A165661 -id:A165661 - 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).

A159335 Triangle read by rows: numerator of n/binomial(n,m).

Original entry on oeis.org

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

Offset: 0

Leroy Quet, Apr 10 2009

This triangle first differs from A109004 (read as a triangle) at T(10, 4) and T(10,6).

T(n,m) is the smallest positive integer such that binomial(n,m)*T(n,m) is a multiple of n.

			Row 10 of Pascal's triangle is: 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1. {a(10,m)} of this sequence (A159335) is: 10, 1, 2, 1, 1, 5, 1, 1, 2, 1,10. Multiplying the corresponding integers, we get multiples of 10: 1*10=10,10*1=10, 45*2=90, 120*1=120, 210*1=210, 252*5=1260, 210*1=210, 120*1=120, 45*2=90, 10*1=10, 1*10=10.

G. C. Greubel, Rows n=0..100 of triangle, flattened

Cf. A165661 (denominators), A007318, A020475, A109004.

/* As triangle */ [[n/GCD(n,Binomial(n, k)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jun 25 2018

Table[n/GCD[n, Binomial[n, k]], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Jun 25 2018 *)

for(n=0, 10, for(k=0,n, print1(n/gcd(n, binomial(n,k)), ", "))) \\ G. C. Greubel, Jun 25 2018

T(n,m) = n/gcd(n,binomial(n,m)).

Extended by Ray Chandler, Jun 19 2009

Edited by Franklin T. Adams-Watters, Sep 24 2009

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

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