A178252 -id:A178252 - cp's OEIS frontend

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

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).

A050169 Triangle read by rows: T(n,k) = gcd(C(n,k), C(n,k-1)), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 5, 10, 5, 1, 1, 3, 5, 5, 3, 1, 1, 7, 7, 35, 7, 7, 1, 1, 4, 28, 14, 14, 28, 4, 1, 1, 9, 12, 42, 126, 42, 12, 9, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 11, 55, 165, 66, 462, 66, 165, 55, 11, 1, 1, 6, 22, 55, 99, 132, 132, 99, 55, 22, 6, 1

Offset: 1

Clark Kimberling

Equivalently, table T(n,k) = gcd(n,k)*(n+k-1)!/(n!*k!) read by antidiagonals. - Michael Somos, Jul 19 2002

Apparently, T(n,k)*gcd(C(n+1,k),n+1) = C(n+1,k). - Thomas Anton, Oct 24 2018

			Triangle starts:
  1;
  1,  1;
  1,  3,  1;
  1,  2,  2,  1;
  1,  5, 10,  5,  1;
  1,  3,  5,  5,  3,  1;
  ...

H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. MR0260659

Muniru A Asiru, Table of n, a(n) for n = 1..1275(Rows n=1..50,flattened)
H. Gupta, On a problem in parity, Indian J. Math., 11 (1969), 157-163. [Annotated scanned copy]

Cf. A002784, A178252.

Flat(List([1..12],n->List([1..n],k->Gcd(Binomial(n,k),Binomial(n,k-1))))); # Muniru A Asiru, Oct 24 2018

/* As triangle */ [[Gcd(Binomial(n,k), Binomial(n,k-1)): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 25 2018

a:=(n,k)->gcd(binomial(n,k),binomial(n,k-1)): seq(seq(a(n,k),k=1..n),n=1..12); # Muniru A Asiru, Oct 24 2018

Table[GCD@@{Binomial[n,k],Binomial[n,k-1]},{n,20},{k,n}]//Flatten (* Harvey P. Dale, Aug 06 2017 *)

T(n,k)=if(n<1 || k<1,0,gcd(n,k)*(n+k-1)!/n!/k!)

T(n,k)=if(k<1 || k>n,0,gcd(n+1,k)*binomial(n,k-1)/k) /* Michael Somos, Mar 03 2004 */

a(2n, n) = n-th Catalan number; see A000108.

Also T(n, k) = gcd(C(n, k), C(n+1, k)).

Offset set to 1 by R. J. Mathar, Dec 21 2010

A178340 Triangle T(n,m) read by rows: denominator of the coefficient [x^m] of the umbral inverse Bernoulli polynomial B^{-1}(n,x).

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 4, 1, 2, 1, 5, 1, 1, 1, 1, 6, 1, 2, 3, 2, 1, 7, 1, 1, 1, 1, 1, 1, 8, 1, 2, 1, 4, 1, 2, 1, 9, 1, 1, 3, 1, 1, 3, 1, 1, 10, 1, 2, 1, 1, 5, 1, 1, 2, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 12, 1, 2, 3, 4, 1, 1, 1, 4, 3, 2, 1, 13

Offset: 0

Paul Curtz, May 25 2010

This is the triangle of denominators associated with the numerators of A178252.
(Unlike the coefficients of the Bernoulli Polynomials, the coefficients of the umbral inverse Bernoulli polynomials are all positive.)
Usually T(n,m) = A003989(n-m+1,m) for m>=1, but since we are tabulating denominators of reduced fractions here, this formula may be wrong by a cancelling integer factor.

			The triangle T(n,m) begins:
n\m  0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 ...
0:   1
1:   2 1
2:   3 1 1
3:   4 1 2 1
4:   5 1 1 1 1
5:   6 1 2 3 2 1
6:   7 1 1 1 1 1 1
7:   8 1 2 1 4 1 2 1
8:   9 1 1 3 1 1 3 1 1
9:  10 1 2 1 1 5 1 1 2 1
10: 11 1 1 1 1 1 1 1 1 1  1
11: 12 1 2 3 4 1 1 1 4 3  2  1
12: 13 1 1 1 1 1 1 1 1 1  1  1  1
13: 14 1 2 1 2 1 2 7 2 1  2  1  2  1
14: 15 1 1 3 1 5 3 1 1 3  5  1  3  1  1
... reformatted. - _Wolfdieter Lang_, Aug 25 2015
-------------------------------------------------
The rational triangle TinvB(n,m):= A178252(n,m) / T(n,m) begins:
n\m    0 1   2    3    4     5    6  7   8  9 10
0:     1
1:   1/2 1
2:   1/3 1   1
3    1/4 1 3/2    1
4:   1/5 1   2    2    1
5:   1/6 1 5/2 10/3  5/2     1
6:   1/7 1   3    5    5     3    1
7:   1/8 1 7/2    7 35/4     7  7/2  1
8:   1/9 1   4 28/3   14    14 28/3  4   1
9:  1/10 1 9/2   12   21 126/5   21 12 9/2  1
10: 1/11 1   5   15   30    42   42 30  15  5  1
... - _Wolfdieter Lang_, Aug 25 2015
Recurrence from the Sheffer a-sequence:
Tinv(3,2) = (3/2)*TinvB(2,1) = (3/2)*1 = 3/2.
From the z-sequence: Tinv(3,0) = 3*Sum_{j=0..2} z_j*TinvB(2,j) = 3*((1/2)*(1/3) -(1/12)*1 + 0*1) = 3*(1/6 - 1/12) = 1/4. - _Wolfdieter Lang_, Aug 25 2015

Cf. A178252.

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 // Denominator (* Jean-François Alcover_, Aug 09 2012 *)

T(n,0) = n+1.
Recurrence for the rational triangle
TinvB(n,m):= A178252(n,m) / T(n,m) from the Sheffer a-sequence, which is 1, (repeat 0), see the comment under A178252: TinvB(n,m) = (n/m)*TinvB(n-1,m-1), for n >= m >= 1. From the z-sequence: TinvB(n,0) = n*Sum_{j=0..n-1} z_j * TinvB(n-1,j), n >= 1, TinvB(0,0) = 1. - Wolfdieter Lang, Aug 25 2015

A178623 Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.

Original entry on oeis.org

1, 2, 1, 3, 1, 1, 5, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 17, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 19, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 23, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 29

Offset: 0

Paul Curtz, May 31 2010

The sequence reflects a conjecture on the denominator of inverse Bernoulli polynomials in A178340: if the row index is one less than one of the primes in A008578, the row of denominators starts with that prime and contains 1's in the remaining entries.

[Row sums in A178252 are A159069(n+1), unless there is a common factor in numerator and denominator. The row sum over columns with index of the same parity as the row index in the table of fractions of the [x^m] B^{-1}(n,x) in A178252 are: 1, 1, 1/3+1=4/3, 1+1=2, 1/5+2+1=16/5, 1+10/3+1=16/3, 1/7+3+5+1=64/7, 16, 256/9, 256/5, 1024/11, 512/3, 496/13, ... =A084623(n+1)/A000265(n+1).]

			1;
2,1;
3,1,1;
5,1,1,1,1;
7,1,1,1,1,1,1;
11,1,1,1,1,1,1,1,1,1,1;
13,1,1,1,1,1,1,1,1,1,1,1,1;
17,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
19,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
23,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;
29,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1;

Cf. A076274 (row sums).

T(n,0) = A008578(n+1). T(n,m) =1, 1<=m<=A008578(n+1)-1.

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

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A050169 Triangle read by rows: T(n,k) = gcd(C(n,k), C(n,k-1)), n >= 1, 1 <= k <= n.

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A178340 Triangle T(n,m) read by rows: denominator of the coefficient [x^m] of the umbral inverse Bernoulli polynomial B^{-1}(n,x).

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A178623 Triangle T(n,m) read by rows: T(n,0)= prime(n); T(n,m)=1 if m>=1.

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