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A257639
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a(n) is the minimal position at which the maximal value of row n appears in row n of triangle A008289.
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3
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1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7
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OFFSET
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1,5
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COMMENTS
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Except rows 3, 4, 10, 11, 21 of triangle A008289, all the other rows up to row number 10^6 contain a single maximal value.
Conjecture: for n >= 22 there is a unique maximal value in row n of triangle A008289.
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LINKS
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FORMULA
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a(n) = min argmax(k->Q(n,k), k=1..m), that is a(n) = min{k, Q(n,k) = max{Q(n,p), p=1..m}}, where m = A003056(n) and Q(n,k) is defined by A008289.
a(n) ~ K*sqrt(n) + O(1), where K = 2*sqrt(3)*log(2)/Pi = 0.76430413884... (A131266).
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EXAMPLE
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For n=9, a(9)=2 because A003056(9)=3 and max{Q(9,p), p=1..3}=4 and Q(9,2)=4.
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PROG
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(PARI)
Q(N) = {
my(q = vector(N)); q[1] = [1, 0, 0, 0];
for (n = 2, N,
my(m = (sqrtint(8*n+1) - 1)\2);
q[n] = vector((1 + (m>>2)) << 2); q[n][1] = 1;
for (k = 2, m, q[n][k] = q[n-k][k] + q[n-k][k-1]));
return(q);
};
seq(N) = {
my(a = vector(N), q = Q(N), vmx = apply(vecmax, q));
for (n = 1, N, a[n] = vecmin(select(v->v==vmx[n], q[n], 1)));
a;
};
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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