A117552 -id:A117552 - cp's OEIS frontend

A109883 Start subtracting from n its divisors beginning from 1 until one reaches a number smaller than the last divisor subtracted or reaches the last nontrivial divisor < n. Define this to be the perfect deficiency of n. Then a(n) = perfect deficiency of n.

0, 1, 2, 1, 4, 0, 6, 1, 5, 2, 10, 2, 12, 4, 6, 1, 16, 6, 18, 8, 10, 8, 22, 0, 19, 10, 14, 0, 28, 3, 30, 1, 18, 14, 22, 11, 36, 16, 22, 10, 40, 9, 42, 4, 12, 20, 46, 12, 41, 7, 30, 6, 52, 15, 38, 20, 34, 26, 58, 2, 60, 28, 22, 1, 46, 21, 66, 10, 42, 31, 70, 9, 72, 34, 26, 12, 58, 27, 78

Offset: 1

Amarnath Murthy, Jul 11 2005

If n is a perfect number then a(n) = 0. But if a(n) = 0, n needs not be perfect, e.g., a(24) = 0, but 24 is not a perfect number. See A064510.

			a(14) = 4: 14-1 = 13, 13-2 = 11, 11-7 = 4.
a(6) = 0: 6-1 = 5, 5-2 = 3, 3-3 = 0. 6 is a perfect number.
a(35) = 22: 35-1 = 34, 34-5 = 29, 29-7 = 22.

Nathaniel Johnston, Table of n, a(n) for n = 1..10000

Cf. A064510, A109884, A109886, A117552.

A109883:=proc(n)local d,j,k,m:if(n=1)then return 0:fi:j:=1:m:=n:d:=divisors(n);k:=nops(d):for j from 1 to k do m:=m-d[j]:if(mNathaniel Johnston, Apr 15 2011

subtract = If[ #1 < #2, Throw[ #1], #1 - #2]&;
a[n_] := Catch @ Fold[subtract, n, Divisors @ n]
Table[ a[n], {n, 80}] (* Bobby R. Treat (DrBob(AT)bigfoot.com), Jul 14 2005 *)

a(n) = {my(r = n); fordiv(n, d, if (r < d, return (r)); r -= d;); 0;} \\ Michel Marcus, Dec 28 2018

from sympy import divisors
def A109883(n):
    if n == 1: return 0
    s = n
    for d in divisors(n)[:-1]:
        if s < d: break
        s -= d
    return s
print([A109883(n) for n in range(1, 80)]) # Michael S. Branicky, Mar 31 2024

a(1) = 0, a(2^n) = 1.
a(p) = p-1, a(p^n) = (p^(n+1) - 2*p^n + 1)/(p-1), if p is a prime.
a(n) = n - A117552(n). - Ridouane Oudra, Jan 25 2024

More terms from Jason Earls and Robert G. Wilson v, Jul 12 2005

A117553 When adding some positive divisors of n in order from lowest divisor to highest divisor, a(n) is lowest sum achievable which is >= n.

Original entry on oeis.org

1, 3, 4, 7, 6, 6, 8, 15, 13, 18, 12, 16, 14, 24, 24, 31, 18, 21, 20, 22, 32, 36, 24, 24, 31, 42, 40, 28, 30, 42, 32, 63, 48, 54, 48, 37, 38, 60, 56, 50, 42, 54, 44, 84, 78, 72, 48, 52, 57, 93, 72, 98, 54, 66, 72, 64, 80, 90, 60, 78, 62, 96, 104, 127, 84, 78, 68, 126, 96, 74, 72

Offset: 1

Leroy Quet, Mar 28 2006

nonn

Often, but not always, a(n)=n+A054024(n). The exceptions to this rule are at n=24, 36, 48, 60, 72, 84,90, 96, 108, ... - R. J. Mathar, Mar 14 2007

			12's divisors are 1,2,3,4,6 and 12. Adding the divisors in order we have:
1 = 1, 1+2 = 3, 1+2+3 = 6, 1+2+3+4 = 10, 1+2+3+4+6 = 16 and 1+2+3+4+6+12 = 28.
Of these sums, 1+2+3+4+6 = 16 is the lowest which is >= 12. So a(12) = 16.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A117552, A125746, A125747.

A117553 := proc(n) local divs,a,i ; divs := numtheory[divisors](n) ; a := op(1,divs) ; i := 1 ; while a < n do i := i+1 ; a := a+op(i,divs) ; od ; RETURN(a) ; end: for n from 1 to 80 do printf("%d, ",A117553(n)) ; od ; # R. J. Mathar, Mar 14 2007

Table[Select[Accumulate[Divisors[n]],#>=n&,1],{n,80}]//Flatten (* Harvey P. Dale, Apr 05 2017 *)

More terms from R. J. Mathar, Mar 14 2007

A377247 a(n) is the largest k such that the sum of the first k divisors of n is at most n.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 1, 3, 2, 3, 1, 4, 1, 3, 3, 4, 1, 4, 1, 4, 3, 3, 1, 6, 2, 3, 3, 5, 1, 6, 1, 5, 3, 3, 3, 6, 1, 3, 3, 6, 1, 6, 1, 5, 5, 3, 1, 7, 2, 5, 3, 5, 1, 6, 3, 6, 3, 3, 1, 9, 1, 3, 5, 6, 3, 6, 1, 5, 3, 6, 1, 9, 1, 3, 5, 5, 3, 6, 1, 8, 4, 3, 1, 9, 3, 3, 3, 6

Offset: 1

David A. Corneth, Oct 21 2024

nonn

			a(1) = 1 as the sum of the first divisor of 1 is 1 <= 1 and 1 has no more divisors.
a(6) = 3 as the sum of the first three divisors is 1+2+3 <= 6 but the sum of the first four divisors is 1 + 2 + 3 + 6 = 12 > 6.

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A081512, A117552 (corresponding sums).

A377247[n_] := LengthWhile[Accumulate[Divisors[n]], # <= n &];
Array[A377247, 100] (* Paolo Xausa, Aug 05 2025 *)

A377247(n) = {my(d = divisors(n), t = 0); for(i = 1, #d, t += d[i]; if(t > n, return(i-1))); 1}

cp's OEIS Frontend

A109883 Start subtracting from n its divisors beginning from 1 until one reaches a number smaller than the last divisor subtracted or reaches the last nontrivial divisor < n. Define this to be the perfect deficiency of n. Then a(n) = perfect deficiency of n.

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A117553 When adding some positive divisors of n in order from lowest divisor to highest divisor, a(n) is lowest sum achievable which is >= n.

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Author

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Maple

Mathematica

Extensions

A377247 a(n) is the largest k such that the sum of the first k divisors of n is at most n.

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Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

PARI