A038148 -id:A038148 - cp's OEIS frontend

A037445 Number of infinitary divisors (or i-divisors) of n.

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4, 8, 2, 8, 4, 4, 4, 4, 4, 8, 2, 4, 4, 4, 2, 8, 2, 8, 8

Offset: 1

Yasutoshi Kohmoto

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.

The smallest number m with exactly 2^n infinitary divisors is A037992(n); for these values m, a(m) increases also to a new record. - Bernard Schott, Mar 09 2023

			For n = 8, n = 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) = 4.
For n = 90, n = 2*5*9 where 2,5,9 are in A050376, so a(90) = 2^3 = 8.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Steven R. Finch, Unitarism and Infinitarism, February 25, 2004. [Cached copy, with permission of the author]
J. O. M. Pedersen, Tables of Aliquot Cycles [Broken link]
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]
Eric Weisstein's World of Mathematics, Infinitary Divisor
Index entries for sequences computed from exponents in factorization of n

Cf. A000120, A001316, A004607, A007358, A007357, A037992, A038148, A049417, A064547, A074848, A077609, A124010, A156552, A286575.

a037445 = product . map (a000079 . a000120) . a124010_row
-- Reinhard Zumkeller, Mar 19 2013

A037445 := proc(n)
    local a,p;
    a := 1 ;
    for p in ifactors(n)[2] do
        a := a*2^wt(p[2]) ;
    end do:
    a ;
end proc: # R. J. Mathar, May 16 2016

Table[Length@((Times @@ (First[it]^(#1 /. z -> List)) & ) /@
Flatten[Outer[z, Sequence @@ bitty /@
Last[it = Transpose[FactorInteger[k]]], 1]]), {k, 2, 240}]
bitty[k_] := Union[Flatten[Outer[Plus, Sequence @@ ({0, #1} & ) /@ Union[2^Range[0, Floor[Log[2, k]]]*Reverse[IntegerDigits[k, 2]]]]]]
y[n_] := Select[Range[0, n], BitOr[n, # ] == n & ] divisors[Infinity][1] := {1}
divisors[Infinity][n_] := Sort[Flatten[Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^y[m])]]] Length /@ divisors[Infinity] /@ Range[105] (* Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 29 2005 *)
a[1] = 1; a[n_] := Times @@ Flatten[ 2^DigitCount[#, 2, 1]&  /@ FactorInteger[n][[All, 2]] ]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Aug 19 2013, after Reinhard Zumkeller *)

A037445(n) = factorback(apply(a -> 2^hammingweight(a), factorint(n)[,2])) \\ Andrew Lelechenko, May 10 2014

from sympy import factorint
def wt(n): return bin(n).count("1")
def a(n):
    f=factorint(n)
    return 2**sum([wt(f[i]) for i in f]) # Indranil Ghosh, May 30 2017

(define (A037445 n) (if (= 1 n) n (* (A001316 (A067029 n)) (A037445 (A028234 n))))) ;; Antti Karttunen, May 28 2017

Multiplicative with a(p^e) = 2^A000120(e). - David W. Wilson, Sep 01 2001
Let n = q_1*...*q_k, where q_1,...,q_k are different terms of A050376. Then a(n) = 2^k (the number of subsets of a set with k elements is 2^k). - Vladimir Shevelev, Feb 19 2011.
a(n) = Product_{k=1..A001221(n)} A000079(A000120(A124010(n,k))). - Reinhard Zumkeller, Mar 19 2013
From Antti Karttunen, May 28 2017: (Start)
a(n) = A286575(A156552(n)). [Because multiplicative with a(p^e) = A001316(e).]
a(n) = 2^A064547(n). (End)
a(A037992(n)) = 2^n. - Bernard Schott, Mar 10 2023

Corrected and extended by Naohiro Nomoto, Jun 21 2001

A006047 Number of entries in n-th row of Pascal's triangle not divisible by 3.

Original entry on oeis.org

1, 2, 3, 2, 4, 6, 3, 6, 9, 2, 4, 6, 4, 8, 12, 6, 12, 18, 3, 6, 9, 6, 12, 18, 9, 18, 27, 2, 4, 6, 4, 8, 12, 6, 12, 18, 4, 8, 12, 8, 16, 24, 12, 24, 36, 6, 12, 18, 12, 24, 36, 18, 36, 54, 3, 6, 9, 6, 12, 18, 9, 18, 27, 6, 12, 18, 12, 24, 36, 18, 36, 54, 9, 18, 27, 18, 36, 54, 27, 54

Offset: 0

Jeffrey Shallit

Fixed point of the morphism a -> a, 2a, 3a, starting from a(1) = 1. - Robert G. Wilson v, Jan 24 2006
This is a particular case of the number of entries in n-th row of Pascal's triangle not divisible by a prime p, which is given by a simple recursion using ⊗, the Kronecker (or tensor) product of vectors. Let v_0=(1,2,...,p). Then v_{n+1}=v_0 ⊗ v_n, where the vector v_n contains the values for the first p^n rows of Pascal's triangle (rows 0 through p^n-1). - William B. Everett (bill(AT)chgnet.ru), Mar 29 2008
a(n) = A206424(n) + A227428(n); number of nonzero terms in row n of triangle A083093. - Reinhard Zumkeller, Jul 11 2013

			15 in base 3 is 120, here r=1 and s=1 so a(15) = 3*2 = 6.
William B. Everett's comment with p=3, n=2: v_0 = (1,2,3), v_1 = (1,2,3) => v_2 = (1*1,1*2,1*3,2*1,2*2,2*3,3*1,3*2,3*3) = (1,2,3,2,4,6,3,6,9), the first 3^2 values of the present sequence. - _Wolfdieter Lang_, Mar 19 2014

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Antti Karttunen, Table of n, a(n) for n = 0..19683 (terms 0..1000 from Reinhard Zumkeller).
J.-P. Allouche and Jeffrey Shallit, The ring of k-regular sequences, Theoretical Computer Sci., 98 (1992), 163-197.
Michael Gilleland, Some Self-Similar Integer Sequences
H. Harborth, Number of Odd Binomial Coefficients, Proc. Amer. Math. Soc. 62.1 (1977), 19-22. (Annotated scanned copy)
Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Periodic minimum in the count of binomial coefficients not divisible by a prime, arXiv:2408.06817 [math.NT], 2024.
Sam Northshield, Sums across Pascal's triangle modulo 2, Congressus Numerantium, 200, pp. 35-52, 2010.
Index entries for sequences that are fixed points of mappings

Cf. A001316, A003586, A038148, A053735, A083093, A089898, A206424, A227428, A286586, A286587, A286633.

a006047 = sum . map signum . a083093_row
-- Reinhard Zumkeller, Jul 11 2013

p:=proc(n) local ct, k: ct:=0: for k from 0 to n do if binomial(n,k) mod 3 = 0 then else ct:=ct+1 fi od: end: seq(p(n),n=0..82); # Emeric Deutsch
f:= proc(n) option remember; ((n mod 3)+1)*procname(ceil((n+1)/3)-1) end proc:
f(0):= 1: f(1):= 2:
seq(f(i), i=0..100); # Robert Israel, Oct 15 2015

Nest[Flatten[ # /. a_Integer -> {a, 2a, 3a}] &, {1}, 4] (* Robert G. Wilson v, Jan 24 2006 *)
Nest[ Join[#, 2#, 3#] &, {1}, 4] (* Robert G. Wilson v, Jul 27 2014 *)

b(n)=if(n<3,n,if(n%3==0,3*b(n/3),if(n%3==1,1*b((n+2)/3),2*b((n+1)/3)))) \\ Ralf Stephan

A006047(n) = b(1+n); \\ (The above PARI-program by Ralf Stephan is for offset-1-version of this sequence.) - Antti Karttunen, May 28 2017

A006047(n) = { my(m=1, d); while(n, d = (n%3); m *= (1+d); n \= 3); m; }; \\ Antti Karttunen, May 28 2017

a(n) = prod(i=1,#d=digits(n, 3), (1+d[i])) \\ David A. Corneth, May 28 2017

upto(n) = my(res = [1], v); while(#res < n, v = concat(2*res, 3*res); res = concat(res, v)); res \\ David A. Corneth, May 29 2017

from sympy.ntheory.factor_ import digits
from sympy import prod
def a(n):
    d=digits(n, 3)
    return n + 1 if n<3 else prod(1 + d[i] for i in range(1, len(d)))
print([a(n) for n in range(51)]) # Indranil Ghosh, Jun 06 2017

from sympy.ntheory import digits
def A006047(n): return 3**(s:=digits(n,3)).count(2)<Chai Wah Wu, Apr 24 2025

(define (A006047 n) (if (zero? n) 1 (let ((d (mod n 3))) (* (+ 1 d) (A006047 (/ (- n d) 3)))))) ;; For R6RS standard. Use modulo instead of mod in older Schemes like MIT/GNU Scheme. - Antti Karttunen, May 28 2017

Write n in base 3; if the representation contains r 1's and s 2's then a(n) = 3^s * 2^r. Also a(n) = Sum_{k=0..n} (C(n, k)^2 mod 3). - Avi Peretz (njk(AT)netvision.net.il), Apr 21 2001
a(n) = b(n+1), with b(1)=1, b(2)=2, b(3n)=3b(n), b(3n+1)=b(n+1), b(3n+2)=2b(n+1). - Ralf Stephan, Sep 15 2003
G.f.: Product_{n>=0} (1+2*x^(3^n)+3*x^(2*3^n)) (Northshield). - Johannes W. Meijer, Jun 05 2011
G.f. g(x) satisfies g(x) = (1 + 2*x + 3*x^2)*g(x^3). - Robert Israel, Oct 15 2015
From Tom Edgar, Oct 15 2015: (Start)
a(3^k) = 2 for k>=0;
a(2*3^k) = 3 for k>=0;
a(n) = Product_{b_j != 0} a(b_j*3^j) where n = Sum_{j>=0} b_j*3^j is the ternary representation of n. (End)
A056239(a(n)) = A053735(n). - Antti Karttunen, Jun 03 2017
a(n) = Sum_{k = 0..n} mod(C(n,k)^2, 3). - Peter Bala, Dec 17 2020

More terms from Ralf Stephan, Sep 15 2003

A318465 The number 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, 2, 2, 2, 2, 4, 2, 2, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 2, 4, 2, 8, 2, 2, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 4, 4, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 4, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4, 4, 2, 8, 4, 4, 4, 4, 4, 4, 2, 4, 4, 4, 2, 8, 2, 4, 8

Offset: 1

Antti Karttunen, Aug 30 2018

Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with Zeckendorf expansion instead of binary expansion. - Amiram Eldar, Jan 09 2020

			a(16) = 4 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.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A007895, A318464, A318469.
Cf. A014417, A037445, A038148, A074848, A077609.
Cf. also A318472, A318474.

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_] := 2^Length@fb[e]; a[1] = 1; a[n_] := Times @@ (Flatten@(f @@@ FactorInteger[n])); Array[a, 100] (* Amiram Eldar, Jan 09 2020 after Robert G. Wilson v at A014417 *)

A072649(n) = { my(m); if(n<1, 0, m=0; until(fibonacci(m)>n, m++); m-2); }; \\ From A072649
A007895(n) = { my(s=0); while(n>0, s++; n -= fibonacci(1+A072649(n))); (s); }
A318465(n) = factorback(apply(e -> 2^A007895(e),factor(n)[,2]));

Multiplicative with a(p^e) = 2^A007895(e), where A007895(n) gives the number of terms in the Zeckendorf representation of n.

a(n) = 2^A318464(n).

Name edited and interpretation in terms of divisors added by Amiram Eldar, Jan 09 2020

A074848 Number of 4-infinitary divisors of n.

Original entry on oeis.org

1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, 6, 2, 4, 4, 2, 2, 6, 2, 6, 4, 4, 2, 8, 3, 4, 4, 6, 2, 8, 2, 4, 4, 4, 4, 9, 2, 4, 4, 8, 2, 8, 2, 6, 6, 4, 2, 4, 3, 6, 4, 6, 2, 8, 4, 8, 4, 4, 2, 12, 2, 4, 6, 6, 4, 8, 2, 6, 4, 8, 2, 12, 2, 4, 6, 6, 4, 8, 2, 4, 2, 4, 2, 12, 4, 4, 4, 8, 2, 12, 4, 6, 4, 4, 4, 8, 2, 6, 6, 9, 2, 8, 2, 8, 8

Offset: 1

Yasutoshi Kohmoto, Sep 10 2002

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.

			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.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n

Cf. A037445, A038148, A074847, A268444.

A074848 := proc(n) if n= 1 then 1; else ifa := ifactors(n)[2] ; a := 1; for f in ifa do e := convert(op(2,f),base,4) ; a := a*mul(d+1,d=e) ; end do: end if; end proc:
seq(A074848(n),n=1..70) ; # R. J. Mathar, Feb 08 2011

f[p_, e_] := Times @@ (IntegerDigits[e, 4] + 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]  (* Amiram Eldar, Sep 09 2020 *)

A268444(n) = { my(m=1, d); while(n, d = (n%4); m *= (1+d); n = (n-d)/4); m; };
A074848(n) = factorback(apply(e -> A268444(e), factorint(n)[, 2])) \\ (After A037445) - Antti Karttunen, May 28 2017

from math import prod
from sympy import factorint
from gmpy2 import digits
def A268444(n):
    s = digits(n,4)
    return prod((int(d)+1)**s.count(d) for d in '123')
def A074848(n): return prod(A268444(e) for e in factorint(n).values()) # Chai Wah Wu, Apr 24 2025

(definec (A074848 n) (if (= 1 n) n (* (A268444 (A067029 n)) (A074848 (A028234 n))))) ;; Antti Karttunen, May 28 2017

Multiplicative: If e = sum d_k 4^k, then a(p^e) = prod (d_k+1). - Christian G. Bower, May 19 2005

a(1) = 1; for n > 1, a(n) = A268444(A067029(n)) * a(A028234(n)). [After Christian G. Bower's 2005 formula.] - _Antti Karttunen, May 28 2017

More terms from Antti Karttunen, May 28 2017

Name shortened by Amiram Eldar, Sep 09 2020

A331109 The number 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, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 8, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 8, 4, 4, 2, 8, 4, 4, 4

Offset: 1

Amiram Eldar, Jan 09 2020

Dual-Zeckendorf-infinitary divisors are analogous to infinitary divisors (A077609) with dual Zeckendorf expansion instead of binary expansion.

First differs from A286324 at n = 32.

			a(32) = 4 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.

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

Cf. A037445, A038148, A074848, A077609, A104326, A112310, A286324, A318465.

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 == {}, 1, 2^Total[v[[i[[1, 1]] ;; -1]]]]];
f[p_, e_] := dualZeck[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

Multiplicative with a(p^e) = 2^A112310(e).

A038182 3-infinitary perfect numbers k: 3-i-sigma(k) = 2*k, where 3-i-sigma = A049418.

Original entry on oeis.org

6, 28, 3024, 6552, 27578880, 49266240, 49095705098695680

Offset: 1

Yasutoshi Kohmoto

Similarly, we have 3-i-sigma(x)/x = r for the following numbers: r = 3 for x = 672, 13104, 4021920, 55157760, 98532480, 459818240, 372667889664, 7267023848448, 1178296922368696320, 5498718971053916160, ...; r = 4 for x = 2178540; r = 3/2 for x = 2, 24, 9192960, 196382820394782720. (Values above 10^7 from Yasutoshi Kohmoto, some terms may be missing.) - M. F. Hasler, Sep 21 2022

			Factorizations: 2*3, 2^2*7, 2^4*3^3*7, 2^3*3^2*7*13, 2^9*3^4*5*7*19, 2^6*3*5*19*37*73, 2^10*3^6*5*19^2*127*379*757.

J. O. M. Pedersen, Tables of Aliquot Cycles: backup on web.archive.org of no more existing web page, as of May 2014.
J. O. M. Pedersen, Tables of Aliquot Cycles. [Cached copy, pdf file only]

Cf. A007357, A037445, A038148, A049418, A074849, A097464.

f[p_, e_] := Module[{d = IntegerDigits[e, 3]}, m = Length[d]; Product[(p^((d[[j]] + 1)*3^(m - j)) - 1)/(p^(3^(m - j)) - 1), {j, 1, m}]]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[7000], s[#] == 2*# &] (* Amiram Eldar, Oct 24 2024 *)

is_A038182(n)=A049418(n)==2*n \\ M. F. Hasler, Sep 21 2022

Definition shortened by R. J. Mathar, Oct 06 2010

A294933 Compound filter related to base-3 expansion of the exponents in prime factorization of n: a(n) = P(A294932(n), A294931(n)), where P(n,k) is sequence A000027 used as a pairing function.

Original entry on oeis.org

1, 2, 2, 3, 2, 7, 2, 4, 3, 7, 2, 5, 2, 7, 7, 16, 2, 5, 2, 5, 7, 7, 2, 16, 3, 7, 4, 5, 2, 29, 2, 8, 7, 7, 7, 10, 2, 7, 7, 16, 2, 29, 2, 5, 5, 7, 2, 67, 3, 5, 7, 5, 2, 16, 7, 16, 7, 7, 2, 12, 2, 7, 5, 6, 7, 29, 2, 5, 7, 29, 2, 8, 2, 7, 5, 5, 7, 29, 2, 67, 16, 7, 2, 12, 7, 7, 7, 16, 2, 12, 7, 5, 7, 7, 7, 23, 2, 5, 5, 10, 2, 29, 2, 16, 29, 7, 2, 8, 2, 29, 7, 67

Offset: 1

Antti Karttunen, Nov 11 2017

nonn

For all i, j: a(i) = a(j) => A038148(i) = A038148(j).

Antti Karttunen, Table of n, a(n) for n = 1..65537
Index entries for sequences computed from exponents in factorization of n

Cf. A294931, A294932.
Cf. also A293225, A293226 and A293442 (analogous filter for base-2).
Cf. A010057, A038148, A240231.

a(n) = (1/2)*(2 + ((A294932(n) + A294931(n))^2) - A294932(n) - 3*A294931(n)).

A376887 The number 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, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 6, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 6, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 2, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 6, 3, 4, 2, 8, 4, 4, 4

Offset: 1

Amiram Eldar, Oct 08 2024

See A376885 for details about this factorization.
If n = Product p_i^e_i is the canonical prime factorization of n, then the divisors that are counted by this function are d = Product p_i^s_i, where all the digits of s_i in factorial base are not larger than the corresponding digits of e_i.
The sum of these divisors is given by A376888(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) = (1+1) * (1+1) = 4, corresponding to the 4 divisors 1, 3, 4 and 12.

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

Cf. A007623, A034968, A376885, A376886, A376888.

Similar sequences: A037445, A038148, A074848, A318465, A331109.

fdigprod[n_] := Module[{k = n, m = 2, r, s = 1}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, s *= (r+1); m++]; s]; f[p_, e_] := fdigprod[e]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100]

fdigprod(n) = {my(k = n, m = 2, r, s = 1); while([k, r] = divrem(k, m); k != 0 || r != 0, s *= (r+1); m++); s;}
a(n) = {my(e = factor(n)[, 2]); prod(i = 1, #e, fdigprod(e[i]));}

Multiplicative with a(p^e) = A227154(e).

cp's OEIS Frontend

A037445 Number of infinitary divisors (or i-divisors) of n.

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A006047 Number of entries in n-th row of Pascal's triangle not divisible by 3.

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A318465 The number 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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A074848 Number of 4-infinitary divisors of n.

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A331109 The number 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).

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