n = 2*m, then evidently the maximum is reached on a C={2*m,1,2*m-1,2,...,m+1,m}; if n=2*m - 1, then the maximum is reached on a C={2*m-1,1,2*m-2,2,...,m-1,m}. In both cases max_{C in S_n}d(C) = m^2 = A008794(n+1). The number of distinct reaches of the maximum is, evidently, floor(n/2)!*floor((n+1)/2)! which is also Avi Peretz's representation (2001) of A010551(n). So, A293857(n) >= A010551(n) and a(n)>=1.
2) Consider two cases: a) there are no C in S_n for which d(C) = k^2 < A008794(n+1). Then A293857(n) = A010551(n) and a(n) = 1; b) there is C for which d(C) = k^2 < A008794(n+1). Then, as in 1) to reach k^2 in case n=2*m consider all (n/2)! permutations of {c_1,c_3,...,c_n} and all (n/2)! permutations of {c_2, c_4, ... , c_(n+1)}, or in case n = 2*m-1, all ((n+1)/2)! permutations of {c_1,c_3,...,c_(2*m-1)} and ((n-1)/2)! permutations of {c_2,c_4,...,c_(2*m-2)}. So we again have A010551(n) distinct reaches. If the same k^2 could be reached by another permutation C_1 (other than above permutations of C), then we again obtain A010551 distinct reaches, etc. So, A293857(n) is always divisible by A010551(n). (End)

Peter J. C. Moses, Table of n, a(n) for n = 0..250

Cf. A008794, A010551, A293783, A293857.

b:= proc(p, m, s) option remember; (n-> `if`(n=0, `if`(issqr(s), 1, 0),
      `if`(p>0, b(p-1, m, s+n), 0)+`if`(m>0, b(p, m-1, s-n), 0)))(p+m)
    end:
a:= n-> (t-> b(n-t, t, 0))(iquo(n, 2)):
seq(a(n), n=0..40);  # Alois P. Heinz, Sep 17 2020

a293984=Table[
possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2,Floor[(n+1)/2]^2];
filteredSums=Select[possibleSums,IntegerQ[Sqrt[#]]&];
positions=Map[Flatten[{#,Position[possibleSums,#,1]-1}]&,filteredSums];
Total[Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#[[2]]/2}]&,positions]],{n,20}] (* Peter J. C. Moses, Nov 05 2017 *)

a(13)-a(30) from David A. Corneth, Oct 21 2017; a(31)-a(38) from Peter J. C. Moses, Nov 02 2017

a(0)=1 prepended by Alois P. Heinz, Sep 17 2020

A293459 Denominator of probability that a permutation of elements of some subset of set {1,2..n} is a permutation of elements of some set of the form 1..k, k <= n.

Original entry on oeis.org

1, 1, 7, 385, 3628921, 1216452311762688721, 10333147966386144929717742279694909445041, 5989285834984945898036392571843137173092920925318860392502631168811983977451725959000900501504040321

Offset: 1

Vladimir Shevelev, Oct 09 2017

Amiram Eldar, Table of n, a(n) for n = 1..10

Cf. A293458(numerators).

a[n_] := Denominator[Sum[k!, {k, 0, n}]/Sum[Binomial[n, k]!, {k, 0, n}]]; Array[a, 8] (* Amiram Eldar, Sep 21 2019 *)

a(n) = denominator(sum(k=0, n, k!)/sum(k=0, n, binomial(n,k)!)); \\ Michel Marcus, Oct 12 2017

More terms from Peter J. C. Moses, Oct 09 2017

A293458 Numerator of probability that a permutation of elements of some subset of set {1,2,...,n} is a permutation of elements of some set of the form 1..k, k <= n.

Original entry on oeis.org

1, 1, 5, 17, 77, 437, 2957, 23117, 204557, 2018957, 21977357, 261478157, 3374988557, 46964134157, 700801318157, 11162196262157, 189005910310157, 3390192763174157, 64212742967590157, 1280663747055910157, 26826134832910630157, 588826498721714470157

Offset: 1

Vladimir Shevelev, Oct 09 2017

The number of all permutations of elements of sets {1..k}, k <= n, is b(n) = Sum_{k=0..n} k! while the number of all permutations of elements of all subsets of set {1,2..n} is c(n) = Sum_{k=0..n} binomial(n,k)!. So the required probability (in a sample space) is b(n)/c(n), n >= 1 (after reduction of the fractions).

Apparently a(n) = A014288(n) for n > 2. - Georg Fischer, Oct 23 2018

Amiram Eldar, Table of n, a(n) for n = 1..30

Denominators are in A293459.

Cf. A014288.

a[n_] := Numerator[Sum[k!, {k, 0, n}]/Sum[Binomial[n, k]!, {k, 0, n}]]; Array[a, 25] (* Amiram Eldar, Sep 21 2019 *)

a(n) = numerator(sum(k=0, n, k!)/sum(k=0, n, binomial(n,k)!)); \\ Michel Marcus, Oct 12 2017

More terms from Peter J. C. Moses, Oct 09 2017

cp's OEIS Frontend

User: Vladimir Shevelev

A294811 Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are triangular numbers (A000217). Then a(n) = b(n)/A010551(n).

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A294812 Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are pentagonal numbers (A000326). Then a(n) = b(n)/A010551(n).

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A297003 a(n) = 2^(n-1), n=1,2,3; for n >= 4, a(n) is the number of the previous terms dividing n.

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A295872 Decimal expansion of the first Ramanujan trigonometric constant (negated).

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A296448 Decimal expansion of the second Ramanujan trigonometric constant r_2.

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A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.

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A293857 a(n) is the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are squares.

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