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.

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A328235 The least k > 0 such that the arithmetic derivative of n+k is a multiple of the arithmetic derivative of n.

Original entry on oeis.org

1, 1, 4, 1, 15, 1, 12, 11, 8, 1, 4, 1, 13, 1, 12, 1, 20, 1, 24, 4, 23, 1, 56, 7, 10, 27, 36, 1, 28, 1, 44, 15, 114, 1, 76, 1, 84, 5, 56, 1, 48, 1, 20, 27, 53, 1, 80, 3, 36, 25, 76, 1, 81, 9, 4, 23, 26, 1, 116, 1, 64, 207, 80, 3, 52, 1, 40, 3, 82, 1, 232, 1, 205, 31, 36, 4, 27, 1, 92, 27, 88, 1, 160, 36, 130, 5, 12, 1, 81, 9, 52, 3
Offset: 2

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Author

Antti Karttunen, Oct 08 2019

Keywords

Examples

			Arithmetic derivative of 6 is A003415(6) = 5. Not until at k=21 we find another number whose arithmetic derivative is a multiple of five (as A003415(21) = 10 = 2*5), thus a(6) = 21-6 = 15.
		

Crossrefs

Cf. A003415, A328236, A328237 (gives the quotient).

Programs

  • PARI
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A328235(n) = { my(d=A003415(n)); for(k=1,oo,if(!(A003415(n+k)%d), return(k))); };

A328238 Quotient A003415(n*m)/A003415(n) for the least m > 1 for which such a quotient is an integer. Here A003415(x) is the arithmetic derivative of x.

Original entry on oeis.org

4, 5, 3, 7, 12, 9, 16, 10, 20, 13, 7, 15, 28, 42, 6, 19, 16, 21, 18, 42, 44, 25, 48, 14, 52, 3, 26, 31, 60, 33, 8, 66, 68, 43, 9, 39, 56, 51, 80, 43, 84, 45, 19, 37, 92, 49, 13, 18, 24, 236, 46, 55, 16, 67, 112, 138, 88, 61, 116, 63, 124, 124, 12, 130, 132, 69, 82, 138, 140, 73, 28, 75, 119, 84, 108, 142, 156, 81, 44, 7, 164
Offset: 2

Views

Author

Antti Karttunen, Oct 08 2019

Keywords

Examples

			Arithmetic derivative of 6 is 6' = A003415(6) = 5. Taking arithmetic derivatives of its successive multiples, we obtain 12' = 16, 18' = 21, 24' = 44, 30' = 31, and not until with A003415(6*6) = 36' = 60 we obtain a multiple of 5. Thus a(6) = 60/5 = 12.
		

Crossrefs

Cf. A328236 (gives the corresponding m).

Programs

  • PARI
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A328238(n) = { my(d=A003415(n),t); for(m=2,oo,if(!((t=A003415(n*m))%d), return(t/d))); };

Formula

a(n) = A003415(n*A328236(n)) / A003415(n).
Showing 1-2 of 2 results.