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%I #30 Apr 29 2019 08:24:41
%S 2,1,1,1,2,0,1,2,12,-2,1729,-2712,1980,-456,36,-103679,399744,-264960,
%T 83520,-11520,576,4603392001,-10890547200,9794649600,-4267987200,
%U 962668800,-107308800,4665600,61326139392001,-149868925747200,139442417049600,-65662061875200,17248412160000
%N Triangle read by rows: coefficients of the polynomial P_n(x) of degree n-1 such that P_n(j) is the j-th prime == 1 (mod A000178(n-1)) for 1<=j<=n.
%C Given n, let p_1, p_2, ..., p_n be the first n primes == 1 (mod A000178(n-1)). Since A000178(n-1) is the determinant of the Vandermonde matrix of the numbers 1..n, the polynomial P(x) of degree <=n-1 with P(j) = p_j for 1<=j<=n has integer coefficients. Row n of the triangle consists of the coefficients of x^i in P(x) for i = 0,1,...,n-1.
%H Robert Israel, <a href="/A279126/b279126.txt">Table of n, a(n) for n = 1..861</a> (rows 1 to 41, flattened)
%H Mathematics StackExchange, <a href="http://math.stackexchange.com/questions/2048884/can-we-always-find-some-polynomial-for-which-this-holds/2048899#comment4205948_2048899">Can we always find some polynomial for which this holds?</a>
%e Triangle begins
%e 2
%e 1, 1
%e 1, 2, 0
%e 1, 2, 12, -2
%e 1729, -2712, 1980, -456, 36
%e -103679, 399744, -264960, 83520, -11520, 576
%e 4603392001, -10890547200, 9794649600, -4267987200, 962668800, -107308800, 4665600
%e 61326139392001, -149868925747200, 139442417049600, -65662061875200, 17248412160000, -2552991436800, 198742118400, -6320332800
%e For n=4 the polynomial P(x) = 1+2x+12x^2-2x^3 has P(1)=13, P(2)=37, P(3)=61, P(4)=73, these values being the first four primes == 1 (mod 12).
%p with(LinearAlgebra):
%p for n from 1 to 10 do
%p V:= VandermondeMatrix([$1..n]);
%p d:= Determinant(V);
%p S:= Vector(n);
%p count:= 0;
%p for k from 1 by d while count < n do
%p if isprime(k) then
%p count:= count+1;
%p S[count]:= k
%p fi
%p od;
%p R[n]:= convert(V^(-1) . S, list);
%p od:
%p seq(op(R[n[),n=1..10);
%t nmax = 10;
%t For[n = 1, n <= nmax, n++, V = Table[Range[n]^k, {k, 0, n-1}] // Transpose; d = Det[V]; S = Table[0, {n}]; count = 0; For[k = 1, count < n, k += d, If[PrimeQ[k], count++; S[[count]] = k]]; R[n] = Inverse[V].S];
%t Array[R, nmax] // Flatten (* _Jean-François Alcover_, Apr 29 2019 *)
%Y Cf. A000178.
%K sign,tabl
%O 1,1
%A _Robert Israel_, Dec 08 2016
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