cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

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

Views

Author

Keywords

Examples

			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
		

Crossrefs

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.

Programs

  • Mathematica
    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}]]
  • PARI
    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

Formula

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

A106617 Numerator of n/(n+16).

Original entry on oeis.org

0, 1, 1, 3, 1, 5, 3, 7, 1, 9, 5, 11, 3, 13, 7, 15, 1, 17, 9, 19, 5, 21, 11, 23, 3, 25, 13, 27, 7, 29, 15, 31, 2, 33, 17, 35, 9, 37, 19, 39, 5, 41, 21, 43, 11, 45, 23, 47, 3, 49, 25, 51, 13, 53, 27, 55, 7, 57, 29, 59, 15, 61, 31, 63, 4, 65, 33, 67, 17, 69, 35, 71, 9, 73, 37, 75, 19, 77, 39, 79
Offset: 0

Views

Author

N. J. A. Sloane, May 15 2005

Keywords

Comments

A multiplicative sequence. Also, a(n) is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for n >= 1, m >= 1. In particular, a(n) is a divisibility sequence: if n divides m then a(n) divides a(m). - Peter Bala, Feb 21 2019

Examples

			From _Peter Bala_, Feb 21 2019: (Start)
Sum_{n >= 1} n*a(n)*x^n = G(x) - 2*G(x^2) - 4*G(x^4) - 8*G(x^8) - 16*G(x^16), where G(x) = x*(1 + x)/(1 - x)^3.
Sum_{n >= 1} n^2*a(n)*x^n = H(x) - 2^2*H(x^2) - 4^2*H(x^4) - 8^2*H(x^8) - 16^2*H(x^16), where H(x) = x*(1 + 4*x + x^2)/(1 - x)^4. In general, the o.g.f. for Sum_{n >= 1} (n^k*a(n))*x^n for positive k involves the Eulerian polynomials.
In the other direction,
Sum_{n >= 1} (a(n)/n)*x^n = J(x) - (1/2)*J(x^2) - (1/4)*J(x^4) - (1/8)*J(x^8) - (1/16)*J(x^16), where J(x) = x/(1 - x).
Sum_{n >= 1} (a(n)/n^2)*x^n = L(x) - (1/2^2)*L(x^2) - (1/4^2)*L(x^4) - (1/8^2)*L(x^8) - (1/16^2)*L(x^16), where L(x) = log(1/(1 - x)). In general, the o.g.f. for Sum_{n >= 0} (a(n)/n^k)*x^n, for k >= 3, involves the polylogarithm Li_(k-1)(x).
Sum_{n >= 1} (1/a(n))*x^n = L(x) + (1/2)*L(x^2) + (1/2)*L(x^4) + (1/2)*L(x^8) + (1/2)*L(x^16). (End)
		

Crossrefs

Cf. Other sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Programs

Formula

a(n) = 2*a(n-16) - a(n-32) for n > 31. - Paul Curtz, Apr 12 2011
Octosections: a(8*n) = A026741(n). a(2+8*n) = 1+4*n. a(4+8*n) = 1+2*n. a(6+8*n) = 3+4*n. Bisection: a(1+2*n) = 1+2*n. - Paul Curtz, Apr 12 2011
Dirichlet g.f.: zeta(s-1)*(1-1/2^s-1/4^s-1/8^s-1/16^s). - R. J. Mathar, Apr 18 2011
a(n) = numerator of n/(2^(2*n+1)). - Ralf Steiner, Feb 09 2017
The previous comment is incorrect, a(n) first differs from the numerator of n/(2^(2*n+1)) at n = 32. - Peter Bala, Feb 27 2019
From Peter Bala, Feb 21 2019: (Start)
a(n) = n/gcd(n,16), where gcd(n,16) = [1, 2, 1, 4, 1, 2, 1, 8, 1, 2, 1, 4, 1, 2, 1, 16, ...] is a periodic sequence of period 16: a(n) is thus quasi_polynomial in n.
O.g.f.: Sum_{n >= 0} a(n)*x^n = F(x) - F(x^2) - F(x^4) - F(x^8) - F(x^16), where F(x) = x/(1 - x)^2.
More generally, Sum_{n >= 0} (a(n)^m)*x^n = F(m,x) + (1 - 2^m)*( F(m,x^2) + F(m,x^4) + F(m,x^8) + F(m,x^16) ), where F(m,x) = A(m,x)/(1 - x)^(m+1) with A(m,x) the m-th Eulerian polynomial: A(1,x) = x, A(2,x) = x*(1 + x), A(3,x) = x*(1 + 4*x + x^2) - see A008292.
Repeatedly applying the Euler operator x*d/dx or its inverse operator to the o.g.f. for the sequence produces generating functions for the sequences ( (n^m)*a(n) )n>=1 for m in Z. Some examples are given below. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(2^e) = 2^max(0,e-4), and a(p^e) = p^e if p>2.
Sum_{k=1..n} a(k) ~ (171/512) * n^2. (End)

A106616 Numerator of n/(n+15).

Original entry on oeis.org

0, 1, 2, 1, 4, 1, 2, 7, 8, 3, 2, 11, 4, 13, 14, 1, 16, 17, 6, 19, 4, 7, 22, 23, 8, 5, 26, 9, 28, 29, 2, 31, 32, 11, 34, 7, 12, 37, 38, 13, 8, 41, 14, 43, 44, 3, 46, 47, 16, 49, 10, 17, 52, 53, 18, 11, 56, 19, 58, 59, 4, 61, 62, 21, 64, 13, 22, 67, 68, 23, 14, 71, 24, 73, 74, 5, 76, 77, 26
Offset: 0

Views

Author

N. J. A. Sloane, May 15 2005

Keywords

Comments

Multiplicative and also a strong divisibility sequence: gcd(a(n),a(m)) = a(gcd(n,m)) for n, m >= 1. - Peter Bala, Feb 24 2019

Crossrefs

Cf. Other sequences given by the formula numerator(n/(n + k)): A026741 (k = 2), A051176 (k = 3), A060819 (k = 4), A060791 (k = 5), A060789 (k = 6), A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Programs

Formula

Dirichlet g.f.: zeta(s-1)*(1-4/5^s-2/3^s+8/15^s). - R. J. Mathar, Apr 18 2011
a(n) = gcd((n-2)*(n-1)*n*(n+1)*(n+2)/15, n) for n>=1. - Lechoslaw Ratajczak, Feb 19 2017
From Peter Bala, Feb 24 2019: (Start)
a(n) = n/gcd(n,15), a quasi-polynomial in n since gcd(n,15) is a purely periodic sequence of period 15.
O.g.f.: F(x) - 2*F(x^3) - 4*F(x^5) + 8*F(x^15), where F(x) = x/(1 - x)^2.
O.g.f. for reciprocals: Sum_{n >= 1} x^n/a(n) = Sum_{d divides 15} (phi(d)/d) * log(1/(1 - x^d)) = log(1/(1 - x)) + (2/3)*log(1/(1 - x^3)) + (4/5)*log(1/(1 - x^5)) + (8/15)*log(1/(1 - x^15)), where phi(n) denotes the Euler totient function A000010. (End)
From Amiram Eldar, Nov 25 2022: (Start)
Multiplicative with a(3^e) = 3^max(0,e-1), a(5^e) = 5^max(0,e-1), and a(p^e) = p^e otherwise.
Sum_{k=1..n} a(k) ~ (49/150) * n^2. (End)
Showing 1-3 of 3 results.