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 1-4 of 4 results.

A248124 Least integer m > 0 such that gcd(m,n) = 1 and (m+n) | (C(m)+C(n)), where C(k) refers to the k-th Catalan number, binomial(2k,k)/(k+1).

Original entry on oeis.org

1, 19, 1, 20, 1, 95, 1, 4, 1, 242, 241, 478, 1, 23, 1, 5, 7, 109, 1, 17, 1, 227, 467, 37, 1, 209, 1, 330, 2077, 17, 1073, 816, 1, 27, 109, 71, 1, 43, 1, 41, 145, 151, 1, 43, 1, 59, 71, 587, 1, 87, 1775, 344, 1773, 1127, 1, 49, 1
Offset: 4

Views

Author

Zhi-Wei Sun, Oct 01 2014

Keywords

Comments

Conjecture: a(n) exists for all n > 3.

Examples

			a(5) = 19 since 5 is relatively prime to 19 and 5 + 19 = 24 divides C(5) + C(19) = 42 + 1767263190 = 1767263232 = 24*73635968.

Links

Crossrefs

Cf. A000108, A247824, A247937, A247940, A248123, A248125.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[GCD[m,n]==1&&Mod[CatalanNumber[m]+CatalanNumber[n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,4,60}]

A248125 Least positive integer m such that m + n divides C(2m,m) + C(2n,n), where C(2k,k) = (2k)!/(k!)^2.

Original entry on oeis.org

1, 2, 5, 16, 3, 6, 2, 22, 101, 6, 21, 86, 43, 16, 15, 4, 3, 6, 21, 20, 11, 8, 49, 48, 7, 22, 29, 28, 27, 26, 25, 49, 11, 29, 133, 20, 19, 22, 71, 70, 7, 18, 13, 46, 11, 14, 25, 24, 23, 93, 45, 80, 43, 67, 29, 286, 171, 102, 97, 38
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 01 2014

Keywords

Comments

Conjecture: a(n) exists for all n > 0. Moreover, for n > 66 we have a(n) < n except for n = 364, 408.
a(n) = n for n = 1, 2, 6, 15, 20, 28, 66, ... The next term, if it exists, is greater than 10^4. - Derek Orr, Oct 01 2014

Examples

			a(3) = 5 since 3 + 5 = 8 divides C(6,3) + C(10,5) = 20 + 252 = 272.

Links

Crossrefs

Cf. A247824, A247937, A247940, A248123, A248124.

Programs

  • Mathematica
    Do[m=1;Label[aa];If[Mod[Binomial[2m,m]+Binomial[2n,n],m+n]==0,Print[n," ",m];Goto[bb]];m=m+1;Goto[aa];Label[bb];Continue,{n,1,60}]
  • PARI
    a(n)=m=1;while((binomial(2*m,m)+binomial(2*n,n))%(m+n),m++);m
vector(100,n,a(n)) \\ Derek Orr, Oct 01 2014

A248189 Least positive integer m such that m*n divides sigma(m^2+n^2), where sigma(k) is the sum of all positive divisors of k.

Original entry on oeis.org

1, 1, 1, 7, 2, 38, 4, 81, 1, 102, 868, 1, 9, 3, 702, 26505, 1554, 14, 3, 243, 1, 650, 108, 1833, 34542, 18, 68, 186, 7252, 39, 58, 736839, 1, 3108, 72, 778, 210, 6, 3, 4830, 267, 2, 567, 5859, 6640, 6363, 3178412, 155771, 4964, 9
Offset: 1

Views

Author

Zhi-Wei Sun, Oct 03 2014

Keywords

Comments

Conjecture: a(n) exists for any n > 0.
See also the comments in A248058.

Examples

			a(6) = 38 since 6*38 = 228 divides sigma(6^2+38^2) = sigma(1480) = 3420 = 15*228.

Links

Crossrefs

Cf. A000203, A248054, A248058, A248123.

Programs

  • Mathematica
    Do[m=1; Label[aa]; If[Mod[DivisorSigma[1,m^2+n^2], m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 50}]
  • PARI
    a(n)=m=1;while(sigma(n^2+m^2)%(m*n),m++);m
n=1;while(n<50,print1(a(n),", ");n++) \\ Derek Orr, Oct 03 2014

A297573 Least positive integer m such that m*n divides F(m+n), where F(k) denotes the k-th Fibonacci number A000045(k).

Original entry on oeis.org

1, 1, 1, 2, 14170, 6, 1, 136, 207, 28340, 979, 12, 1, 322, 385, 368, 1, 306, 17, 19780, 3, 68, 1, 24, 524975, 58, 2889, 92, 13, 3570, 12749, 736, 7, 2, 165, 612, 1, 34, 633, 13160, 339, 6, 1, 1846, 5355, 2, 1, 336, 8183, 509950, 21, 116, 1, 918, 4895, 184, 51, 26, 10207, 7140
Offset: 1

Views

Author

Zhi-Wei Sun, Jan 01 2018

Keywords

Comments

If p is a prime congruent to 2 or 3 modulo 5, then a(p) = 1 since it is known that p divides F(p+1).
Conjecture: a(n) exists for any n > 0.
See also A297574 for a similar conjecture.

Examples

			a(2) = 1 since 1*2 divides F(1+2) = F(3) = 2.
a(4) = 2 since 2*4 divides F(2+4) = 8.
a(5) = 14170 since 5*14170 = 70850 divides F(5+14170) = F(14175).
a(6) = 6 since 6*6 = 36 divides F(6+6) = F(12) = 144.

Links

Crossrefs

Cf. A000045, A247937, A248123, A297574.

Programs

  • Mathematica
    Do[m=1; Label[aa]; If[Mod[Fibonacci[m+n], m*n]==0, Print[n, " ", m]; Goto[bb]]; m=m+1; Goto[aa]; Label[bb]; Continue, {n, 1, 60}]
lpi[n_]:=Module[{k=1},While[!Divisible[Fibonacci[k+n],k*n],k++];k]; Array[ lpi,60] (* Harvey P. Dale, May 05 2018 *)
  • PARI
    a(n) = my(m=1); while(1, if(Mod(fibonacci(m+n), m*n)==0, return(m)); m++) \\ Felix Fröhlich, Jan 01 2018
Showing 1-4 of 4 results.

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