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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.

User: Giosuè Cavallo

Giosuè Cavallo's wiki page.

Giosuè Cavallo has authored 1 sequences.

A356594 Numbers k for which there exists at least one pair of positive integers (x,y) such that k = x + y and k' = x' + y', and every such pair is coprime.

Original entry on oeis.org

3, 25, 55, 82, 85, 95, 116, 121, 145, 194, 226, 245, 253, 289, 295, 301, 305, 332, 335, 343, 362, 391, 407, 418, 422, 446, 455, 493, 529, 535, 548, 583, 611, 671, 731, 745, 749, 754, 778, 779, 781, 785, 799, 805, 815, 817, 818, 833, 838, 845, 866, 869, 899, 917, 931, 943, 955, 959, 985, 995, 998
Offset: 1

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Author

Giosuè Cavallo, Aug 14 2022

Keywords

Comments

Subsequence of A212662.
a(1)=3 is the only prime term.
It is not known if this sequence is finite.
Every term in A212662 is a multiple of a term in this sequence (this could be considered its primitive sequence), and no term in this sequence divides another term of this sequence.
In general, we do not have (i+j)'=i'+j'; this is in contrast with the normal derivative, because the derivative of the sum of two functions is equal to the sum of the derivatives of the individual functions. The terms in this sequence are the ones for which there exist positive integers i and j with the properties illustrated above, thus building another common point between the arithmetic derivative and the normal derivative.

Crossrefs

Cf. A003415 (arithmetic derivative), A212662.

Programs

  • PARI
    D(n)={x=factor(n);n*sum(i=1,matsize(x)[1],x[i,2]/x[i,1])}
Base(n)={for(i=1,n\2,if(D(n-i)+D(i)==D(n)&&gcd(i,n-i)==1,return(1)))}
Impure(n)={for(i=1,n\2,if(D(n-i)+D(i)==D(n)&&gcd(i,n-i)!=1,return(1)))}
IsTerm(n)={Base(n)&&!Impure(n)}
Bp(n)={for(i=2,n,if(IsTerm(i),print1(i, ", ")))}
Bp(1000)
  • PARI
    ard(n)=vecsum([n/f[1]*f[2]|f<-factor(n+!n)~]); \\ A003415
isok1(m) = for (k=1, m\2, if (ard(m-k)+ard(k) == ard(m), return(1))); \\ A212662
isok(m) = if (isok1(m), my(d=divisors(m)); for (k=1, #d, if((d[k]!=m) && isok1(d[k]), return(0))); return(1)); \\ Michel Marcus, Aug 28 2022

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