@Israel2.3.2

3 years ago

I enjoyed this.Reminded me of the early days just staring at tables of numbers trying to understand the cosmos. 3:43 The reason that successive box/fence entries have differences of 4 in the even case and fail to have differences of 4 in the odd case is an easy consequence of prime factorization and properties of finite difference equations. Even squares have the form 4×1, 4×4, 4×9, 4×16, etc. After division this yields box/fence sequence in the even case as 4×1, 4×2, 4×3, 4×4, etc. Hence we have a sequence whose differences are constant with value 4. In order for a sequence to have constant difference 4 it is necessary that the nth term have the linear form 4n + c for some constant c. Forming box/fence in the odd case yields (2n-1)×(2n-1)÷n for the general term. Equating this with the desired form 4n + c and simplifying yields n(c+4) = 1 But n is a variable quantity while c is constant so this identity cannot be satisfied in general, hence we cannot have divisibility by 4 in the odd case. 3:58 The modified sequence replaces odd squares with odd squares shifted down by 1. If a number is odd it is of the form 4n + 1 or 4n + 3. So the odd squares have the forms 16nn + 8n + 1 and 16nn + 24n + 9 That is 4kn + 1 and 4sn + 9 for integers k = 4n + 2 and s = 4n + 6 Looking closely at these forms shows that odd squares are always 1 more than a multiple of 4. Damn that took too long to type. This fact about odd squares is an elementary but important result in elementary number theory and is easily proved using the language of congruences. 5:05 the constant differences of 8 given the slope 2 sequence 13, 58, 135, etc. can be shown by noticing that these values are displaced from the corners of the square in a regular manner, namely 13 = 4×4 - 3×1, 58 = 8×8 - 3×2, 135 = 12×12 - 3×3, etc. The general form of the nth value is 4n×4n - 3×n Forming the box/fence quotient gives the sequence 13÷2, 58÷4, 135÷6, etc Or 4×2 - 3÷2, 8×2 - 3÷2, 12×2 - 3÷2, etc. A sequence which has the general term 4n×2 - 3÷2 Or 8n - 1.5 For the general term in the box/fence sequence. Taking finite differences yields the constant difference 8 as desired. Now for your interpolation. If in our general term 4n×4n - 3n we substitute 4n + 2 for 4n then divide by 2n + 1 we get the box/fence sequence of the intermediate terms which have the form (16nn + 13n + 2.5) / (2n + 1) = 8n + 2.5 This lies between 8n - 1.5 and 8(n+1) - 1.5 in our interpolated box/fence sequence. Taking differences yields 4 as a constant difference. You asked whether sequences of larger slope will exhibit constant differences. The answer is yes. Every such sequence corresponding to slope m will have the form 2mn×2mn - (m+1)n for the nth term To get box/frame sequence we divide by frame mn which yields 4mn - (m+1)/m This is a linear sequence in n so that its constant difference is 4m. For slope 3 constant difference is 12, for slope 4 its 16, etc. One can derive your interpolation formulas with similar substitutions.

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