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A338432 Triangle read by rows: T(n, k) = (n - k + 1)^2 + 2*k^2, for n >= 1, and k = 1, 2, ..., n. 3
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 (list; table; graph; refs; listen; history; text; internal format)
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.)

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

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

LINKS

Table of n, a(n) for n=1..55.

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(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.

Sequence in context: A189302 A086883 A154777 * A288821 A343306 A267881

Adjacent sequences:  A338429 A338430 A338431 * A338433 A338434 A338435

KEYWORD

nonn,tabl,easy

AUTHOR

Wolfdieter Lang, Dec 09 2020

STATUS

approved

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Last modified June 22 21:29 EDT 2021. Contains 345393 sequences. (Running on oeis4.)