A050169 -id:A050169 - cp's OEIS frontend

A107711 Triangle read by rows: T(0,0)=1, T(n,m) = binomial(n,m) * gcd(n,m)/n.

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 5, 10, 5, 1, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1

Offset: 0

Leroy Quet, Jun 10 2005

T(0,0) is an indeterminate, but 1 seems a logical value to assign it. T(n,0) = T(n,1) = T(n,n-1) = T(n,n) = 1.

T(2n,n) = A001700(n-1) (n>=1). - Emeric Deutsch, Jun 13 2005

			T(6,2)=5 because binomial(6,2)*gcd(6,2)/6 = 15*2/6 = 5.
The triangle T(n,m) begins:
n\m 0  1  2   3   4    5   6   7  8  9  10...
0:  1
1:  1  1
2:  1  1  1
3:  1  1  1   1
4:  1  1  3   1   1
5:  1  1  2   2   1    1
6:  1  1  5  10   5    1   1
7:  1  1  3   5   5    3   1   1
8:  1  1  7   7  35    7   7   1  1
9:  1  1  4  28  14   14  28   4  1  1
10: 1  1  9  12  42  126  42  12  9  1   1
n\m 0  1  2   3   4    5   6   7  8  9  10...
... reformatted - _Wolfdieter Lang_, Feb 23 2014

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened
Wolfdieter Lang, On Collatz' Words, Sequences and Trees, arXiv:1404.2710 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.11.7.

Row sums A159554.
Cf. A007318, A001700, A000108, A050169, A234040, A026741, A234042.
Cf. A109004.

a107711 n k = a107711_tabl !! n !! k
a107711_row n = a107711_tabl !! n
a107711_tabl = [1] : zipWith (map . flip div) [1..]
               (tail $ zipWith (zipWith (*)) a007318_tabl a109004_tabl)
-- Reinhard Zumkeller, Feb 28 2014

a:=proc(n,k) if n=0 and k=0 then 1 elif k<=n then binomial(n,k)*gcd(n,k)/n else 0 fi end: for n from 0 to 13 do seq(a(n,k),k=0..n) od; # yields sequence in triangular form. - Emeric Deutsch, Jun 13 2005

T[0, 0] = 1; T[n_, m_] := Binomial[n, m] * GCD[n, m]/n;
Table[T[n, m], {n, 1, 13}, {m, 1, n}] // Flatten (* Jean-François Alcover, Nov 16 2017 *)

From Wolfdieter Lang, Feb 28 2014 (Start)
T(n, m) = T(n-1,m)*(n-1)*gcd(n,m)/((n-m)*gcd(n-1,m)), n > m >= 1, T(n, 0) = 1, T(n, n) = 1, otherwise 0.
T(n, m) = binomial(n-1,m-1)*gcd(n,m)/m for n >= m >= 1, T(n,0) = 1, otherwise 0 (from iteration of the preceding recurrence).
T(n, m) = T(n-1, m-1)*(n-1)*gcd(n,m)/(m*gcd(n-1,m-1)) for n >= m >= 2, T(n, 0) = 1, T(n, 1) = 0, otherwise 0 (from the preceding formula).
T(2*n, n) = A001700(n-1) (n>=1) (see the Emeric Deutsch comment above), T(2*n, n-1) = A234040(n), T(2*n+1,n) = A000108(n), n >= 0 (Catalan numbers).
Column sequences: T(n+2, 2) = A026741(n+1), T(n+3, 3) = A234041(n), T(n+4, 4) = A208950(n+2), T(n+5, 5) = A234042, n >= 0. (End)

More terms from Emeric Deutsch, Jun 13 2005

A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 2, 2, 1, 1, 1, 5, 10, 5, 1, 1, 1, 3, 5, 5, 3, 1, 1, 1, 7, 7, 35, 7, 7, 1, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 1, 9, 12, 21, 126, 21, 12, 9, 1, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 1, 11, 55, 165, 66, 77, 66, 165, 55, 11, 1, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1

Offset: 0

Paul Curtz, May 24 2010

The fractions A053382(n,m)/A053383(n,m) give the triangle of the coefficients of the Bernoulli polynomials:
    1;
  -1/2,   1;
   1/6,  -1,    1;
    0,   1/2, -3/2,  1;
  -1/30,  0,    1,  -2,    1;
    0,  -1/6,   0,  5/3, -5/2,  1;
   1/42,  0,  -1/2,  0,   5/2, -3, 1;
The matrix inverse of this triangle defines coefficients of the umbral inverse Bernoulli polynomials B^{-1}(n,x) in row n:
   1;
  1/2,  1;
  1/3,  1,  1;
  1/4,  1, 3/2,   1;
  1/5,  1,  2,    2,   1;
  1/6,  1, 5/2, 10/3, 5/2,   1;
  1/7,  1,  3,    5,   5,    3,   1;
  1/8,  1, 7/2,   7, 35/4,   7,  7/2,  1;
  1/9,  1,  4,  28/3, 14,   14, 28/3,  4,  1;
  1/10, 1, 9/2,  12,  21, 126/5, 21,  12, 9/2, 1;
  1/11, 1,  5,   15,  30,   42,  42,  30, 15,  5, 1;
The current triangle T(n,m) is the numerator of the entry in row n and column m.
In the majority of cases, T(n,m) = A050169(n,m), but since we use the numerators of the reduced fractions, an integer factor may be missing in this equation.
Umbral composition (e.g., B(.,x)^k = B(k,x)) gives B^(-1)(n,B(.,x)) = x^n = B(n,B^(-1)(.,x)). - Tom Copeland, Aug 25 2015

			The triangle T(n,k) begins:
n\k 0 1  2  3   4   5   6    7   8   9  10 11 12 13
0:  1
1:  1 1
2:  1 1  1
3:  1 1  3  1
4:  1 1  2  2   1
5:  1 1  5 10   5   1
6:  1 1  3  5   5   3   1
7:  1 1  7  7  35   7   7    1
8:  1 1  4 28  14  14  28    4   1
9:  1 1  9 12  21 126  21   12   9   1
10: 1 1  5 15  30  42  42   30  15   5   1
11: 1 1 11 55 165  66  77   66 165  55  11  1
12: 1 1  6 22  55  99 132  132  99  55  22  6  1
13: 1 1 13 26 143 143 429 1716 429 143 143 26 13  1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

Cf. A178340 (denominators).

Cf. A074909, A258820, A003989, A053382, A053383.

nm := 15 : eM := Matrix(nm,nm) :
for n from 0 to nm-1 do for m from 0 to n do eM[n+1,m+1] :=coeff(bernoulli(n,x),x,m) ; end do: for m from n+1 to nm-1 do eM[n+1,m+1] := 0 ; end do: end do:
eM := LinearAlgebra[MatrixInverse](eM) :
for n from 1 to nm do for m from 1 to n do printf("%a,", numer(eM[n,m])) ; end do: end do: # R. J. Mathar, Dec 21 2010

max = 13; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes]; Table[ Take[inv[[n]], n], {n, 1, max}] // Flatten // Numerator (* Jean-François Alcover, Aug 09 2012 *)

tabl(nn) = {for (n=0, nn, for (k=0, n, print1(numerator(binomial(n+1,k)/(n+1)), ", ");); print(););} \\ after Tom Copeland comment; Michel Marcus, Jul 25 2015

"Palindromic:" T(n,m+1) = T(n,n-m). T(n,0)=1.
From Tom Copeland, Jun 18 2015: (Start)
The umbral inverse Bernoulli polynomials are Binv(n,x) = [(1+x)^(n+1)-x^(n+1)]/(n+1) with the e.g.f. e^(t*x) * (e^t-1)/t. (See A074909 for more details.) Therefore, T(n,k) is the numerator of the reduced fraction C(n+1,k)/(n+1) for 0 <= k < (n+1).
The reversed rows are presented as the diagonals of A258820.
T(n,k) = A258820(2n-k,n-k) = A003989(n+1,n+1-k) * n! / [ k! (n+1-k)! ], where A003989(j,k) = gcd(j,k). (End)
From Wolfdieter Lang, Aug 26 2015: (Start)
The following refers to the rational triangle TBinv with entries T(n,k)/A178340(n, m), n >= m >= 0.
The inverse of the Bernoulli triangle TB(n, m) with entries A196838(n,m)/A196839(n,m), n >= m >= 0, is the Sheffer triangle (z/(exp(z)-1),z). Therefore, the inverse triangle TBinv is the Sheffer triangle ((exp(z)-1)/z, z). This means that the e.g.f. of the sequence of column m of TBinv ((exp(x)-1)/x)*x^m/m! for m >= 0.
The e.g.f. of the row polynomials of TBinv, called Binv(n, x) = Sum_{m=0..n} TBinv(n,m)*x^m, is gBinv(z,x) = ((exp(z)-1)/z)*exp(x*z) (of the so-called Appell type).
The e.g.f. of the row sums is gBinv(x,1).
The e.g.f. of the alternating row sums is gBinv(x,-1) = (1 - exp(-x))/x.
The e.g.f. of the a-sequence of this Sheffer triangle is 1, and the e.g.f. of the z-sequence is (exp(x) - x -1)/((exp(x) -1)*x). This is the sequence 1/2, -1/12, 0, 1/120, 0, -1/252, 0, 1/240, 0, -1/132, .... For a- and z-sequences of Sheffer triangles and the corresponding recurrences see A006232.
The convolution property of the row polynomials Binv(n, x) is Binv(n, x+y) = Sum_{k=0..n} binomial(n, k)*Binv(n-k, x)*y^n (or with x and y exchanged).
The row polynomials satisfy (d/dx)Binv(n, x) = n*Binv(n-1, x), with Binv(0, x) = 1 (from Meixner's identity).
(End)

Redefined based on reduced fractions by R. J. Mathar, Dec 21 2010

The term umbral was added by Tom Copeland, Aug 25 2015

A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 5, 2, 1, 1, 3, 10, 1, 1, 7, 5, 5, 1, 1, 4, 7, 5, 1, 1, 9, 28, 35, 3, 1, 1, 5, 12, 14, 7, 1, 1, 11, 15, 21, 14, 7, 1, 1, 6, 55, 30, 126, 28, 1, 1, 13, 22, 165, 42, 21, 4, 1

Offset: 0

Tom Copeland, Jun 18 2015

The diagonals of T are the reversed rows of A178252. E.g., the fifth diagonal of T is (1,2,2,1,1) from the example below, which is the fifth reversed row of A178252.
Factoring out the greatest common divisor (gcd) of the coefficients of the sub-polynomials in the indeterminate q of the polynomials in s presented on p. 12 of the Alexeev et al. link and then evaluating the sub-polynomials at q=1 gives the first nine rows of T given in the example below. E.g., for k=6 (the seventh row), q*s^6 + (6*q + 9*q^2) s^4 + (15*q + 15*q^2) s^2 + 5  = q*s^6 + 3*(2*q + 3*q^2)*s^4 + 15*(q + q^2)*s^2 + 5 generates (1,2+3,1+1,1)=(1,5,2,1).
The row length sequence of this irregular triangle is A008619(n) = 1 + floor(n/2). - Wolfdieter Lang, Aug 25 2015

			The irregular triangle T(n,k) starts
n\k  0 1  2  3 4 5 ...
0:   1
1:   1
2:   1 1
3:   1 1
4:   1 3  1
5:   1 2  1
6:   1 5  2  1
7:   1 3 10  1
8:   1 7  5  5 1
9:   1 4  7  5 1
10:  1 9 28 35 3 1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015

N. Alexeev, J. Andersen, R. Penner, P. Zograf, Enumeration of chord diagrams on many intervals and their non-orientable analogs, arXiv:1307.0967 [math.CO], 2013-2014.

Cf. A050169, A161642, A055151, A088617, A178252, A107711, A165661, A003989, A008619.

max = 15; coes = Table[ PadRight[ CoefficientList[ BernoulliB[n, x], x], max], {n, 0, max-1}]; inv = Inverse[coes] // Numerator; t[n_, k_] := inv[[n, k]]; t[n_, k_] /; k == n+1 = 1; Table[t[n-k+1, k], {n, 2, max+1}, {k, 2, Floor[n/2]+1}] // Flatten (* Jean-François Alcover, Jul 22 2015 *)

T(n,k) = A178252(n-k,n-2k) = A055151(n,k) / A161642(n,k) = A007318(n,2k) * A000108(k) / A161642(n,k) = n! / [(n-2k)! k! (k+1)! A161642(n,k)] = A003989(n-k+1,k+1) * (n-k)! / [ (n-2k)! (k+1)! ], where A003989(j,k) = gcd(j,k).

A002784 A problem in parity.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 1, 4, 5, 1, 3, 1, 3, 1, 1, 8, 15, 3, 7, 4, 5, 2, 3, 3, 6, 2, 3, 2, 3, 1, 1, 16, 19, 7, 10, 5, 15, 4, 5, 7, 15, 3, 7, 4, 5, 2, 3, 5, 13, 3, 5, 4, 7, 1, 3, 3, 5, 2, 3, 1, 3, 1, 1, 32, 47, 11, 31, 14, 21, 6, 15, 11, 31, 7, 9, 7, 12, 3, 7, 12, 21, 7, 15, 4, 11, 4, 5, 7, 15, 2, 7, 4, 5

Offset: 1

N. J. A. Sloane

nonn

H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. MR0260659
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. [Annotated scanned copy]

Cf. A050169, A073166.

a(n)=if(n<1,0,sum(k=1,n,if(gcd(n,k) == 1,((n+k-1)!/(n!*k!))%2))) /* Michael Somos, Jul 19 2002 */

More terms from Michael Somos, Jul 19 2002

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A073166 Duplicate of A050169.

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A107711 Triangle read by rows: T(0,0)=1, T(n,m) = binomial(n,m) * gcd(n,m)/n.

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A178252 Triangle T(n,m) read by rows: the numerator of the coefficient [x^m] of the umbral inverse Bernoulli polynomials B^{-1}(n,x), 0 <= m <= n.

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A258820 Reversed rows of A178252 presented as diagonals of an irregular triangle.

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A002784 A problem in parity.

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