A099302 Number of integer solutions to x' = n, where x' is the arithmetic derivative of x.
0, 0, 1, 1, 1, 1, 1, 1, 2, 0, 2, 1, 2, 1, 3, 0, 2, 1, 2, 2, 3, 0, 4, 1, 3, 1, 2, 0, 3, 2, 4, 1, 4, 0, 4, 0, 2, 2, 3, 1, 4, 1, 4, 2, 4, 0, 6, 1, 4, 1, 3, 0, 5, 2, 4, 0, 4, 1, 7, 2, 3, 1, 5, 0, 6, 0, 3, 1, 5, 2, 7, 1, 5, 3, 5, 1, 7, 0, 6, 2, 5, 0, 8, 1, 5, 2, 4, 0, 9, 3, 6, 0, 5, 1, 8, 0, 3, 1, 6, 2, 8, 2, 5, 1, 6
Offset: 2
References
- See A003415
Links
- Antti Karttunen, Table of n, a(n) for n = 2..100000 (terms 2..40000 from T. D. Noe)
- Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.
Crossrefs
Programs
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Haskell
a099302 n = length $ filter (== n) $ map a003415 [1 .. a002620 n] -- Reinhard Zumkeller, Mar 18 2014
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Mathematica
dn[0]=0; dn[1]=0; dn[n_]:=Module[{f=Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Plus@@(n*f[[2]]/f[[1]])]]; d1=Table[dn[n], {n, 40000}]; Table[Count[d1, n], {n, 2, 400}]
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PARI
up_to = 100000; \\ A002620(10^5) = 2500000000 A002620(n) = ((n^2)>>2); A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1])); A099302list(up_to) = { my(d,c,v=vector(up_to)); for(i=1, A002620(up_to), d = A003415(i); if(d>1 && d<=up_to, v[d]++)); (v); }; v099302 = A099302list(up_to); A099302(n) = v099302[n]; \\ Antti Karttunen, Jan 21 2024
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Python
from sympy import factorint def A099302(n): return sum(1 for m in range(1,(n**2>>2)+1) if sum((m*e//p for p,e in factorint(m).items())) == n) # Chai Wah Wu, Sep 12 2022
Formula
From Antti Karttunen, Jan 21 2024: (Start)
(End)
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