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.

Previous Showing 11-17 of 17 results.

A252489 Index of the largest prime which divides n*(n+1).

Original entry on oeis.org

1, 2, 2, 3, 3, 4, 4, 2, 3, 5, 5, 6, 6, 4, 3, 7, 7, 8, 8, 4, 5, 9, 9, 3, 6, 6, 4, 10, 10, 11, 11, 5, 7, 7, 4, 12, 12, 8, 6, 13, 13, 14, 14, 5, 9, 15, 15, 4, 4, 7, 7, 16, 16, 5, 5, 8, 10, 17, 17, 18, 18, 11, 4, 6, 6, 19, 19, 9, 9, 20, 20, 21, 21, 12, 8, 8, 6
Offset: 1

Views

Author

M. F. Hasler, Jan 16 2015

Keywords

Comments

Yields the row of A145605 in which n appears, and also the first row of A138180 in which n appears.

Crossrefs

Programs

  • Maple
    A061395:= [1, seq(numtheory:-pi(max(numtheory:-factorset(n))), n=2..101)]:
    zip(max,A061395[1..-2],A061395[2..-1]); # Robert Israel, Feb 12 2021
  • Mathematica
    a[n_] := PrimePi[Max[FactorInteger[n][[-1, 1]], FactorInteger[n+1][[-1, 1]]]];
    Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Dec 05 2023 *)
  • PARI
    a(n)=primepi(vecmax(factor(n*(n+1))[,1]))

Formula

a(n) = pi(A074399(n)), where pi = A000720.
a(n) = max(A061395(n),A061395(n+1)). - Robert Israel, Feb 12 2021

A117582 The number of ratios t/(t-1), where t is a square number, which factor into primes less than or equal to prime(n).

Original entry on oeis.org

0, 2, 5, 10, 15, 24, 34, 46, 57, 74, 90, 114, 141, 174, 208, 244, 287, 334, 387
Offset: 1

Views

Author

Gene Ward Smith, Apr 02 2006

Keywords

Comments

By a theorem of Størmer, the number of ratios m/(m-1) factoring into primes only up to p is finite. Some of these have square numerators.
Equivalently, a(n) is the number of triples of consecutive prime(n)-smooth numbers. - Lucas A. Brown, Oct 04 2022

Examples

			The ratios counted by a(3) are 4/3, 9/8, 16/15, 25/24, and 81/80.
The ratios counted by a(4) are 4/3, 9/8, 16/15, 25/24, 36/35, 49/48, 64/63, 81/80, 225/224, and 2401/2400.
		

Crossrefs

Extensions

Offset 1 and a(14)-a(18) by Lucas A. Brown, Oct 04 2022
a(19) from Lucas A. Brown, Oct 16 2022

A117583 The number of ratios t/(t-1), where t is a triangular number, which factor into primes less than or equal to prime(n).

Original entry on oeis.org

0, 1, 3, 7, 9, 16, 22, 29, 35, 39, 50, 57, 68, 84, 100, 112, 127, 151, 167
Offset: 1

Views

Author

Gene Ward Smith, Apr 02 2006

Keywords

Comments

As in the case of square numerators, triangular numerators of superparticular ratios m/(m-1) factorizable only up to a relatively small prime p are relatively common.
Equivalently, a(n) is the number of quadruples of consecutive prime(n)-smooth numbers. - Lucas A. Brown, Oct 04 2022

Examples

			The ratios counted by a(3) are 3/2, 6/5, and 10/9.
The ratios counted by a(4) are 3/2, 6/5, 10/9, 15/14, 21/20, 28/27, and 36/35.
		

Crossrefs

Extensions

a(14)-a(18) by Lucas A. Brown, Oct 04 2022
a(19) from Lucas A. Brown, Oct 16 2022

A362593 Number of coprime positive integer S-unit solutions to a + b = c where a <= b < c, and where S = {prime(1), ..., prime(n)}.

Original entry on oeis.org

0, 1, 4, 17, 63, 190, 545, 1433, 3649, 8828, 20015, 44641, 95358, 199081, 412791, 839638, 1663449
Offset: 0

Views

Author

Robin Visser, Apr 26 2023

Keywords

Comments

Let S = {p_1, p_2, ..., p_n} be a finite set of prime numbers. A positive integer S-unit is a positive integer x such that x = p_1^k_1 * p_2^k_2 * ... * p_n^k_n for some nonnegative integers k_1, k_2, ..., k_n.
Thus a(n) is the number of positive integer triples (a,b,c) such that a + b = c, gcd(a,b,c) = 1, a <= b < c and v_p(a) = v_p(b) = v_p(c) = 0 for all primes p greater than prime(n), i.e., the primes dividing a, b or c are some subset of the first n prime numbers.
Mahler (1933) first proved that a(n) is finite for all n, with effective bounds first given by Györy (1979).

Examples

			For n = 2, the a(2) = 4 solutions are 1 + 1 = 2, 1 + 2 = 3, 1 + 3 = 4, and 1 + 8 = 9.
For n = 3, the a(3) = 17 solutions are 1 + 1 = 2, 1 + 2 = 3, 1 + 3 = 4, 1 + 4 = 5, 1 + 5 = 6, 1 + 8 = 9, 1 + 9 = 10, 1 + 15 = 16, 1 + 24 = 25, 1 + 80 = 81, 2 + 3 = 5, 2 + 25 = 27, 3 + 5 = 8, 3 + 125 = 128, 4 + 5 = 9, 5 + 27 = 32, and 9 + 16 = 25.
		

Crossrefs

Cf. A002071 (Case a = 1), A361661, A362567.

Programs

  • SageMath
    from sage.rings.number_field.S_unit_solver import solve_S_unit_equation
    def a(n):
        Q = CyclotomicField(1)
        S = Q.primes_above(prod([p for p in Primes()[:n]]))
        sols = len(solve_S_unit_equation(Q, S))
        return (sols + 1)/3

Formula

a(n) = (A362567(n) + 3)/6 if n > 0.

A275156 The 108 numbers n such that n(n+1) is 17-smooth.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 20, 21, 24, 25, 26, 27, 32, 33, 34, 35, 39, 44, 48, 49, 50, 51, 54, 55, 63, 64, 65, 77, 80, 84, 90, 98, 99, 104, 119, 120, 125, 135, 143, 153, 168, 169, 175, 195, 220, 224, 242, 255, 272, 288, 324, 350, 351, 363, 374, 384, 440, 441, 539, 560, 594, 624, 675, 714, 728, 832, 935, 1000, 1088, 1155, 1224, 1274, 1700, 1715, 2057, 2079, 2400, 2430, 2499, 2600, 3024, 4095, 4224, 4374, 4913, 5831, 6655, 9800, 10647, 12375, 14399, 28560, 31212, 37179, 123200, 194480, 336140
Offset: 1

Views

Author

Jean-François Alcover, Nov 13 2016

Keywords

Comments

This is the 7th row of the table A138180.

References

Crossrefs

Programs

  • Mathematica
    pMax = 17; smoothMax = 10^12; smoothNumbers[p_?PrimeQ, max_] := Module[{a, aa, k, pp, iter}, k = PrimePi[p]; aa = Array[a, k]; pp = Prime[Range[k]]; iter = Table[{a[j], 0, PowerExpand@Log[pp[[j]], max/Times @@ (Take[pp, j - 1]^Take[aa, j - 1])]}, {j, 1, k}]; Table[Times @@ (pp^aa), Sequence @@ iter // Evaluate] // Flatten // Sort]; Select[(Sqrt[1 + 4*smoothNumbers[pMax, smoothMax]] - 1)/2, IntegerQ]
  • PARI
    is(n)=my(t=510510); n*=n+1; while((t=gcd(n,t))>1, n/=t); n==1 \\ Charles R Greathouse IV, Nov 13 2016

A275164 The 167 numbers n such that n(n+1) is 19-smooth.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 24, 25, 26, 27, 32, 33, 34, 35, 38, 39, 44, 48, 49, 50, 51, 54, 55, 56, 63, 64, 65, 75, 76, 77, 80, 84, 90, 95, 98, 99, 104, 119, 120, 125, 132, 135, 143, 152, 153, 168, 169, 170, 175, 189, 195, 208, 209, 220, 224, 242, 255, 272, 285, 288, 323, 324, 342, 350, 351, 360, 363, 374, 384, 399, 440
Offset: 1

Views

Author

Jean-François Alcover, Nov 14 2016

Keywords

Comments

See A002071.
The full list of 167 terms is given in the b-file (this is the 8th row of the table A138180).

Crossrefs

Programs

  • Mathematica
    pMax = 19; smoothMax = 10^15; smoothNumbers[p_?PrimeQ, max_] := Module[{a, aa, k, pp, iter}, k = PrimePi[p]; aa = Array[a, k]; pp = Prime[Range[k]]; iter = Table[{a[j], 0, PowerExpand@Log[pp[[j]], max/Times @@ (Take[pp, j - 1]^Take[aa, j - 1])]}, {j, 1, k}]; Table[Times @@ (pp^aa), Sequence @@ iter // Evaluate] // Flatten // Sort]; Select[(Sqrt[1 + 4*smoothNumbers[pMax, smoothMax]] - 1)/2, IntegerQ]

A362567 Number of rational solutions to the S-unit equation x + y = 1, where S = {prime(i): 1 <= i <= n}.

Original entry on oeis.org

0, 3, 21, 99, 375, 1137, 3267, 8595, 21891, 52965, 120087, 267843, 572145, 1194483, 2476743, 5037825, 9980691
Offset: 0

Views

Author

Robin Visser, Apr 25 2023

Keywords

Comments

Let S = {p_1, p_2, ..., p_n} be a finite set of prime numbers. A rational S-unit is a rational number x such that abs(x) = p_1^k_1 * p_2^k_2 * ... * p_n^k_n for some integers k_1, k_2, ..., k_n.
Thus a(n) is the number of ordered pairs (x,y) of rational numbers such that x+y=1 and v_p(x) = v_p(y) = 0 for all primes p greater than prime(n), i.e., the primes dividing the numerator or denominator of x or y are some subset of the first n prime numbers.
Mahler (1933) first proved that a(n) is finite for all n, with effective bounds first given by Györy (1979).

Examples

			For n = 1, the a(1) = 3 solutions are -1 + 2 = 1, 1/2 + 1/2 = 1, and 2 + -1 = 1.
For n = 2, the a(2) = 21 solutions are -8 + 9 = 1, -3 + 4 = 1, -2 + 3 = 1, -1 + 2 = 1, -1/2 + 3/2 = 1, -1/3 + 4/3 = 1, -1/8 + 9/8 = 1, 1/9 + 8/9 = 1, 1/4 + 3/4 = 1, 1/3 + 2/3 = 1, 1/2 + 1/2 = 1, 2/3 + 1/3 = 1, 3/4 + 1/4 = 1, 8/9 + 1/9 = 1, 9/8 + -1/8 = 1, 4/3 + -1/3 = 1, 3/2 + -1/2 = 1, 2 + -1 = 1, 3 + -2 = 1, 4 + -3 = 1, and 9 + -8 = 1.
		

Crossrefs

Programs

  • SageMath
    from sage.rings.number_field.S_unit_solver import solve_S_unit_equation
    def a(n):
        Q = CyclotomicField(1)
        S = Q.primes_above(prod([p for p in Primes()[:n]]))
        sols = len(solve_S_unit_equation(Q, S))
        return 2*sols - 1

Formula

a(n) = 6*A362593(n) - 3 if n > 0.
Previous Showing 11-17 of 17 results.