OFFSET
1,1
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.)
The author met this representation problem in connection with special triples of integer curvatures in the Descartes-Steiner five circle problem.
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.
EXAMPLE
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(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
KEYWORD
AUTHOR
Wolfdieter Lang, Dec 09 2020
STATUS
approved