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A279126
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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.
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1
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2, 1, 1, 1, 2, 0, 1, 2, 12, -2, 1729, -2712, 1980, -456, 36, -103679, 399744, -264960, 83520, -11520, 576, 4603392001, -10890547200, 9794649600, -4267987200, 962668800, -107308800, 4665600, 61326139392001, -149868925747200, 139442417049600, -65662061875200, 17248412160000
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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Triangle begins
2
1, 1
1, 2, 0
1, 2, 12, -2
1729, -2712, 1980, -456, 36
-103679, 399744, -264960, 83520, -11520, 576
4603392001, -10890547200, 9794649600, -4267987200, 962668800, -107308800, 4665600
61326139392001, -149868925747200, 139442417049600, -65662061875200, 17248412160000, -2552991436800, 198742118400, -6320332800
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).
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MAPLE
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with(LinearAlgebra):
for n from 1 to 10 do
V:= VandermondeMatrix([$1..n]);
d:= Determinant(V);
S:= Vector(n);
count:= 0;
for k from 1 by d while count < n do
if isprime(k) then
count:= count+1;
S[count]:= k
fi
od;
R[n]:= convert(V^(-1) . S, list);
od:
seq(op(R[n[), n=1..10);
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MATHEMATICA
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nmax = 10;
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];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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