A338432 Triangle read by rows: T(n, k) = (n - k + 1)^2 + 2*k^2, for n >= 1, and k = 1, 2, ..., n.

3, 6, 9, 11, 12, 19, 18, 17, 22, 33, 27, 24, 27, 36, 51, 38, 33, 34, 41, 54, 73, 51, 44, 43, 48, 59, 76, 99, 66, 57, 54, 57, 66, 81, 102, 129, 83, 72, 67, 68, 75, 88, 107, 132, 163, 102, 89, 82, 81, 86, 97, 114, 137, 166, 201

Offset: 1

Author

Wolfdieter Lang, Dec 09 2020

Keywords

• nonn
• tabl
• easy

Comments

This triangle is obtained from the array A(m, k) = m^2 + 2*k^2, for k and m >= 1, read by upwards antidiagonals. This array A is of interest for representing numbers as a sum of three non-vanishing squares with two squares coinciding.

For the numbers represented this way, see A154777. To find the actual values for m and k (taken positive), given a representable number from A154777, one can also use the number triangle T(n, k) = A(n-k+1, k).

To find the number of representations of value N (from A154777), it is sufficient to consider the rows n >= 1 not exceeding n_{max} = Floor(N, Min), where the sequence Min gives the minima of the numbers in each row: Min = {min(n)}_{n>=1} with min(n) = min(T(n, 1), T(n, 2), ..., T(n, n)) and Floor(N, Min) is the greatest member of Min not exceeding N.

Conjecture: min(n) = T(n, ceiling(n/3)), n >= 1. This is the sequence (n+1)^2 - ceiling(n/3)*(2*(n+1) - 3*ceiling(n/3)) = A071619(n+1) = ceiling((2/3)*(n+1)^2) = (n+1)^2 - floor((1/3)*(n+1)^2) = 3, 6, 11, 17, 24, 33, 43, .... (Proof of these identities by considering the three n (mod 3) cases.)

For the multiplicities of the representable values A154777(n), see A339047.

The author met this representation problem in connection with special triples of integer curvatures in the Descartes-Steiner five circle problem.

Examples

			The triangle T(n, k) begins:
n \ k  1   2   3   4   5   6   7   8   9  10  11  12 ...
1:     3
2:     6   9
3:    11  12  19
4:    18  17  22  33
5:    27  24  27  36  51
6:    38  33  34  41  54  73
7:    51  44  43  48  59  76  99
8:    66  57  54  57  66  81 102 129
9:    83  72  67  68  75  88 107 132 163
10:  102  89  82  81  86  97 114 137 166 201
11:  123 108  99  96  99 108 123 144 171 204 243
12:  146 129 118 113 114 121 134 153 178 209 246 289
...
----------------------------------------------------
T(5, 1) = 5^2 + 2*1^2 = 27 = T(5, 3) = 3^2 + 2*3^2. A338433(11) = 2 for A154777(11) = 27.
T(4, 4) = 1^2 + 2*4^2 = 33 = T(6, 2) = 5^2 + 2*2^2. A338433(12) = 2 for A154777(12) = 33.
T(5, 5) = 1^2 + 2*5^2 = 51 = T(7, 1) = 7^2 + 2*1^2. A338433(20) = 2 for A154777(20) = 51.
T(7, 7) = 1^1 - 2*7^2 = 99 = T(11, 3) = 9^2 + 2*3^2 = 99 = T(11, 5) = 7^2 + 2*5^2. A338433(39) = 3 for A154777(39) = 99.
The first multiplicity 4 appears for 297.
		

Crossrefs

Cf. A154777, A339047.

Cf. Columns k = 1..3: A059100, A189833, A241848.

Cf. Diagonals m = 1..4: A058331, A255843, A339048, A255847.

Formula

T(n, k) = A(n - k + 1, k), with the array A(m, k) = m^2 + 2*k^2, for n >= 1 and k = 1, 2, ..., n, and 0 otherwise.

G.f. of T and A column k (offset 0): G(k, x) = (1 + x + 2*(1 - x)^2*k^2)/(1-x)^3, for k >= 1.

G.f. of T diagonal m (A row m) (offset 0): D(m, x) = ((2*(1+x) + (1-x)^2*m^2)/(1-x)^3), for m >= 1.

G.f. of row polynomials in x (that is, g.f. of the triangle): G(z,x) = (3 - 3*z + (2 - 6*x + x^2)*z^2 + (2 + x)*x*z^3)*x*z / ((1 - z)*(1 - x*z))^3.