A126213 a(n) = sum of the d(k)'s, where 1<=k<=n and d(k) is equal to any divisor of n, where d(k) is the number of positive divisors of k.
1, 3, 1, 5, 1, 10, 1, 17, 7, 9, 1, 35, 1, 13, 7, 33, 1, 33, 1, 42, 7, 17, 1, 79, 6, 19, 10, 55, 1, 59, 1, 75, 10, 23, 6, 119, 1, 25, 10, 110, 1, 66, 1, 85, 24, 29, 1, 174, 1, 46, 13, 95, 1, 108, 6, 149, 13, 33, 1, 204, 1, 37, 22, 165, 6, 109, 1, 123, 13, 61, 1, 304, 1, 43, 18, 135, 8
Offset: 1
Examples
The number of divisors of the integers 1 through 10 form the sequence 1,2,2,3,2,4,2,4,3,4. The divisors of 10 are 1,2,5,10. The terms of the sequence of the first ten d(k)'s which equal any divisor of 10 are the five terms 1,2,2,2,2. So a(10) = 1+2+2+2+2 = 9.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16384
Programs
-
Mathematica
f[n_] := Plus @@ Select[Table[Length@Divisors[k], {k, n}], MemberQ[Divisors[n], # ] &];Table[f[n], {n, 78}] (* Ray Chandler, Dec 21 2006 *) -
PARI
A126213(n) = sumdiv(n, d, d*sum(k=1, n, (numdiv(k)==d))); \\ Antti Karttunen, Apr 01 2021
-
PARI
first(n) = { n = min(n, 245044799); qdivs = vector(960); res = vector(n); for(i = 1, n, nd = numdiv(i); qdivs[nd]++; d = select(x -> x <= #qdivs, divisors(i)); res[i] = sum(j = 1, #d, d[j]*qdivs[d[j]]) ); res } \\ David A. Corneth, Apr 01 2021
Extensions
Extended by Ray Chandler, Dec 21 2006
Comments