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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A159077 a(n) = A008475(n) + 1.

Original entry on oeis.org

1, 3, 4, 5, 6, 6, 8, 9, 10, 8, 12, 8, 14, 10, 9, 17, 18, 12, 20, 10, 11, 14, 24, 12, 26, 16, 28, 12, 30, 11, 32, 33, 15, 20, 13, 14, 38, 22, 17, 14, 42, 13, 44, 16, 15, 26, 48, 20, 50, 28, 21, 18, 54, 30, 17, 16, 23, 32, 60, 13, 62, 34, 17, 65, 19, 17, 68, 22, 27, 15, 72, 18, 74
Offset: 1

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Author

Jaroslav Krizek, Apr 04 2009

Keywords

Comments

If n = Product (p_i^k_i) for i = 1, …, j then a(n) is sum of divisor d from set of divisors{1, p_1^k_1, p_2^k_2, …, p_j^k_j}.

Examples

			For n = 12, set of divisors {1, p_1^k_1, p_2^k_2, …, p_j^k_j}: {1, 3, 4}. a(12) = 1+3+4=8.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := 1 + Plus @@ Power @@@ FactorInteger@ n; f[1] = 1; Array[f, 60]
  • PARI
    a(n)=local(t); if(n<1, 0, t=factor(n); 1+sum(k=1, matsize(t)[1], t[k, 1]^t[k, 2])) /* Anton Mosunov, Jan 05 2017 */

Formula

a(n) = [Sum_(i=1,…, j) p_i^k_i] + 1 = A000203(n) - A178636(n).
a(1) = 1, a(p) = p+1, a(pq) = p+q+1, a(pq...z) = p+q+...+z+1, a(p^k) = p^k+1, for p, q = primes, k = natural numbers, pq...z = product of k (k > 2) distinct primes p, q, ..., z.

Extensions

Edited by N. J. A. Sloane, Apr 07 2009

A167515 The sum over the divisors of n, except the maximum-prime-power divisors collected in A008475.

Original entry on oeis.org

1, 1, 1, 3, 1, 7, 1, 7, 4, 11, 1, 21, 1, 15, 16, 15, 1, 28, 1, 33, 22, 23, 1, 49, 6, 27, 13, 45, 1, 62, 1, 31, 34, 35, 36, 78, 1, 39, 40, 77, 1, 84, 1, 69, 64, 47, 1, 105, 8, 66, 52, 81, 1, 91, 56, 105, 58, 59, 1, 156, 1, 63, 88, 63, 66, 128, 1, 105, 70
Offset: 1

Views

Author

Jaroslav Krizek, Dec 15 2010

Keywords

Comments

If n = Product (p_j^k_j) is the standard prime power decomposition of n, there is a set of size A001221(n) which contains the divisors which are largest powers of primes, {p_1^k_1, p_2^k_2, ..., p_j^k_j}. a(n) sums all the divisors not in this set. If p, q, ..., z are distinct primes, k are natural numbers (A000027), p^k prime powers (A000961), the following formulas hold: a(p) = 1. a(pq) = pq+1. a(pq...z) = (p+1)* (q+1)* ... *(z+1) - (p+q+ ...+z). a(p^k) = (p^k-1)/(p-1).

Examples

			For n = 12, set of prime-power-factor divisors of 12: {3, 4}, set of non-(prime-power-factor) divisors on 12: {1, 2, 6, 12}. a(12) = 1+2+6+12=21.
		

Crossrefs

Programs

Formula

a(n) = A000203(n) - A008475(n).
a(n) = A178636(n) + 1.
Showing 1-2 of 2 results.