A097863 -id:A097863 - cp's OEIS frontend

A049417 a(n) = isigma(n): sum of infinitary divisors of n.

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 17, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 51, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 68, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 85, 84, 144, 68, 90

Offset: 1

Yasutoshi Kohmoto, Dec 11 1999

A divisor of n is called infinitary if it is a product of divisors of the form p^{y_a 2^a}, where p^y is a prime power dividing n and sum_a y_a 2^a is the binary representation of y.
This sequence is an infinitary analog of the Dedekind psi function A001615. Indeed, a(n) = Product_{q in Q_n}(q+1) =  n*Product_{q in Q_n} (1+1/q), where {q} are terms of A050376 and Q_n is the set of distinct q's whose product is n. - Vladimir Shevelev, Apr 01 2014
1/a(n) is the asymptotic density of numbers that are infinitarily divided by n (i.e., numbers whose set of infinitary divisors includes n). - Amiram Eldar, Jul 23 2025

			If n = 8: 8 = 2^3 = 2^"11" (writing 3 in binary) so the infinitary divisors are 2^"00" = 1, 2^"01" = 2, 2^"10" = 4 and 2^"11" = 8; so a(8) = 1+2+4+8 = 15.
n = 90 = 2*5*9, where 2, 5, 9 are in A050376; so a(n) = 3*6*10 = 180. - _Vladimir Shevelev_, Feb 19 2011

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..7417 from R. J. Mathar)
Graeme L. Cohen, On an integer's infinitary divisors, Math. Comp. 54 (189) (1990) 395-411.
Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]
Steven R. Finch, Mathematical Constants II, Encyclopedia of Mathematics and Its Applications, Cambridge University Press, Cambridge, 2018, pp. 49-56.
J. O. M. Pedersen, Tables of Aliquot Cycles.
J. O. M. Pedersen, Tables of Aliquot Cycles. [Via Internet Archive Wayback-Machine]
J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]
Tomohiro Yamada, Infinitary superperfect numbers, Annales Mathematicae et Informaticae, Vol. 47 (2017), pp. 211-218; arXiv preprint, arXiv:1705.10933 [math.NT], 2017.

Cf. A004607, A037445, A050376, A077609, A091732, A127661, A293355.
Cf. A049418 (3-infinitary), A074847 (4-infinitary), A097863 (5-infinitary).
Cf. A000079, A001615, A027748, A030308, A124010, A126168.

a049417 1 = 1
a049417 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f p e = product $ zipWith div
           (map (subtract 1 . (p ^)) $
                zipWith (*) a000079_list $ map (+ 1) $ a030308_row e)
           (map (subtract 1 . (p ^)) a000079_list)
-- Reinhard Zumkeller, Sep 18 2015

isidiv := proc(d, n)
    local n2, d2, p, j;
    if n mod d <> 0 then
        return false;
    end if;
    for p in numtheory[factorset](n) do
        padic[ordp](n,p) ;
        n2 := convert(%, base, 2) ;
        padic[ordp](d,p) ;
        d2 := convert(%, base, 2) ;
        for j from 1 to nops(d2) do
            if op(j, n2) = 0 and op(j, d2) <> 0 then
                return false;
            end if;
        end do:
    end do;
    return true;
end proc:
idivisors := proc(n)
    local a, d;
    a := {} ;
    for d in numtheory[divisors](n) do
        if isidiv(d, n) then
            a := a union {d} ;
        end if;
    end do:
    a ;
end proc:
A049417 := proc(n)
    local d;
    add(d, d=idivisors(n)) ;
end proc:
seq(A049417(n),n=1..100) ; # R. J. Mathar, Feb 19 2011

bitty[k_] := Union[Flatten[Outer[Plus, Sequence @@ ({0, #1} & ) /@ Union[2^Range[0, Floor[Log[2, k]]]*Reverse[IntegerDigits[k, 2]]]]]]; Table[Plus@@((Times @@ (First[it]^(#1 /. z -> List)) & ) /@ Flatten[Outer[z, Sequence @@ bitty /@ Last[it = Transpose[FactorInteger[k]]], 1]]), {k, 2, 120}]
(* Second program: *)
a[n_] := If[n == 1, 1, Sort @ Flatten @ Outer[ Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^Select[Range[0, m], BitOr[m, #] == m &])]] // Total;
Array[a, 100] (* Jean-François Alcover, Mar 23 2020, after Paul Abbott in A077609 *)

A049417(n) = {my(b, f=factorint(n)); prod(k=1, #f[,2], b = binary(f[k,2]); prod(j=1, #b, if(b[j], 1+f[k,1]^(2^(#b-j)), 1)))} \\ Andrew Lelechenko, Apr 22 2014

isigma(n)=vecprod([vecprod([f[1]^2^k+1|k<-[0..exponent(f[2])], bittest(f[2],k)])|f<-factor(n)~]) \\ M. F. Hasler, Oct 20 2022

from math import prod
from sympy import factorint
def A049417(n): return prod(p**(1<Chai Wah Wu, Jul 11 2024

Multiplicative: If e = Sum_{k >= 0} d_k 2^k (binary representation of e), then a(p^e) = Product_{k >= 0} (p^(2^k*{d_k+1}) - 1)/(p^(2^k) - 1). - Christian G. Bower and Mitch Harris, May 20 2005 [This means there is a factor p^2^k + 1 if d_k = 1, otherwise the factor is 1. - M. F. Hasler, Oct 20 2022]
Let n = Product(q_i) where {q_i} is a set of distinct terms of A050376. Then a(n) = Product(q_i + 1). - Vladimir Shevelev, Feb 19 2011
If n is squarefree, then a(n) = A001615(n). - Vladimir Shevelev, Apr 01 2014
a(n) = Sum_{k>=1} A077609(n,k). - R. J. Mathar, Oct 04 2017
a(n) = A126168(n)+n. - R. J. Mathar, Oct 05 2017
Multiplicative with a(p^e) = Product{k >= 0, e_k = 1} p^2^k + 1, where e = Sum e_k 2^k, i.e., e_k is bit k of e. - M. F. Hasler, Oct 20 2022
a(n) = iphi(n^2)/iphi(n), where iphi(n) = A091732(n). - Amiram Eldar, Sep 21 2024

More terms from Wouter Meeussen, Sep 02 2001

A074847 Sum of 4-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its 4-ary expansion everywhere that the corresponding r(i) has a digit b, then d is a 4-infinitary-divisor of n.

Original entry on oeis.org

1, 3, 4, 7, 6, 12, 8, 15, 13, 18, 12, 28, 14, 24, 24, 17, 18, 39, 20, 42, 32, 36, 24, 60, 31, 42, 40, 56, 30, 72, 32, 51, 48, 54, 48, 91, 38, 60, 56, 90, 42, 96, 44, 84, 78, 72, 48, 68, 57, 93, 72, 98, 54, 120, 72, 120, 80, 90, 60, 168, 62, 96, 104, 119, 84, 144, 68, 126, 96

Offset: 1

Yasutoshi Kohmoto, Sep 10 2002

If we group the exponents e in the Bower-Harris formula into the sets with d_k=0, 1, 2 and 3, we see that every n has a unique representation of the form n=prod q_i *prod (r_j)^2 *prod (s_k)^3, where each of q_i, r_j, s_k is a prime power of the form p^(k^4), p prime, k>=0. Using this representation, a(n)=prod (q_i+1)prod ((r_j)^2+r_j+1)prod ((s_k)^3+(s_k)^2+s_k+1) by simple expansion of the quotient on the right hand side of the Bower-Harris formula. - Vladimir Shevelev, May 08 2013

			2^4*3 is a 4-infinitary-divisor of 2^5*3^2 because 2^4*3 = 2^10*3^1 and 2^5*3^2 = 2^11*3^2 in 4-ary expanded power. All corresponding digits satisfy the condition. 1<=1, 0<=1, 1<=2.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A049417 (2-infinitary), A049418 (3-infinitary), A097863 (5-infinitary).

Cf. A000302, A030386, A027748, A074848, A124010.

following Bower and Harris, cf. A049418:
a074847 1 = 1
a074847 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f p e = product $ zipWith div
           (map (subtract 1 . (p ^)) $
                zipWith (*) a000302_list $ map (+ 1) $ a030386_row e)
           (map (subtract 1 . (p ^)) a000302_list)
-- Reinhard Zumkeller, Sep 18 2015

A074847 := proc(n) option remember; ifa := ifactors(n)[2] ; a := 1 ; if nops(ifa) = 1 then p := op(1,op(1,ifa)) ; e := op(2,op(1,ifa)) ; d := convert(e,base,4) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1,d))*4^k)-1)/(p^(4^k)-1) ; end do: else for d in ifa do a := a*procname( op(1,d)^op(2,d)) ; end do: return a; end if; end proc:
seq(A074847(n),n=1..100) ; # R. J. Mathar, Oct 06 2010

f[p_, e_] := Module[{d = IntegerDigits[e, 4]}, m = Length[d]; Product[(p^((d[[j]] + 1)*4^(m - j)) - 1)/(p^(4^(m - j)) - 1), {j, 1, m}]]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Sep 09 2020 *)

Multiplicative. If e = sum_{k >= 0} d_k 4^k (base 4 representation), then a(p^e) = prod_{k >= 0} (p^(4^k*{d_k+1}) - 1)/(p^(4^k) - 1). - Christian G. Bower and Mitch Harris, May 20 2005

More terms from R. J. Mathar, Oct 06 2010

A049418 3-i-sigma(n): sum of 3-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its ternary expansion everywhere that the corresponding r(i) has a digit b, then d is a 3-i-divisor of n.

Original entry on oeis.org

1, 3, 4, 7, 6, 12, 8, 9, 13, 18, 12, 28, 14, 24, 24, 27, 18, 39, 20, 42, 32, 36, 24, 36, 31, 42, 28, 56, 30, 72, 32, 63, 48, 54, 48, 91, 38, 60, 56, 54, 42, 96, 44, 84, 78, 72, 48, 108, 57, 93, 72, 98, 54, 84, 72, 72, 80, 90, 60, 168, 62, 96, 104, 73, 84, 144, 68, 126, 96

Offset: 1

Yasutoshi Kohmoto

			Let n = 28 = 2^2*7. Then a(n) = (2^2 + 2 + 1)*(7 + 1) = 56. - _Vladimir Shevelev_, May 07 2013

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
J. O. M. Pedersen, Tables of Aliquot Cycles, backup on web.archive.org of no more exiting page, as of May 2014
J. O. M. Pedersen, Tables of Aliquot Cycles [Cached copy, pdf file only]

Cf. A049417 (2-infinitary), A074847 (4-infinitary), A097863 (5-infinitary).

Cf. A000244, A030341, A027748, A124010.

following Bower and Harris:
a049418 1 = 1
a049418 n = product $ zipWith f (a027748_row n) (a124010_row n) where
   f p e = product $ zipWith div
           (map (subtract 1 . (p ^)) $
                zipWith (*) a000244_list $ map (+ 1) $ a030341_row e)
           (map (subtract 1 . (p ^)) a000244_list)
-- Reinhard Zumkeller, Sep 18 2015

A049418 := proc(n) option remember; local ifa,a,p,e,d,k ; ifa := ifactors(n)[2] ; a := 1 ; if nops(ifa) = 1 then p := op(1,op(1,ifa)) ; e := op(2,op(1,ifa)) ; d := convert(e,base,3) ; for k from 0 to nops(d)-1 do a := a*(p^((1+op(k+1,d))*3^k)-1)/(p^(3^k)-1) ; end do: else for d in ifa do a := a*procname( op(1,d)^op(2,d)) ; end do: return a; end if; end proc:
seq(A049418(n),n=1..40) ; # R. J. Mathar, Oct 06 2010

A049418[n_] := Module[{ifa = FactorInteger[n], a = 1, p, e, d, k}, If[ Length[ifa] == 1, p = ifa[[1, 1]]; e = ifa[[1, 2]]; d = Reverse[ IntegerDigits[e, 3] ]; For[k = 1, k <= Length[d], k++, a = a*(p^((1 + d[[k]])*3^(k - 1)) - 1)/(p^(3^(k - 1)) - 1)], Do[ a = a*A049418[ d[[1]]^d[[2]] ], {d, ifa}]]; Return[a] ]; A049418[1] = 1; Table[ A049418[n] , {n, 1, 69}] (* Jean-François Alcover, Jan 03 2012, after R. J. Mathar *)

apply( {A049418(n)=vecprod([prod(k=1,#n=digits(f[2],3),(f[1]^(3^(#n-k)*(n[k]+1))-1)\(f[1]^3^(#n-k)-1))|f<-factor(n)~])}, [1..99]) \\ M. F. Hasler, Sep 21 2022

Multiplicative with a(p^e) = prod_{k >= 0} (p^(3^k*{d_k+1}) - 1)/(p^(3^k) - 1), where e = sum_{k >= 0} d_k 3^k (base 3 representation). - Christian G. Bower and Mitch Harris, May 20 2005. [Edited by M. F. Hasler, Sep 21 2022]

Denote P_3 = {p^3^k}, k = 0, 1, ..., p runs primes. Then every n has a unique representation of the form n = prod q_i prod (r_j)^2, where q_i, r_j are distinct elements of P_3. Using this representation, we have a(n) = prod (q_i+1)*prod ((r_j)^2+r_j+1). - Vladimir Shevelev, May 07 2013

More terms from Naohiro Nomoto, Sep 10 2001

A097464 5-infinitary perfect numbers: numbers k such that 5-infinitary-sigma(k) = 2*k.

Original entry on oeis.org

6, 28, 496, 47520, 288288, 308474880

Offset: 1

Yasutoshi Kohmoto

Here 5-infinitary-sigma(k) means sum of 5-infinitary-divisors of k. If k = Product p_i^r_i and d = Product p_i^s_i, each s_i has a digit a <= b in its 5-ary expansion everywhere that the corresponding r_i has a digit b, then d is a 5-infinitary-divisor of k.
Is it certain that 308474880 is the 6th term? M. F. Hasler, Nov 20 2010
Data is verified. a(7) > 10^11, if it exists. - Amiram Eldar, Oct 24 2024

			Factorizations: 2*3, 2^2*7, 2^4*31, 2^5*3^3*5*11, 2^5*3^2*7*11*13, 2^10*3*5*7*19*151.

Cf. A007357, A038182, A074849, A097863.

f[p_, e_] := Module[{d = IntegerDigits[e, 5]}, m = Length[d]; Product[(p^((d[[j]] + 1)*5^(m - j)) - 1)/(p^(5^(m - j)) - 1), {j, 1, m}]]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[300000], s[#] == 2*# &] (* Amiram Eldar, Oct 24 2024 *)

{k: A097863(k) = 2*k}.

Missing a(4) inserted by R. J. Mathar, Nov 20 2010

A331107 The sum of Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the Zeckendorf expansion (A014417) of each s(i) contains only terms that are in the Zeckendorf expansion of r(i).

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 9, 10, 18, 12, 20, 14, 24, 24, 27, 18, 30, 20, 30, 32, 36, 24, 36, 26, 42, 28, 40, 30, 72, 32, 33, 48, 54, 48, 50, 38, 60, 56, 54, 42, 96, 44, 60, 60, 72, 48, 108, 50, 78, 72, 70, 54, 84, 72, 72, 80, 90, 60, 120, 62, 96, 80, 99, 84, 144, 68

Offset: 1

Amiram Eldar, Jan 09 2020

First differs from A034448 at n = 16.

			a(16) = 27 since 16 = 2^4 and the Zeckendorf expansion of 4 is 101, i.e., its Zeckendorf representation is a set with 2 terms: {1, 3}. There are 4 possible exponents of 2: 0, 1, 3 and 4, corresponding to the subsets {}, {1}, {3} and {1, 3}. Thus 16 has 4 Zeckendorf-infinitary divisors: 2^0 = 1, 2^1 = 2, 2^3 = 8, and 2^4 = 16, and their sum is 1 + 2 + 8 + 16 = 27.

Amiram Eldar, Table of n, a(n) for n = 1..10000

The number of Zeckendorf-infinitary divisors of n is in A318465.

Cf. A014417, A034448, A049417, A049418, A074847, A097863.

fb[n_] := Block[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; Fibonacci[1 + Position[Reverse@fr, ?(# == 1 &)]]]; f[p, e_] := p^fb[e]; a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100] (* after Robert G. Wilson v at A014417 *)

Multiplicative with a(p^e) = Product_{i} (p^s(i) + 1), where s(i) are the terms in the Zeckendorf representation of e (A014417).

A376888 The sum of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 21, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 63, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 84, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 65, 84, 144

Offset: 1

Amiram Eldar, Oct 08 2024

See A376885 for details about this factorization.
First differs from A188999 at n = 16.
The number of these divisors is given by A376887(n).

			For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) is the sum of the 4 divisors 1 + 3 + 4 + 12 = 20.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A188999, A376885, A376886, A376887.

Similar sequences: A049417, A049418, A074847, A097863, A331107, A331110.

ff[q_, s_] := (q^(s + 1) - 1)/(q - 1); f[p_, e_] := Module[{k = e, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, AppendTo[s, {p^(m - 1)!, r}];]; m++]; Times @@ ff @@@ s]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

fdigits(n) = {my(k = n, m = 2, r, s = []); while([k, r] = divrem(k, m); k != 0 || r != 0, s = concat(s, r); m++); s;}
a(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], d); prod(i = 1, #p, prod(j = 1, #d=fdigits(e[i]), (p[i]^(j!*(d[j]+1)) - 1)/(p[i]^j! - 1)));}

Multiplicative: if e = Sum_{k>=1} d_k * k! (factorial base representation), then a(p^e) = Product_{k>=1} (p^(k!*{d_k+1}) - 1)/(p^(k!) - 1).

A331110 The sum of dual-Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the dual Zeckendorf expansion (A104326) of each s(i) contains only terms that are in the dual Zeckendorf expansion of r(i).

Original entry on oeis.org

1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 27, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 45, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 108, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 135, 84, 144

Offset: 1

Amiram Eldar, Jan 09 2020

First differs from A188999 at n = 32.

			a(32) = 45 since 32 = 2^5 and the dual Zeckendorf expansion of 5 is 110, i.e., its dual Zeckendorf representation is a set with 2 terms: {2, 3}. There are 4 possible exponents of 2: 0, 2, 3 and 5, corresponding to the subsets {}, {2}, {3} and {2, 3}. Thus 32 has 4 dual-Zeckendorf-infinitary divisors: 2^0 = 1, 2^2 = 4, 2^3 = 8, and 2^5 = 32, and their sum is 1 + 4 + 8 + 32 = 45.

Amiram Eldar, Table of n, a(n) for n = 1..10000

The number of dual-Zeckendorf-infinitary divisors of n is in A331109.

Cf. A049417, A049418, A074847, A097863, A104326, A188999, A331107.

fibTerms[n_] := Module[{k = Ceiling[Log[GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k--]; fr];
dualZeck[n_] := Module[{v = fibTerms[n]}, nv = Length[v]; i = 1; While[i <= nv - 2, If[v[[i]] == 1 && v[[i + 1]] == 0 && v[[i + 2]] == 0, v[[i]] = 0; v[[i + 1]] = 1; v[[i + 2]] = 1; If[i > 2, i -= 3]]; i++]; i = Position[v, _?(# > 0 &)]; If[i == {}, {}, v[[i[[1, 1]] ;; -1]]]];
f[p_, e_] := p^Fibonacci[1 + Position[Reverse@dualZeck[e], _?(# == 1 &)]];
a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n]) + 1); Array[a, 100]

Multiplicative with a(p^e) = Product_{i} (p^s(i) + 1), where s(i) are the terms in the dual Zeckendorf representation of e (A104326).

cp's OEIS Frontend

A049417 a(n) = isigma(n): sum of infinitary divisors of n.

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A074847 Sum of 4-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its 4-ary expansion everywhere that the corresponding r(i) has a digit b, then d is a 4-infinitary-divisor of n.

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A049418 3-i-sigma(n): sum of 3-infinitary divisors of n: if n=Product p(i)^r(i) and d=Product p(i)^s(i), each s(i) has a digit a<=b in its ternary expansion everywhere that the corresponding r(i) has a digit b, then d is a 3-i-divisor of n.

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A097464 5-infinitary perfect numbers: numbers k such that 5-infinitary-sigma(k) = 2*k.

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A331107 The sum of Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the Zeckendorf expansion (A014417) of each s(i) contains only terms that are in the Zeckendorf expansion of r(i).

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A331110 The sum of dual-Zeckendorf-infinitary divisors of n = Product_{i} p(i)^r(i): divisors d = Product_{i} p(i)^s(i), such that the dual Zeckendorf expansion (A104326) of each s(i) contains only terms that are in the dual Zeckendorf expansion of r(i).