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 1-7 of 7 results.

A226357 Number of ordered triples (i,j,k) with |i|,|j|,|k|,|i*j*k| <= n and gcd(i,j,k) <= 1.

Original entry on oeis.org

1, 27, 75, 147, 243, 363, 483, 651, 819, 1011, 1179, 1443, 1683, 1995, 2211, 2475, 2763, 3171, 3459, 3915, 4251, 4611, 4923, 5475, 5883, 6411, 6771, 7275, 7707, 8403, 8811, 9555, 10059, 10611, 11067, 11715, 12291, 13179, 13683, 14331, 14931, 15915, 16419
Offset: 0

Views

Author

Robert Price, Jun 04 2013

Keywords

Comments

Note that gcd(0,m) = m for any m.

Crossrefs

|i| + |j| + |k| <= n instead of |i*j*k| <= n: A100450.
This sequence (A226357) without the GCD qualifier: A226359.
Distinct sums i+j+k with the GCD qualifier: A222947.
Distinct sums i+j+k without the GCD qualifier: A222945.
Distinct products i*j*k with or without the GCD qualifier is 2n+1: A005408.
With the further restriction i,j,k >= 0 ...
Distinct sums i+j+k <= n with the GCD qualifier: A223133.
Distinct sums i+j+k <= n without the GCD qualifier: A223134.
Distinct products i*j*k with or without the GCD qualifier is n+1: A000217(n+1).
Distinct sums i+j+k with i*j*k = n with the GCD qualifier: A223135.
Distinct sums i+j+k with i*j*k = n without the GCD qualifier: A226378.
Distinct products i*j*k with i*j*k = n with or without the GCD qualifier is trivial and always 1: A000012.
Ordered triples with the product <= n with the GCD qualifier: A226001.
Ordered triples with the product <= n without the GCD qualifier: A226600.
Ordered triples with the product = n with the GCD qualifier: A226602.
Ordered triples with the product = n without the GCD qualifier: A007425.

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[Abs[i*j*k] <=  n && GCD[i, j, k] <= 1, {i, j, k}], {i, -n, n}, {j, -n, n}, {k, -n, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]

A222947 Number of distinct sums i+j+k with |i|, |j|, |k|, |i*j*k| <= n and gcd(i,j,k) <= 1.

Original entry on oeis.org

1, 7, 9, 11, 15, 19, 21, 27, 29, 35, 37, 43, 45, 51, 53, 59, 61, 67, 69, 75, 77, 83, 85, 91, 93, 99, 101, 107, 109, 115, 117, 123, 125, 131, 133, 139, 141, 147, 149, 155, 157, 163, 165, 171, 173, 179, 181, 187, 189, 195, 197, 203, 205, 211, 213, 219, 221
Offset: 0

Views

Author

Robert Price, Jun 12 2013

Keywords

Comments

Note that gcd(0,m) = m for any m.

Crossrefs

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[Abs[i*j*k] ≤ n && GCD[i, j, k] <= 1, {i + j +  k}], {i, -n, n}, {j, -n, n}, {k, -n, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]

A223133 Number of distinct sums i+j+k with i,j,k >= 0, i*j*k <= n and gcd(i,j,k) <= 1.

Original entry on oeis.org

1, 4, 5, 6, 8, 10, 11, 14, 15, 18, 19, 22, 23, 26, 27, 30, 31, 34, 35, 38, 39, 42, 43, 46, 47, 50, 51, 54, 55, 58, 59, 62, 63, 66, 67, 70, 71, 74, 75, 78, 79, 82, 83, 86, 87, 90, 91, 94, 95, 98, 99, 102, 103, 106, 107, 110, 111, 114, 115, 118, 119, 122, 123
Offset: 0

Views

Author

Robert Price, Jun 12 2013

Keywords

Comments

Note that gcd(0,m) = m for any m.

Crossrefs

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[i*j*k <= n && GCD[i, j, k] <= 1, {i + j + k}], {i, 0, n}, {j, 0, n}, {k, 0, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]

A223134 Number of distinct sums i+j+k with i,j,k >= 0, i*j*k <= n.

Original entry on oeis.org

1, 4, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125
Offset: 0

Views

Author

Robert Price, Jun 12 2013

Keywords

Comments

Appears to be essentially the same as A176271, A140139, A130773, A062545. - R. J. Mathar, Aug 23 2024

Crossrefs

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[i*j*k <= n, {i + j + k}], {i, 0, n}, {j, 0, n}, {k, 0, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]

A226378 Number of distinct sums i+j+k with i,j,k >= 0, i*j*k = n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 4, 1, 2, 2, 4, 1, 4, 1, 4, 2, 2, 1, 6, 2, 2, 3, 4, 1, 5, 1, 5, 2, 2, 2, 7, 1, 2, 2, 5, 1, 5, 1, 4, 4, 2, 1, 9, 2, 4, 2, 4, 1, 6, 2, 6, 2, 2, 1, 10, 1, 2, 4, 7, 2, 5, 1, 4, 2, 5, 1, 11, 1, 2, 4, 4, 2, 5, 1, 9, 4, 2, 1, 10
Offset: 0

Views

Author

Robert Price, Jun 12 2013

Keywords

Examples

			From _Antti Karttunen_, Aug 30 2017: (Start)
For n = 4 = 1*1*4 = 1*2*2, 1+1+4 = 6 and 1+2+2 = 5, so there are two distinct sums, and a(4) = 2.
For n = 6 = 1*1*6 = 1*2*3, 1+1+6 = 8 and 1+2+3 = 6, so there are two distinct sums, and a(6) = 2.
For n = 36, of its A034836(36) = 8 factorizations as x*y*z with 1 <= x <= y <= z: 1*1*36 = 1*2*18 = 1*3*12 = 1*4*9 = 1*6*6 = 2*2*9 = 2*3*6 = 3*3*4, sums 1+6+6 and 2+2+9 are both 13, while other triples yield unique sums, thus a(36) = 8-1 = 7. (End)
		

Crossrefs

Cf. A008578 (gives the positions of 1's after a(0)=1).
Differs from A034836 for the first time at n=36.

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[i*j*k == n, {i + j + k}], {i, 0, n}, {j, 0, n}, {k, 0, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]
    (* Second program, more efficient: *)
    {1}~Join~Table[With[{D = Divisors@ n}, Length@ Union@ Reap[Map[Function[a, Map[Function[b, Map[Function[c, If[a b c == n, Sow[a + b + c]]], Select[D, # <= n/a b &]]], Select[D, # <= n/a &]]], D]][[-1, 1]] ], {n, 100}] (* Michael De Vlieger, Aug 24 2017 *)
  • PARI
    A226378(n) = { my(sums=Set()); if(!n,1,fordiv(n, i, for(j=i, (n/i), if(!(n%j),for(k=j, n/(i*j), if(i*j*k==n, sums = Set(concat(sums, (i+j+k)))))))); length(sums)); }; \\ Antti Karttunen, Aug 30 2017

Formula

For n >= 1, a(n) <= A034836(n). - Antti Karttunen, Aug 30 2017

A223135 Number of distinct sums i + j + k with i, j, k >= 0, i*j*k = n and gcd(i,j,k) <= 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 5, 2, 2, 2, 4, 1, 5, 1, 3, 2, 2, 2, 7, 1, 2, 2, 5, 1, 5, 1, 4, 4, 2, 1, 7, 2, 4, 2, 4, 1, 5, 2, 5, 2, 2, 1, 10, 1, 2, 4, 4, 2, 5, 1, 4, 2, 5, 1, 10, 1, 2, 4, 4, 2, 5, 1, 7, 3, 2, 1, 10, 2, 2, 2, 5, 1, 8, 2, 4, 2, 2, 2, 7, 1, 4, 4, 8, 1, 5, 1, 5, 5
Offset: 0

Views

Author

Robert Price, Jun 12 2013

Keywords

Comments

Note that gcd(0,m) = m for any m.

Crossrefs

Programs

  • Mathematica
    f[n_] := Length[Complement[Union[Flatten[Table[If[i*j*k == n && GCD[i, j, k] ≤ 1, {i + j + k}], {i, 0, n}, {j, 0, n}, {k, 0, n}], 2]], {Null}]]; Table[f[n], {n, 0, 100}]
  • PARI
    A223135(n) = { my(sums=Set()); if(!n,1,fordiv(n, i, for(j=i, (n/i), if(!(n%j),for(k=j, n/(i*j), if((i*j*k==n)&&(gcd(i,gcd(j,k))<=1), sums = Set(concat(sums, (i+j+k)))))))); length(sums)); }; \\ Antti Karttunen, Oct 21 2017

Extensions

More terms from Antti Karttunen, Oct 21 2017

A270741 a(n) = (Fibonacci(n+2)-1) mod Fibonacci(floor(n/2)).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 0, 1, 3, 2, 0, 1, 11, 10, 0, 1, 32, 31, 0, 1, 87, 86, 0, 1, 231, 230, 0, 1, 608, 607, 0, 1, 1595, 1594, 0, 1, 4179, 4178, 0, 1, 10944, 10943, 0, 1, 28655, 28654, 0, 1, 75023, 75022, 0, 1, 196416, 196415, 0, 1, 514227, 514226, 0, 1, 1346267, 1346266, 0, 1, 3524576, 3524575, 0, 1
Offset: 2

Views

Author

Altug Alkan, Mar 22 2016

Keywords

Comments

It appears that a(n)=1 gives same terms as A222945. - Michel Marcus, Mar 23 2016

Examples

			a(10) = 3 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55) mod 5 = 143 mod 5 = 3.
a(11) = 2 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89) mod 5 = 2.
a(12) = 0 since (1 + 1 + 2 + 3 + 5 + 8 + 13 + 21 + 34 + 55 + 89 + 144) mod 8 = 0.
		

Crossrefs

Programs

  • Mathematica
    Table[Mod[Fibonacci[n + 2] - 1, Fibonacci[Floor[n/2]]], {n, 2, 69}] (* Michael De Vlieger, Mar 23 2016 *)
  • PARI
    for(n=2, 1e2, print1((fibonacci(n+2)-1) % fibonacci(n\2), ", "));

Formula

Empirical g.f.: x^7*(1-x+2*x^2+x^3-3*x^4+3*x^5-6*x^6+5*x^7+x^8-x^9+2*x^10-2*x^11) / ((1-x)*(1+x^2)*(1+x^2-x^4)*(1-x^2-x^4)). - Colin Barker, Mar 23 2016
Showing 1-7 of 7 results.