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.

A216023 Fermat pseudoprimes to base 2 divisible by 5.

Original entry on oeis.org

645, 1105, 1905, 2465, 10585, 11305, 16705, 18705, 34945, 39865, 41665, 55245, 62745, 72885, 74665, 83665, 107185, 121465, 208465, 215265, 223345, 266305, 278545, 449065, 451905, 464185, 493885, 588745, 743665, 757945, 800605, 825265, 831405, 898705, 1050985
Offset: 1

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Author

Marius Coman, Aug 30 2012

Keywords

Comments

Many Fermat pseudoprimes to base 2 divisible by 5 have one of the following four properties:
(1) the sum of their prime factors is divisible by the sum of their digits:
for 1105 = 3*5*17 we have 35 divisible by 15;
for 1905 = 3*5*127 we have 135 divisible by 15;
for 2465 = 5*17*29 we have 51 divisible by 17;
for 34945 = 5*29*241 we have 275 divisible by 25;
for 62745 = 3*5*47*89 we have 144 divisible by 24;
for 107185 = 3*5*47*89 we have 132 divisible by 22;
for 223345 = 5*19*2351 we have 2375 divisible by 19;
for 451905 = 3*5*47*641 we have 696 divisible by 24.
(2) the sum of their prime factors is divisible by 5 (1105, 1905, 16705, 18705, 34945, 223345, 757945, 800605).
(3) the sum of their digits is divisible by 5 (645, 1905, 11305, 34945, 72885, 208465, 72885);
(4) they are Harshad numbers (645, 1905, 2465, 223345, 757945).
Interesting is that the first property is found to other squarefree numbers, not Fermat pseudoprimes, divisible by 5 (e.g., for 1505 = 5*7*43 we have 55 divisible by 11, for 2555 = 5*7*73 we have 85 divisible by 17). It looks like it's a property which deserves further study.
Note: the four properties from above are also found to other Fermat pseudoprimes to base 2, but not in this high density (taking, for the second and third properties, a prime factor beside 5 and not considering for the third property the prime factor 3, because would be obviously satisfied).

Crossrefs

Cf. A001567.

Programs

  • Mathematica
    Select[5*Range[2, 210200], PowerMod[2, # - 1, #] == 1 &] (* T. D. Noe, Aug 31 2012 *)
  • PARI
    Korselt(n)=my(f=factor(n)); for(i=1, #f[, 1], if(f[i, 2]>1||(n-1)%(f[i, 1]-1), return(0))); 1
    list(lim)=my(v=List()); forstep(n=645, lim, 20, if(Korselt(n), listput(v, n))); Vec(v) \\ Charles R Greathouse IV, Jun 30 2017