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%I #16 Feb 01 2021 02:19:27
%S 3,6,9,11,12,19,18,17,22,33,27,24,27,36,51,38,33,34,41,54,73,51,44,43,
%T 48,59,76,99,66,57,54,57,66,81,102,129,83,72,67,68,75,88,107,132,163,
%U 102,89,82,81,86,97,114,137,166,201
%N Triangle read by rows: T(n, k) = (n - k + 1)^2 + 2*k^2, for n >= 1, and k = 1, 2, ..., n.
%C 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.
%C 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).
%C 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.
%C 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.)
%C For the multiplicities of the representable values A154777(n), see A339047.
%C The author met this representation problem in connection with special triples of integer circle curvatures in the Descartes-Steiner five circle problem.
%F 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.
%F 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.
%F 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.
%F 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.
%e The triangle T(n, k) begins:
%e n \ k 1 2 3 4 5 6 7 8 9 10 11 12 ...
%e 1: 3
%e 2: 6 9
%e 3: 11 12 19
%e 4: 18 17 22 33
%e 5: 27 24 27 36 51
%e 6: 38 33 34 41 54 73
%e 7: 51 44 43 48 59 76 99
%e 8: 66 57 54 57 66 81 102 129
%e 9: 83 72 67 68 75 88 107 132 163
%e 10: 102 89 82 81 86 97 114 137 166 201
%e 11: 123 108 99 96 99 108 123 144 171 204 243
%e 12: 146 129 118 113 114 121 134 153 178 209 246 289
%e ...
%e ----------------------------------------------------
%e T(5, 1) = 5^2 + 2*1^2 = 27 = T(5, 3) = 3^2 + 2*3^2. A338433(11) = 2 for A154777(11) = 27.
%e T(4, 4) = 1^2 + 2*4^2 = 33 = T(6, 2) = 5^2 + 2*2^2. A338433(12) = 2 for A154777(12) = 33.
%e T(5, 5) = 1^2 + 2*5^2 = 51 = T(7, 1) = 7^2 + 2*1^2. A338433(20) = 2 for A154777(20) = 51.
%e 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.
%e The first multiplicity 4 appears for 297.
%Y Cf. A154777, A339047.
%Y Cf. Columns k = 1..3: A059100, A189833, A241848.
%Y Cf. Diagonals m = 1..4: A058331, A255843, A339048, A255847.
%K nonn,tabl,easy
%O 1,1
%A _Wolfdieter Lang_, Dec 09 2020
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