A074848 -id:A074848 - 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

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

A038148 Number 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-infinitary-divisor of n.

Original entry on oeis.org

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

Offset: 1

Yasutoshi Kohmoto

Multiplicative: If e = sum d_k 3^k, then a(p^e) = prod (d_k+1). - Christian G. Bower, May 19 2005

			2^3*3 is a 3-infinitary-divisor of 2^5*3^2 because 2^3*3 = 2^10*3^1 and 2^5*3^2 = 2^12*3^2 in ternary expanded power. All corresponding digits satisfy the condition. 1 <= 1, 0 <= 2, 1 <= 2.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Frédéric Chyzak, Ivan Gutman, and Peter Paule, Predicting the number of hexagonal systems with 24 and 25 hexagons, Communications in Mathematical and Computer Chemistry (1999) No. 40, 139-151. See p. 141.
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]
Index entries for sequences computed from exponents in factorization of n

Cf. A006047, A037445, A038182, A074848.

A038148 := 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,3) ; a := a*mul(d+1,d=e) ; end do: end if; end proc:
seq(A038148(n),n=1..50) ; # R. J. Mathar, Feb 08 2011

a[1] = 1; a[n_] := (k = 1; Do[k = k * Times @@ (IntegerDigits[f, 3] + 1), {f, FactorInteger[n][[All, 2]]}]; k); Table[a[n], {n, 1, 102}](* Jean-François Alcover, Feb 03 2012, after R. J. Mathar *)

A006047(n) = { my(m=1, d); while(n, d = (n%3); m *= (1+d); n = (n-d)/3); m; };
A038148(n) = factorback(apply(e -> A006047(e), factorint(n)[, 2])); \\ (After A037445) - Antti Karttunen, May 28 2017

(define (A038148 n) (if (= 1 n) n (* (A006047 (A067029 n)) (A038148 (A028234 n))))) ;; Antti Karttunen, May 28 2017

a(1) = 1; for n > 1, a(n) = A006047(A067029(n)) * a(A028234(n)). [After Christian G. Bower's 2005 comment.] - Antti Karttunen, May 28 2017

More terms from Naohiro Nomoto, Jun 21 2001

Data section further extended to 105 terms by Antti Karttunen, May 28 2017

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

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).

A268444 a(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*4^i is the base-4 representation of n.

Original entry on oeis.org

1, 2, 3, 4, 2, 4, 6, 8, 3, 6, 9, 12, 4, 8, 12, 16, 2, 4, 6, 8, 4, 8, 12, 16, 6, 12, 18, 24, 8, 16, 24, 32, 3, 6, 9, 12, 6, 12, 18, 24, 9, 18, 27, 36, 12, 24, 36, 48, 4, 8, 12, 16, 8, 16, 24, 32, 12, 24, 36, 48, 16, 32, 48, 64, 2, 4, 6, 8, 4, 8, 12, 16, 6, 12, 18, 24

Offset: 0

Tom Edgar, Feb 04 2016

a(n) gives the number of 1's in row n of A243756.

			The base-4 representation of 10 is (2,2) so a(10) = (2+1)*(2+1) = 9.

Antti Karttunen, Table of n, a(n) for n = 0..16384
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Cf. A001316, A006047, A074848.

a(n) = my(d=digits(n,4)); prod(k=1, #d, d[k]+1); \\ Michel Marcus, Feb 05 2016

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

[prod(x+1 for x in n.digits(4)) for n in [0..75]]

(define (A268444 n) (if (zero? n) 1 (let ((d (mod n 4))) (* (+ 1 d) (A268444 (/ (- n d) 4)))))) ;; For R6RS standard. Use modulo instead of mod in older Schemes like MIT/GNU Scheme. - Antti Karttunen, May 28 2017

a(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*4^i.

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Scheme

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A038148 Number 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-infinitary-divisor of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Scheme

Formula

Extensions

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

Formula

Extensions

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).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A268444 a(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*4^i is the base-4 representation of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs