A111615 -id:A111615 - cp's OEIS frontend

A111616 n divided by the third upper diagonal of the array in A111615.

1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1

Offset: 1

Paul D. Hanna and Robert G. Wilson v, Jul 30 2005

nonn

a(n)=2 for n's: 6,10,18,20,22,30,34,36,38,44,50,52,56,58,62,... =A111617.

Cf. A111615.

f[n_] := f[n] = Block[{a}, a[0] = 1; a[l_] := a[l] = Block[{k = 1, s = Sum[ a[i]*x^i, {i, 0, l - 1}]}, While[ IntegerQ[ Last[ CoefficientList[ Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[a[j], {j, 0, 128}]]; g[n_, m_] := f[n][[m]]; Table[ n / g[n, n + 3], {n, 105}]

A109626 Consider the array T(n,m) where the n-th row is the sequence of integer coefficients of A(x), where 1<=a(n)<=n, such that A(x)^(1/n) consists entirely of integer coefficients and where m is the (m+1)-th coefficient. This is the antidiagonal read from lower left to upper right.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 4, 3, 2, 1, 1, 5, 2, 1, 2, 1, 1, 6, 5, 4, 3, 2, 1, 1, 7, 3, 5, 3, 3, 1, 1, 1, 8, 7, 2, 5, 4, 3, 2, 1, 1, 9, 4, 7, 3, 1, 4, 3, 2, 1, 1, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 11, 5, 3, 2, 7, 6, 5, 1, 3, 1, 1, 1, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 1, 13, 6, 11, 10, 9, 4, 1, 3, 5

Offset: 1

Paul D. Hanna and Robert G. Wilson v, Aug 01 2005

			Table begins:
\k...0...1...2...3...4...5...6...7...8...9..10..11..12..13
n\
 1|  1   1   1   1   1   1   1   1   1   1   1   1   1   1
 2|  1   2   1   2   2   2   1   2   2   2   1   2   1   2
 3|  1   3   3   1   3   3   3   3   3   3   3   3   1   3
 4|  1   4   2   4   3   4   4   4   1   4   4   4   3   4
 5|  1   5   5   5   5   1   5   5   5   5   4   5   5   5
 6|  1   6   3   2   3   6   6   6   3   4   6   6   6   6
 7|  1   7   7   7   7   7   7   1   7   7   7   7   7   7
 8|  1   8   4   8   2   8   4   8   7   8   8   8   4   8
 9|  1   9   9   3   9   9   3   9   9   1   9   9   6   9
10|  1  10   5  10  10   2   5  10  10  10   3  10   5  10
11|  1  11  11  11  11  11  11  11  11  11  11   1  11  11
12|  1  12   6   4   9  12   4  12  12   8   6  12   6  12
13|  1  13  13  13  13  13  13  13  13  13  13  13  13   1
14|  1  14   7  14   7  14  14   2   7  14  14  14  14  14
15|  1  15  15   5  15   3  10  15  15  10  15  15   5  15
16|  1  16   8  16   4  16   8  16  10  16   8  16  12  16

G. C. Greubel, Antidiagonals n = 1..100, flattened

Rows: A000012, A083952, A083953, A083954, A083945, A083946, A083947, A083948, A083949, A083950, A084066, A084067.
Columns: A000012, A111627, A026741, A051176, A111607, A060791, A111608, A106608, A111609, A111610, A111611, A106612, A106614, A106618, A106620.
Diagonals: A000027 (main), A111614 (first upper), A111627 (2nd), A111615 (3rd), A111618 (first lower), A111623 (2nd).
Other diagonals: A005408 (T(2*n-1, n)), A111626, A111627, A111628, A111629, A111630.
Cf. A111603, A111604.

f[n_]:= f[n]= Block[{a}, a[0] = 1; a[l_]:= a[l]= Block[{k = 1, s = Sum[ a[i]*x^i, {i,0,l-1}]}, While[ IntegerQ[Last[CoefficientList[Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[a[j], {j,0,32}]];
T[n_, m_]:= f[n][[m]];
Flatten[Table[T[i,n-i], {n,15}, {i,n-1,1,-1}]]

A109626_row(n, len=40)={my(A=1, m); vector(len, k, if(k>m=1, while(denominator(polcoeff(sqrtn(O(x^k)+A+=x^(k-1), n), k-1))>1, m++); m, 1))} \\ M. F. Hasler, Jan 27 2025

When m is prime, column m is T(n,m) = n/gcd(m, n) = numerator of n/(n+m). - M. F. Hasler, Jan 27 2025

A111617 Where A111616(n)=2.

Original entry on oeis.org

6, 10, 18, 20, 22, 30, 34, 36, 38, 44, 50, 52, 56, 58, 62, 72, 78, 84, 86, 90, 92, 96, 98, 100, 102, 104, 110, 116, 134

Offset: 1

Paul D. Hanna and Robert G. Wilson v, Jul 30 2005

nonn

Cf. A109626, A111615.

f[n_] := f[n] = Block[{a}, a[0] = 1; a[l_] := a[l] = Block[{k = 1, s = Sum[ a[i]*x^i, {i, 0, l - 1}]}, While[ IntegerQ[ Last[ CoefficientList[ Series[(s + k*x^l)^(1/n), {x, 0, l}], x]]] != True, k++ ]; k]; Table[a[j], {j, 0, 144}]]; g[n_, m_] := f[n][[m]]; Select[ Range[141], # / g[ #, # + 3] == 2 &]

cp's OEIS Frontend

A111616 n divided by the third upper diagonal of the array in A111615.

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A109626 Consider the array T(n,m) where the n-th row is the sequence of integer coefficients of A(x), where 1<=a(n)<=n, such that A(x)^(1/n) consists entirely of integer coefficients and where m is the (m+1)-th coefficient. This is the antidiagonal read from lower left to upper right.

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A111617 Where A111616(n)=2.

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