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Number of digits in the decimal expansion of the regular unit fractions 1/ A003592.
+20
4
1, 2, 1, 3, 1, 4, 2, 2, 5, 3, 2, 6, 4, 2, 3, 7, 5, 3, 3, 8, 6, 4, 3, 9, 4, 7, 5, 3, 10, 4, 8, 6, 4, 11, 4, 9, 5, 7, 5, 12, 4, 10, 5, 8, 6, 13, 4, 11, 5, 9, 6, 7, 14, 5, 12, 5, 10, 6, 8, 15, 6, 13, 5, 11, 6, 9, 16, 7, 7, 14, 5, 12, 6, 10, 17, 7, 8, 15, 6, 13, 6, 11, 18, 7, 9, 16, 8, 7, 14, 6, 12, 19
EXAMPLE
1/ A003592(2) = 1/2 = 0.5, with 1 digit after the decimal point, so a(2) = 1.
MATHEMATICA
digNum[n_] := Length[(dig = RealDigits[1/n, 10])[[1]]] - dig[[2]]; s = {}; m = 12; Do[n = 5^k; While[n <= 5^m, AppendTo[s, n]; n *= 2], {k, 0, m}]; Rest[digNum /@ Union[s]] (* Amiram Eldar, Feb 08 2020 *)
The number of terms of the form 2^i*5^j ( A003592) less than or equal to 10^n.
+20
2
1, 6, 15, 29, 48, 72, 100, 134, 172, 214, 262, 314, 371, 433, 500, 571, 647, 728, 813, 904, 999, 1099, 1204, 1313, 1427, 1546, 1670, 1798, 1932, 2070, 2212, 2359, 2511, 2668, 2829, 2996, 3167, 3342, 3523, 3708, 3898, 4093, 4293, 4497, 4706, 4920, 5138
MATHEMATICA
f[n_] := Sum[1 + Floor@ Log[2, 10^n/5^k], {k, 0, Floor@ Log[5, 10^n]}]; Array[f, 47, 0]
a(n) = least k such that A003592(n) | 20^k.
+20
2
0, 1, 1, 1, 2, 1, 2, 1, 2, 3, 2, 2, 3, 2, 2, 3, 4, 3, 2, 3, 4, 3, 2, 3, 5, 4, 4, 3, 3, 5, 4, 4, 3, 3, 6, 4, 5, 5, 4, 3, 6, 4, 5, 5, 4, 3, 7, 4, 6, 5, 5, 6, 4, 7, 4, 6, 5, 5, 6, 4, 8, 4, 7, 5, 6, 6, 5, 8, 7, 4, 7, 5, 6, 6, 5, 9, 7, 4, 8, 5, 7, 6, 6, 9, 7, 5, 8
COMMENTS
Also, number of digits in the vigesimal (base 20) expansion of terminating unit fractions 1/ A003592.
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.
LINKS
Eric Weisstein's World of Mathematics, Vigesimal
EXAMPLE
a(1) = 0 since A003592(1) = 1 | 20^0. a(4) = 1 since A003592(4) = 5 | 20^1; 1/5 in base 20 = .4. a(5) = 2 since A003592(5) = 8 | 20^2; 1/8 in base 20 = .2a, where "a" is digit 10), etc.
MATHEMATICA
With[{nn = 360000}, Sort[Join @@ Table[{2^a*5^b, Max[Ceiling[a/2], b]}, {a, 0, Log2[nn]}, {b, 0, Log[5, nn/(2^a)]}]][[All, -1]] ]
1, 2, 1, 3, 2, 6, 4, 5, 7, 8, 10, 14, 16, 20, 25, 3, 32, 40, 50, 6, 64, 80, 100, 12, 113, 15, 160, 200, 24, 226, 30, 320, 400, 48, 452, 60, 565, 75, 800, 96, 904, 120, 1130, 150, 1600, 192, 1808, 240, 2260, 300, 2825, 375, 384, 3616, 480, 4520, 600, 5650
COMMENTS
The gaps between the natural numbers whose reciprocals are terminating decimals.
EXAMPLE
The first few terms of A003592 are 1, 2, 4, 5, 8, 10, 16 because 1/1, 1/2, 1,4, 1/5, 1/8 etc. terminate and 1/3, 1/6, 1/7, 1/9 etc. repeat. Hence the denominators of the first few terminating decimals are 1, 2, 4, 5, 8, 10, 16 and the first differences between these gives the sequence 1,2,1,3,2,4
AUTHOR
Barry Wells (wells.barry(AT)gmail.com), Aug 26 2009, Sep 24 2009
Least positive integer m > 1 such that 1 - m^k + m^(2k) - m^(3k) + m^(4k) is prime, where k = A003592(n).
+20
1
2, 4, 2, 6, 2, 20, 20, 26, 25, 10, 14, 5, 373, 4, 65, 232, 56, 2, 521, 911, 1156, 1619, 647, 511, 34, 2336, 2123, 1274, 2866, 951, 2199, 1353, 4965, 7396, 13513, 3692, 14103, 32275, 2257, 86, 3928, 2779, 18781, 85835, 820, 16647, 2468, 26677, 1172, 38361, 40842
COMMENTS
1 - m^k + m^(2*k) - m^(3^k) + m^(4*k) equals Phi(10*k,m).
First 15 terms were generated by the provided Mathematica program. All other terms found using OpenPFGW as Fermat and Lucas PRP. Term 16-20, 22-24, 27 have N^2-1 factored over 33.3% and proved using OpenPFGW;
terms 21, 25, 29-33, 36, 37, 39, 41, 42, 45, 48, 51 are proved using CHG pari script;
terms 26, 28, 34, 40 are proved using kppm PARI script;
terms 35, 38, 43, 44, 46, 47, 49, 50 do not yet have a primality certificate.
The corresponding prime number of term 51 (40842) has 236089 digits.
The corresponding prime numbers for the following terms are equal:
p(3) = p(2) = Phi(10, 2^4),
p(12) = p(9) = Phi(10, 5^50),
p(18) = p(14) = Phi(10, 2^160),
p(25) = p(21) = Phi(10, 34^512),
p(40) = p(34) = Phi(10, 86^4000).
EXAMPLE
n=1, A003592(1) = 1, when a=2, 1 - 2^1 + 2^2 - 2^3 + 2^4 = 11 is prime, so a(1)=2; n=2, A003592(2) = 2, when a=4, 1 - 4^2 + 4^4 - 4^6 + 4^8 = 61681 is prime, so a(2)=4; ...
n=13, A003592(13) = 64, when a=373, PrimeQ(1 - 373^64 + 373^128 - 373^192 + 373^256) = True, while for a = 2..372, PrimeQ(1 - a^64 + a^128 - a^192 + a^256) = False, so a(13)=373.
MATHEMATICA
fQ[n_] := PowerMod[10, n, n] == 0; a = Select[10 Range@100, fQ]/10; l = Length[a]; Table[m = a[[j]]; i = 1; While[i++; cp = 1 - i^m + i^(2*m)-i^(3*m)+i^(4*m); ! PrimeQ[cp]]; i, {j, 1, l}]
PROG
(PARI) do(k)=my(m=1); while(!ispseudoprime(polcyclo(10*k, m++)), ); m
list(lim)=my(v=List(), N); for(n=0, log(lim)\log(5), N=5^n; while(N<=lim, listput(v, N); N<<=1)); apply(do, vecsort(Vec(v))) \\ Charles R Greathouse IV, Apr 04 2012
EXTENSIONS
Term 50 added and comments updated by Lei Zhou, Jul 27 2012 Term 51 added and comments updated by Lei Zhou, Oct 10 2012
a(n) = the fractional part of 1 / A003592(n) read backwards.
+20
1
0, 5, 52, 2, 521, 1, 5260, 50, 40, 52130, 520, 20, 526510, 5210, 10, 800, 5218700, 52600, 500, 400, 52609300, 521300, 5200, 200, 521359100, 6100, 5265100, 52100, 100, 5265679000, 8000, 52187000, 526000, 5000, 52182884000, 4000, 526093000, 23000, 5213000, 52000
COMMENTS
Numbers in this sequence that also appear in A003592, sorted, include the product of numbers k | 10^e with integer e >= 0 and 10^m with m >= e. For instance, the proper divisors of 10 {1, 2, 5} appear and {10, 20, 40, 50} follow, finally {100, 200, 400, 500, 800} followed by any product k 10^m with k = {1, 2, 4, 5, 8} and m >= 3. - Michael De Vlieger, Feb 03 2018
FORMULA
a( A180953(n)) = 10^(n-1) for any n > 0.
EXAMPLE
The first terms, alongside A003592(n) and the fractional part of 1/ A003592(n), are: -- ---- ---------- ------------------
1 0 1 0
2 5 2 0.5
3 52 4 0.25
4 2 5 0.2
5 521 8 0.125
6 1 10 0.1
7 5260 16 0.0625
8 50 20 0.05
9 40 25 0.04
10 52130 32 0.03125
11 520 40 0.025
12 20 50 0.02
13 526510 64 0.015625
14 5210 80 0.0125
15 10 100 0.01
16 800 125 0.008
17 5218700 128 0.0078125
18 52600 160 0.00625
19 500 200 0.005
20 400 250 0.004
MATHEMATICA
With[{e = 12}, Table[FromDigits@ Reverse@ PadLeft[#1, Length@ #1 + Abs@ #2] - 10 Boole[n == 1] & @@ RealDigits[1/n], {n, Sort@ Flatten@ Table[2^i*5^j, {i, 0, e}, {j, 0, Log[5, 2^(e - i)]}]}]] (* Michael De Vlieger, Feb 03 2018, after Robert G. Wilson v at A003592 *)
PROG
(PARI) mx = 4000; A003592 = vecsort(concat(vector(1+logint(mx, 2), i, vector(1+logint(floor(mx/2^(i-1)), 5), j, 2^(i-1) * 5^(j-1))))) backward(n) = my (v=0, i=frac(1/n), r=1/10); while (i, v += r*floor(i); i=frac(i)*10; r*=10); v
Irregular triangle T(n,k) with row n listing A003592(j) not divisible by 20 such that A352218( A003592(j)) = n.
+20
1
1, 2, 4, 5, 10, 8, 16, 25, 50, 32, 64, 125, 250, 128, 256, 625, 1250, 512, 1024, 3125, 6250, 2048, 4096, 15625, 31250, 8192, 16384, 78125, 156250, 32768, 65536, 390625, 781250, 131072, 262144, 1953125, 3906250, 524288, 1048576, 9765625, 19531250, 2097152, 4194304, 48828125, 97656250
COMMENTS
All terms in A003592 are products T(n,k)*20^j, j >= 0. When expressed in base 20, T(n,k) does not end in zero, yet 1/T(n,k) is a terminating fraction, regular to 20.
The first 5 terms are the proper divisors of 20.
For these reasons, the terms may be called vigesimal "proper regular" numbers.
REFERENCES
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, Chapter IX: The Representation of Numbers by Decimals, Theorem 136. 8th ed., Oxford Univ. Press, 2008, 144-145.
LINKS
Eric Weisstein's World of Mathematics, Vigesimal
FORMULA
Row 0 contains the empty product, thus row length = 1.
Row n sorts {2^(2n-1), 5^n, 2^(2n), 2*5^n}, thus row length = 4.
EXAMPLE
Row 0 contains 1 since 1 is the empty product.
Row 1 contains 2, 4, 5, and 10 since these divide 20 and are not divisible by 20.
Row 2 contains 8, 16, 25, and 50 since these divide 20^2 but not 20. The other divisors of 20^2 either divide smaller powers of 20 or they are divisible by 20 and do not appear.
Row 3 contains 32, 64, 125, and 250 since these divide 20^3 but not 20^2. The other divisors of 20^3 either divide smaller powers of 20 or they are divisible by 20 therefore do not appear.
MATHEMATICA
{{1}}~Join~Array[Union@ Flatten@ {#, 2 #} &@ {2^(2 # - 1), 5^#} &, 11] // Flatten
5-smooth numbers, i.e., numbers whose prime divisors are all <= 5.
+10
114
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 30, 32, 36, 40, 45, 48, 50, 54, 60, 64, 72, 75, 80, 81, 90, 96, 100, 108, 120, 125, 128, 135, 144, 150, 160, 162, 180, 192, 200, 216, 225, 240, 243, 250, 256, 270, 288, 300, 320, 324, 360, 375, 384, 400, 405
COMMENTS
Sometimes called the Hamming sequence, since Hamming asked for an efficient algorithm to generate the list, in ascending order, of all numbers of the form 2^i*3^j*5^k for i,j,k >= 0. The problem was popularized by Edsger Dijkstra.
Numbers k such that 8*k = EulerPhi(30*k). - Artur Jasinski, Nov 05 2008 Also called "harmonic whole numbers", see Howard and Longair, 1982, Table I, page 121. - Hugo Pfoertner, Jul 16 2020 Also called ugly numbers, although it is not clear why. - Gus Wiseman, May 21 2021 Some woody bamboo species have extraordinarily long and stable flowering intervals that belong to this sequence. The model by Veller, Nowak & Davis justifies this observation from the evolutionary point of view. - Andrey Zabolotskiy, Jun 27 2021 Also those integers k for which, for every prime p > 5, p^(4*k) - 1 == 0 (mod 240*k). - Federico Provvedi, May 23 2022 As noted in the comments to A085152, Størmer's theorem implies that the only pairs of consecutive integers that appear as consecutive terms of this sequence are (1,2), (2,3), (3,4), (4,5), (5,6), (8,9), (9,10), (15,16), (24,25), and (80,81). These all represent significant musical intervals. - Hal M. Switkay, Dec 05 2022
FORMULA
Let s(n) = Card(k | a(k)<n) and f(n) = log(n*sqrt(30))^3/(6*log(2)*log(3)*log(5)). Then s(n) = f(n) + O(log(n)). Conjecture: s(n)=f(n) + O(log log n). For example, s(10000000) = 768 is well approximated by f(10000000) = 769.3... (see graphic given as link). - Benoit Cloitre, Dec 30 2001 The characteristic function of this sequence is given by:
Sum_{n>=1} x^a(n) = Sum_{n>=1} -Möbius(30*n)*x^n/(1-x^n). - Paul D. Hanna, Sep 18 2011 Sum_{n>=1} 1/a(n) = Product_{primes p <= 5} p/(p-1) = (2*3*5)/(1*2*4) = 15/4. - Amiram Eldar, Sep 22 2020
EXAMPLE
The sequence of terms together with their prime indices begins:
1: {} 25: {3,3}
2: {1} 27: {2,2,2}
3: {2} 30: {1,2,3}
4: {1,1} 32: {1,1,1,1,1}
5: {3} 36: {1,1,2,2}
6: {1,2} 40: {1,1,1,3}
8: {1,1,1} 45: {2,2,3}
9: {2,2} 48: {1,1,1,1,2}
10: {1,3} 50: {1,3,3}
12: {1,1,2} 54: {1,2,2,2}
15: {2,3} 60: {1,1,2,3}
16: {1,1,1,1} 64: {1,1,1,1,1,1}
18: {1,2,2} 72: {1,1,1,2,2}
20: {1,1,3} 75: {2,3,3}
24: {1,1,1,2} 80: {1,1,1,1,3}
(End)
MAPLE
option remember;
local a;
if n = 1 then
1;
else
for a from procname(n-1)+1 do
numtheory[factorset](a) minus {2, 3, 5 } ;
if % = {} then
return a;
end if;
end do:
end if;
end proc:
MATHEMATICA
mx = 405; Sort@ Flatten@ Table[ 2^a*3^b*5^c, {a, 0, Log[2, mx]}, {b, 0, Log[3, mx/2^a]}, {c, 0, Log[5, mx/(2^a*3^b)]}] (* Or *)
With[{nn=10}, Select[Union[Times@@@Flatten[Table[Tuples[{2, 3, 5}, n], {n, 0, nn}], 1]], #<=2^nn&]] (* Harvey P. Dale, Feb 28 2022 *)
PROG
(PARI) test(n)= {m=n; forprime(p=2, 5, while(m%p==0, m=m/p)); return(m==1)}
for(n=1, 500, if(test(n), print1(n", ")))
(PARI) a(n)=local(m); if(n<1, 0, n=a(n-1); until(if(m=n, forprime(p=2, 5, while(m%p==0, m/=p)); m==1), n++); n)
(PARI) list(lim)=my(v=List(), s, t); for(i=0, logint(lim\=1, 5), t=5^i; for(j=0, logint(lim\t, 3), s=t*3^j; while(s<=lim, listput(v, s); s<<=1))); Set(v) \\ Charles R Greathouse IV, Sep 21 2011; updated Sep 19 2016 (PARI) smooth(P:vec, lim)={ my(v=List([1]), nxt=vector(#P, i, 1), indx, t);
while(1, t=vecmin(vector(#P, i, v[nxt[i]]*P[i]), &indx);
if(t>lim, break); if(t>v[#v], listput(v, t)); nxt[indx]++);
Vec(v)
};
(Magma) [n: n in [1..500] | PrimeDivisors(n) subset [2, 3, 5]]; // Bruno Berselli, Sep 24 2012 (Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a051037 n = a051037_list !! (n-1)
a051037_list = f $ singleton 1 where
f s = y : f (insert (5 * y) $ insert (3 * y) $ insert (2 * y) s')
where (y, s') = deleteFindMin s
(Python)
def isok(n):
while n & 1 == 0: n >>= 1
while n % 3 == 0: n //= 3
while n % 5 == 0: n //= 5
(Python)
from sympy import integer_log
def bisection(f, kmin=0, kmax=1):
while f(kmax) > kmax: kmax <<= 1
while kmax-kmin > 1:
kmid = kmax+kmin>>1
if f(kmid) <= kmid:
kmax = kmid
else:
kmin = kmid
return kmax
def f(x):
c = n+x
for i in range(integer_log(x, 5)[0]+1):
for j in range(integer_log(y:=x//5**i, 3)[0]+1):
c -= (y//3**j).bit_length()
return c
(Python) # faster for initial segment of sequence
import heapq
from itertools import islice
def A051037gen(): # generator of terms
v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3, 5]
while True:
v = heapq.heappop(h)
if v != oldv:
yield v
oldv = v
for p in psmooth_primes:
heapq.heappush(h, v*p)
CROSSREFS
The partitions with these Heinz numbers are counted by A001399. Requiring the sum of prime indices to be even gives A344297.
Monotonic ordering of set S generated by these rules: if x and y are in S then x^2 + y^2 is in S, and 1 is in S.
+10
49
1, 2, 5, 8, 26, 29, 50, 65, 68, 89, 128, 677, 680, 701, 740, 842, 845, 866, 905, 1352, 1517, 1682, 2501, 2504, 2525, 2564, 3176, 3341, 4226, 4229, 4250, 4289, 4625, 4628, 4649, 4688, 4901, 5000, 5066, 5300, 5465, 6725, 7124, 7922, 7925, 7946, 7985
COMMENTS
Let N denote the positive integers, and suppose that f(x,y): N x N->N. Let "start" denote a subset of N, and let S be the set of numbers defined by these rules: if x and y are in S, then f(x,y) is in S, and "start" is a subset of S. The monotonic increasing ordering of S is a sequence:
A051037: f(x,y)=x*y, start={1,2,3,5}
A002473: f(x,y)=x*y, start={1,2,3,5,7}
A192518: f(x,y)=(x+1)(y+1), start={2}
A192519: f(x,y)=floor(x*y/2), start={3}
A192520: f(x,y)=floor(x*y/2), start={5}
A192521: f(x,y)=floor((x+1)(y+1)/2), start={2}
A192522: f(x,y)=floor((x-1)(y-1)/2), start={5}
A192523: f(x,y)=2x*y-x-y, start={2}
A192525: f(x,y)=2x*y-x-y, start={3}
A192524: f(x,y)=4x*y-x-y, start={1}
A192528: f(x,y)=5x*y-x-y, start={1}
A192529: f(x,y)=3x*y-x-y, start={2}
A192531: f(x,y)=3x*y-2x-2y, start={2}
A192533: f(x,y)=x^2+y^2-x*y, start={2}
A192535: f(x,y)=x^2+y^2+x*y, start={1}
A192536: f(x,y)=x^2+y^2-floor(x*y/2), start={1}
A192537: f(x,y)=x^2+y^2-x*y/2, start={2}
A192539: f(x,y)=2x*y+floor(x*y/2), start={1}
EXAMPLE
1^2+1^2=2, 1^2+2^2=5, 2^2+2^2=8, 1^2+5^2=26.
MATHEMATICA
start = {1}; f[x_, y_] := x^2 + y^2 (* start is a subset of t, and if x, y are in t then f(x, y) is in t. *)
b[z_] := Block[{w = z}, Select[Union[Flatten[AppendTo[w, Table[f[w[[i]], w[[j]]], {i, 1, Length[w]}, {j, 1, i}]]]], # < 30000 &]];
t = FixedPoint[b, start] (* A192476 *) Differences[t]
PROG
(Haskell)
import Data.Set (singleton, deleteFindMin, insert)
a192476 n = a192476_list !! (n-1)
a192476_list = f [1] (singleton 1) where
f xs s =
m : f xs' (foldl (flip insert) s' (map (+ m^2) (map (^ 2) xs')))
where xs' = m : xs
(m, s') = deleteFindMin s
Period of decimal representation of 1/n.
+10
47
1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 2, 1, 6, 6, 1, 1, 16, 1, 18, 1, 6, 2, 22, 1, 1, 6, 3, 6, 28, 1, 15, 1, 2, 16, 6, 1, 3, 18, 6, 1, 5, 6, 21, 2, 1, 22, 46, 1, 42, 1, 16, 6, 13, 3, 2, 6, 18, 28, 58, 1, 60, 15, 6, 1, 6, 2, 33, 16, 22, 6, 35, 1, 8, 3, 1, 18, 6, 6, 13, 1, 9, 5, 41, 6, 16, 21, 28, 2, 44, 1
COMMENTS
When 1/n has a finite decimal expansion (namely, when n = 2^a*5^b), a(n) = 1 while A051626(n) = 0. - M. F. Hasler, Dec 14 2015
REFERENCES
J. H. Conway and R. K. Guy, The Book of Numbers, Copernicus Press, NY, 1996, pp. 159 etc.
FORMULA
Note that if n=r*s where r is a power of 2 and s is odd then a(n)=a(s). Also if n=r*s where r is a power of 5 and s is not divisible by 5 then a(n) = a(s). So we just need a(n) for n not divisible by 2 or 5. This is the smallest number m such that n divides 10^m - 1; m is a divisor of phi(n), where phi = A000010. phi(n) = n-1 only if n is prime and since a(n) divides phi(n), a(n) can only equal n-1 if n is prime. - Scott Hemphill (hemphill(AT)alumni.caltech.edu), Nov 23 2006
MAPLE
a132740 := 1 ;
for pe in ifactors(n)[2] do
if not op(1, pe) in {2, 5} then
a132740 := a132740*op(1, pe)^op(2, pe) ;
end if;
end do:
if a132740 = 1 then
1 ;
else
numtheory[order](10, a132740) ;
end if;
end proc:
MATHEMATICA
Table[r = n/2^IntegerExponent[n, 2]/5^IntegerExponent[n, 5]; MultiplicativeOrder[10, r], {n, 100}] (* T. D. Noe, Oct 17 2012 *)
PROG
(Sage)
def a(n):
n = ZZ(n)
rad = 2**n.valuation(2) * 5**n.valuation(5)
return Zmod(n // rad)(10).multiplicative_order()
[a(n) for n in range(1, 20)]
(Python)
from sympy import n_order, multiplicity
def A007732(n): return n_order(10, n//2**multiplicity(2, n)//5**multiplicity(5, n)) # Chai Wah Wu, Feb 07 2022
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