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 1-5 of 5 results.

A129283 a(n) = (arithmetic derivative of n) + n.

Original entry on oeis.org

0, 1, 3, 4, 8, 6, 11, 8, 20, 15, 17, 12, 28, 14, 23, 23, 48, 18, 39, 20, 44, 31, 35, 24, 68, 35, 41, 54, 60, 30, 61, 32, 112, 47, 53, 47, 96, 38, 59, 55, 108, 42, 83, 44, 92, 84, 71, 48, 160, 63, 95, 71, 108, 54, 135, 71, 148, 79, 89, 60, 152, 62, 95, 114, 256, 83, 127, 68
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 07 2007

Keywords

Crossrefs

Programs

Formula

a(n) = A003415(n) + n.
a(n) = A003415(n*A051674(k)) / A051674(k).
a(A129284(n)) > 1, a(A129285(n)) > 1, a(A129286(n)) > 1.
Sum_{k=1..n} a(k) ~ c * n^2 / 2, where c = 1 + Sum_{p prime} 1/(p*(p-1)) = 1 + A136141 = 1.773156... . - Amiram Eldar, May 14 2025

A129152 The n-th arithmetic derivative of 5^6.

Original entry on oeis.org

15625, 18750, 34375, 37500, 87500, 187500, 475000, 1212500, 2437500, 6362500, 12737500, 25487500, 50987500, 101987500, 206975000, 530037500, 1060087500, 3890025000, 15175012500, 45525375000, 177026512500, 596222100000, 2708984250000, 12765250350000
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 01 2007

Keywords

Comments

In general, the trajectory of p^(p+1) under A003415 is equal to p^p times the trajectory of p under A129283: n -> n + n'. Here we have the case p = 5 (see A129286 for a(n)/5^5), see A129150 and A129151 for p = 2 and 3. - M. F. Hasler, Nov 28 2019

Crossrefs

Programs

  • Haskell
    a129152 n = a129152_list !! n
    a129152_list = iterate a003415 15625  -- Reinhard Zumkeller, Apr 29 2012
    
  • Mathematica
    dn[0] = 0; dn[1] = 0; dn[n_?Negative] := -dn[-n]; dn[n_] := Module[{f = Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Total[n*f[[2]]/f[[1]]]]]; s = 5^6; Join[{s}, Table[s = dn[s], {18}]] (* T. D. Noe, Mar 07 2013 *)
  • PARI
    A129152_upto(N)=vector(N,n,N=if(n>1,A003415(N),5^6)) \\ gives a(0..N-1). To get a(1..N) put A003415() around if() instead inside.  M. F. Hasler, Nov 28 2019

Formula

a(n+1) = A003415(a(n)), a(0) = 5^6 = 15625.
a(n) = A129286(n)*5^5; A129251(a(n)) > 0. - Reinhard Zumkeller, Apr 07 2007

A090635 Trajectory of 63 under the map k -> A003415(k) (taking the arithmetic derivative).

Original entry on oeis.org

63, 51, 20, 24, 44, 48, 112, 240, 608, 1552, 3120, 8144, 16304, 32624, 65264, 130544, 264928, 678448, 1356912, 4979232, 19424016, 58272480, 226593936, 763164288, 3467499840, 16339520448, 65370077568, 295266178368, 1223245608192, 6931725175296, 40582548986112
Offset: 1

Views

Author

N. J. A. Sloane, Dec 14 2003

Keywords

References

  • A. M. Gleason et al., The William Lowell Putnam Mathematical Competition: Problems and Solutions 1938-1964, Math. Assoc. America, 1980, p. 295.

Crossrefs

Programs

  • PARI
    A090635(n, a=63)={if(n<0, vector(-n, n, if(n>1, a=A003415(a), a)), for(n=2, n, a=A003415(a)); a)}  \\ For n<0 return the vector a[1..-n]. - M. F. Hasler, Nov 27 2019

Formula

a(n+1) = A003415(a(n)), a(1) = 63. a(n) = 4*A129286(n-3) for n > 2. - M. F. Hasler, Nov 27 2019

Extensions

More explicit name from M. F. Hasler, Nov 27 2019

A129284 a(n) = A129150(n) / 4, where A129150(n) = n-th arithmetic derivative of 2^3.

Original entry on oeis.org

2, 3, 4, 8, 20, 44, 92, 188, 380, 856, 2148, 5024, 17616, 58768, 176320, 755904, 3305920, 13885184, 69634816, 348174336, 2385273856, 14652403712, 102566830080, 849285738496, 6035962949632, 44017806979072, 308166534991872, 2380768960708608, 23410894780694528
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 07 2007

Keywords

Comments

In general, the trajectory of p^(p+1) under A003415 has a common factor p^p, and divided by p^p it gives the trajectory of p under A129283: n -> n + n'. Here we have the case p = 2, see A129151 and A129152 for p = 3 and 5. - M. F. Hasler, Nov 28 2019

Crossrefs

Programs

Formula

a(n+1) = A129283(a(n)), a(0) = 2.

Extensions

a(18)-a(28) from Paolo P. Lava, Apr 16 2012
Edited by M. F. Hasler, Nov 27 2019

A129285 a(n) = A129151(n) / 27.

Original entry on oeis.org

3, 4, 8, 20, 44, 92, 188, 380, 856, 2148, 5024, 17616, 58768, 176320, 755904, 3305920, 13885184, 69634816, 348174336, 2385273856, 14652403712, 102566830080, 849285738496, 6035962949632, 44017806979072, 308166534991872, 2380768960708608, 23410894780694528
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 07 2007

Keywords

Crossrefs

Cf. A129284 (essentially the same), A129283 (n + n'), A129286, A051674, A129150, A090636 (trajectory of 15 under arithmetic derivative A003415).

Programs

Formula

a(n+1) = A129283(a(n)), a(0) = 3.
a(n) = A129284(n+1). - Eric M. Schmidt, Oct 22 2013
Thus a(n) = A129150(n+1) / 4 = A090636(n+3) / 4. - M. F. Hasler, Nov 27 2019
Showing 1-5 of 5 results.