Proof of ∏(√n) Increment for Centered Polygonal Numbers w/ Prime Index

[JeremyEbert]

Is there a proof that ∏(√n) increments only when n is a centered polygonal number with a prime index?
∏(n) is the prime counting function
n=p^2-p+1 for a prime p
3, 7, 21, 43, 111, 157, 273, 343, 507, 813, 931, 1333...
http://oeis.org/A119959

 

[Norwegian]

Is there a proof that ∏(√n) increments only when n is a centered polygonal number with a prime index?

The statement in your question is not very precise. Obviously, ∏(√n) increments exactly when n=p², where p is a prime, and for no other n. But I suppose that is not what you are looking for, and I am curious to know what your centered polygonal numbers have to do with this.

 

[JeremyEbert]

The statement in your question is not very precise. Obviously, ∏(√n) increments exactly when n=p², where p is a prime, and for no other n. But I suppose that is not what you are looking for, and I am curious to know what your centered polygonal numbers have to do with this.

Norwegian,
Sorry for the extreme lack of precision. I definitely forgot to include a key part while typing this up.
Obviously, ∏(√n) increments exactly when n=p². However ∏( Floor(√n + 0.5) ) increments only when n is a centered polygonal number with a prime index. It seems like a proof of this should be pretty straight forward.

 

[Norwegian]

Hi Jeremy, if we instead look at ∏(1/2 + √n), this function jumps exactly when n=(p-1/2)²=p²-p+1/4. If we require n to be an integer, the jumps will indeed happen when n=p²-p+1.

 

[JeremyEbert]

Hi Jeremy, if we instead look at ∏(1/2 + √n), this function jumps exactly when n=(p-1/2)²=p²-p+1/4. If we require n to be an integer, the jumps will indeed happen when n=p²-p+1.

Ah yes, the floor function is a bit redundant. Thanks for that.
This obviously relates then to the square root of a oblong (pronic) number + 1/4 having a half-integer value. The name "oblong" indicating they are analogous to polygonal numbers.

 

[Anti-Crackpot]

Ah yes, the floor function is a bit redundant. Thanks for that.
This obviously relates then to the square root of a oblong (pronic) number + 1/4 having a half-integer value. The name "oblong" indicating they are analogous to polygonal numbers.

I'm not sure what you mean by "analogous to polygonal numbers." Oblong numbers map 1:1 to triangular numbers (= 2 times a triangular number) and triangular numbers are one kind of polygonal number, but Polygonal numbers in general?

It's true that the polygonal numbers can be constructed from the counting numbers + multiples of triangular numbers, but that involves first differences between polygonal numbers of equal index (e.g. the 5th square is 25 and 25 + 10, the fourth triangular number, = 35, the 5th Pentagonal number), not polygonal numbers themselves.

- AC

 

[JeremyEbert]

I'm not sure what you mean by "analogous to polygonal numbers." Oblong numbers map 1:1 to triangular numbers (= 2 times a triangular number) and triangular numbers are one kind of polygonal number, but Polygonal numbers in general?

It's true that the polygonal numbers can be constructed from the counting numbers + multiples of triangular numbers, but that involves first differences between polygonal numbers of equal index (e.g. the 5th square is 25 and 25 + 10, the fourth triangular number, = 35, the 5th Pentagonal number), not polygonal numbers themselves.

- AC

A pronic + 1 is a centered polygonal number. ∏(1/2 + √n) gives no information about ∏ itseft, it is trivial.