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

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

0,4

Comments

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.

Links

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

Formula

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

Example

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.

Mathematica

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

Prog

(PARI) for(n=0, 10, for(k=0, n, print1(n/gcd(n, binomial(n, k)), ", "))) \\ G. C. Greubel, Jun 25 2018
(Magma) /* As triangle */ [[n/GCD(n, Binomial(n, k)): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Jun 25 2018

Crossrefs

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

Sequence in context: A133232 A137152 A328902 * A109004 A103823 A355246

Adjacent sequences: A159332 A159333 A159334 * A159336 A159337 A159338

Keyword

frac,nonn,tabl

Author

Leroy Quet, Apr 10 2009

Extensions

Extended by Ray Chandler, Jun 19 2009

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

Status

approved