- #1
Raphie
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CONJECTURE:
Subtract the Absolute Values of the Stirling Triangle (of the first kind) from those of the Eulerian Triangle. When row number is equal to one less than a prime number, then all entries in that row are divisible by that prime number.
Take for instance, row 6 (see below). The differences between Stirling and Euler Entries are:
0, 42, 217,77,-217,-119
Divide each value by 7 and you get...
0, 6, 31, 11, -31, -17
Note: Row numbers designations are callibrated to n!/(n-1)!, where n! is the row sum...
Stirling Triangle of First Kind (positive and negative signs not shown...)
http://oeis.org/A094638
http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind
1 --> Row 1
1 01 --> Row 2
1 03 002 --> Row 3
1 06 011 006 --> Row 4
1 10 035 050 024 --> Row 5
1 15 085 225 274 120 --> Row 6
Euler's Triangle (without 0-th row = 1 = 0!)
http://oeis.org/A008292
http://noticingnumbers.net/230EULERStriangle.htm
1 --> Row 1
1 01 --> Row 2
1 04 001 --> Row 3
1 11 011 001 --> Row 4
1 26 066 026 001 --> Row 5
1 57 302 302 057 001 --> Row 6
I have only checked this (by hand, not by computer) to Row 11 (more than a year ago). Why? Because I have been trying to look at number progressions (and matrices) as if I were living in the time of Euler, Gauss, etc.. The general hypothesis is that A) one can "discover" meaningful mathematics via observation, a general understanding of how various number progressions relate to one another, and a healthy dose of inductive logic backed by "mathematical facts," even if that "one" be a non-mathematician; and B) that such observations may be based upon very small sample sizes.
A few relevant points:
I. Both triangles are generated via recourse to Binomial Coefficients.
II. All entries in row p Pascal's Triangle of Pascal's Triangle, save the first and and last entries (both 1's), are divisible by p (for p a prime number).
III. The form p-1 figures prominently in both the Euler Totient Function and Wilson's Theorem.
A counter-example or lower bound to this conjecture, or better yet, a proof, would be most welcome. And I am not tied here to being "right." In fact, I would be far more surprised and intrigued should this conjecture prove false.Raphie
P.S. The Stirling Triangle of the First Kind is quite well known as it gives the coefficients of n-hedral generating polynomials. Euler's Triangle is less well known, but conceivably important if Frampton & Kephart were on the right track, even if not "right," in their 1999 paper:
Mersenne Primes, Polygonal Anomalies and String Theory Classification
http://arxiv.org/abs/hep-th/9904212
Subtract the Absolute Values of the Stirling Triangle (of the first kind) from those of the Eulerian Triangle. When row number is equal to one less than a prime number, then all entries in that row are divisible by that prime number.
Take for instance, row 6 (see below). The differences between Stirling and Euler Entries are:
0, 42, 217,77,-217,-119
Divide each value by 7 and you get...
0, 6, 31, 11, -31, -17
Note: Row numbers designations are callibrated to n!/(n-1)!, where n! is the row sum...
Stirling Triangle of First Kind (positive and negative signs not shown...)
http://oeis.org/A094638
http://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind
1 --> Row 1
1 01 --> Row 2
1 03 002 --> Row 3
1 06 011 006 --> Row 4
1 10 035 050 024 --> Row 5
1 15 085 225 274 120 --> Row 6
Euler's Triangle (without 0-th row = 1 = 0!)
http://oeis.org/A008292
http://noticingnumbers.net/230EULERStriangle.htm
1 --> Row 1
1 01 --> Row 2
1 04 001 --> Row 3
1 11 011 001 --> Row 4
1 26 066 026 001 --> Row 5
1 57 302 302 057 001 --> Row 6
I have only checked this (by hand, not by computer) to Row 11 (more than a year ago). Why? Because I have been trying to look at number progressions (and matrices) as if I were living in the time of Euler, Gauss, etc.. The general hypothesis is that A) one can "discover" meaningful mathematics via observation, a general understanding of how various number progressions relate to one another, and a healthy dose of inductive logic backed by "mathematical facts," even if that "one" be a non-mathematician; and B) that such observations may be based upon very small sample sizes.
A few relevant points:
I. Both triangles are generated via recourse to Binomial Coefficients.
II. All entries in row p Pascal's Triangle of Pascal's Triangle, save the first and and last entries (both 1's), are divisible by p (for p a prime number).
III. The form p-1 figures prominently in both the Euler Totient Function and Wilson's Theorem.
A counter-example or lower bound to this conjecture, or better yet, a proof, would be most welcome. And I am not tied here to being "right." In fact, I would be far more surprised and intrigued should this conjecture prove false.Raphie
P.S. The Stirling Triangle of the First Kind is quite well known as it gives the coefficients of n-hedral generating polynomials. Euler's Triangle is less well known, but conceivably important if Frampton & Kephart were on the right track, even if not "right," in their 1999 paper:
Mersenne Primes, Polygonal Anomalies and String Theory Classification
http://arxiv.org/abs/hep-th/9904212
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