A055642 -id:A055642 - cp's OEIS frontend

A224927 Denominator of lexicographically least fraction f satisfying floor(f*10^A055642(n)) = n.

6, 4, 3, 5, 2, 3, 4, 5, 10, 10, 9, 8, 15, 7, 13, 6, 17, 11, 21, 5, 14, 9, 13, 25, 4, 15, 11, 7, 17, 10, 16, 25, 3, 23, 14, 11, 8, 13, 23, 5, 12, 7, 16, 9, 11, 13, 17, 25, 51, 2, 27, 17, 13, 11, 9, 16, 7, 12, 22, 5, 13, 8, 11, 14, 20, 3, 28, 16, 13, 10, 7, 11, 15

Offset: 1

Paul Tek, Apr 20 2013

1 < a(n) <= 10^A055642(n).

For any reduced fraction u/v in the interval [1/10..1[, a(floor(u/v*10^k))=v for k sufficiently large.

			The fractions f satisfying floor(f*100)=42, are, in lexicographical order: 3/7, 6/14, 8/19, 9/21, 11/26, 12/28, 14/33, 15/35, 16/38, 17/40, 18/42, 19/45, 20/47, 21/49...
Hence, a(42)=denominator(3/7)=7.

Paul Tek, Table of n, a(n) for n = 1..11000

Cf. A224926 (numerators), A055642, A002487.

a224927(n) =\
local(a=0, b=1, c, d, e=1, f=0, x=1); \
while(x<=n, x=x*10); \
while(1, c=a+e; d=b+f; \
if(c/d < n/x, a=c; b=d, \
if(c/d >= (n+1)/x, e=c; f=d, \
return(d))))

A308314 Decimal expansion of Sum_{k>=1} (1/A055642(k)^A055642(k)) where A055642(k) is the number of digits of the integer k.

Original entry on oeis.org

1, 6, 8, 0, 5, 2, 4, 5, 3, 7, 5, 2, 6, 2, 1, 6, 8, 9, 4, 9, 0, 8, 5, 6, 7, 3, 3, 2, 0, 5, 5, 6, 7, 2, 4, 5, 2, 1, 9, 6, 5, 2, 6, 7, 9, 9, 7, 1, 9, 8, 4, 9, 5, 0, 4, 9, 1, 5, 5, 7, 0, 3, 5, 9, 8, 1, 4, 3, 7, 9, 8, 3, 4, 8, 1, 7, 5, 7, 0, 8, 8, 9, 4, 8, 3, 4, 6, 1, 6, 4, 4, 4, 5, 0, 7, 8, 4, 8, 6, 4

Offset: 3

Bernard Schott, May 19 2019

With summation by parts to obtain 1st formula:
Sum_{k>=1} (1/length(k)^length(k)) =
Sum_{m=1..9} (1/1^1) + Sum_{m=10..99} (1/2^2) + Sum_{m=100...999} (1/3^3) + Sum_{m=1000...9999} (1/4^4) + ... =
9*(1/1^1) + 90*(1/2^2) + 900*(1/3^3) + 9000*(1/4^4) + 90000*(1/5^5) + ... =
9 ( 1/1^1 + 10^1/2^2 + 10^2/3^3 + 10^3/4^4 + 10^4/5^5 + ... =
(9/10) * (10^1/1^1 + 10^2/2^2 + 10^3/3^3 + 10^4/4^4 + 10^5/5^5 + ... =
(9/10) * ( (10/1)^1 + (10/2)^2 + (10/3)^3 + (10/4)^4 + (10/5)^5 + ... =
(9/10) * Sum_{m>=1} (10/m)^m.

			168.05245375262168949085673320556724...

Xavier Merlin, Methodix Analyse, Ellipses, 1997, Exercice 22 p. 120.
J.-M. Monier, Analyse, Tome 3, 2ème année, MP.PSI.PC.PT, Dunod, 1997, Exercice 3.2.1.h" p. 248.

Cf. A055642, A138908.

evalf((9/10) * Sum((10/n)^n, n=1..infinity), 100);

(9/10) * suminf(k=1, (10/k)^k) \\ Michel Marcus, Jun 08 2019

Equals (9/10) * Sum_{k>=1} (10/k)^k.

Equals Sum_{n>=1} (1/A138908(n)).

A330814 a(1) = 1; a(n+1) = Sum_{k=1..n} {q(a(k)): q(a(k)) = q(a(n))}, where q(n) = A007953(n) + A055642(n).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 6, 14, 21, 10, 9, 20, 8, 18, 11, 12, 15, 16, 27, 22, 12, 20, 16, 36, 33, 24, 32, 28, 12, 25, 45, 44, 30, 30, 35, 40, 18, 55, 24, 40, 24, 48, 14, 35, 50, 42, 56, 13, 30, 40, 36, 66, 28, 36, 77, 16, 54

Offset: 1

David James Sycamore, Jan 01 2020

a(n+1) = k(n)*q(a(n)), where k(n) is the number of times (up to and including a(n)) that a term having the same q-value as a(n) has occurred in the sequence so far.

			a(2) is q(a(1))=a(1)=2; a(10)=q(10)=3, and 3=q(a(2)) has been seen once before, so a(11)=3+3=6.

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

Cf. A279818, A330807, A007953, A055642.

q:=func; a:=[1,2]; for n in [3..70] do Append( ~a,&+[ q(a[k-1]):k in [2..n]| q(a[k-1]) eq q(a[n-1])]); end for; a; // Marius A. Burtea, Jan 02 2020

s[n_] := Plus @@(d = IntegerDigits[n]) + Length[d]; a[1] = 1; a[n_] := a[n] = (s1 = s[a[n - 1]])*(1 + Sum[Boole[s[a[k]] == s1], {k, 1, n - 2}]); Array[a, 100] (* Amiram Eldar, Jan 01 2020 *)

A345672 a(n) is the least m such that the decimal expansion of n appears in the concatenation of m, ..., m+k for some k >= 0 and Sum_{i = m..m+k-1} A055642(i) < A055642(n) <= Sum_{i = m..m+k} A055642(i).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 13, 14, 15, 16, 17, 18, 19, 20, 12, 22, 2, 24, 25, 26, 27, 28, 29, 30, 13, 23, 33, 3, 35, 36, 37, 38, 39, 40, 14, 24, 34, 44, 4, 46, 47, 48, 49, 50, 15, 25, 35, 45, 55, 5, 57, 58, 59, 60, 16, 26, 36, 46, 56, 66, 6, 68

Offset: 0

Rémy Sigrist, Jun 22 2021

This sequence is similar to A055642; here we consider decimal expansions, there binary expansions.

			For n = 21:
- the decimal expansion of 21 first appears in the concatenation of 12 and 13,
- so a(21) = 12.

Rémy Sigrist, PARI program for A345672

Cf. A007376, A055642, A083653 (binary analog), A252043.

PARI
```
See Links section.
```

a(n) <= n.

a(n) = 1 iff n belongs to A252043.

A369861 The orbit of n under iterations of x -> concatenate(A048762(x), A055400(x)) enters a pseudo-loop x(k) = a^3 * 10^((k-k0)*A055642(b)) + b for k > k0. This sequence lists the b-value.

Original entry on oeis.org

586, 587, 588, 589, 590, 591, 592, 584, 585, 586, 587, 588, 589, 590, 591, 592, 593, 748673, 748674, 748675, 748676, 748677, 748678, 748679, 748680, 748681, 709030, 709031, 709032, 709033, 709034, 709035, 709036, 709037, 709038, 709039, 513, 514, 515, 516, 517, 518, 519, 520

Offset: 1

M. F. Hasler, Apr 03 2024

The iterated function can also be defined as x -> c(x)*(10^L(x-c(x))-1) + x, where c(x) = A048762(x) = floor(x^(1/3))^3 is the largest perfect cube <= x; A055400(x) = x-c(x) is the "cube excess" of x, and L(x) = A055642(x) = floor(log_10(max(x,1))+1) is the number of decimal digits of x.

Often a(n+1) = a(n) + 1, especially when c(n+1) = c(n), in which case it is probable that all elements of the orbit of n+1 are just one larger than the elements of the orbit of n.

			Starting with 1, we get 1 -> 10 -> 82 (since 8 is the largest cube <= 10, at distance 2) -> 6418 (since the cube 64 is at distance 18) -> 5832586 (since 5832 = 18^3 is at distance 586) -> 5832000586 (since 180^3 is again at distance 586) -> ...: Each time 3 '0's will be inserted in front of the remainder which remains always the same, a(1) = 586, as does the cube root up to an additional factor of 10.
Starting with 2, we get 2 -> 11 (since the largest cube <= 2 is 1, at distance 1) -> 83 (since largest cube <= 11 is 8, at distance 2) -> 6419(since the cube 64 is at distance 19) -> 5832587 (since 5832 = 18^3 is at distance 587) -> 5832000587 (since 180^3 is again at distance 587) -> ... We see that in this sequence each term is one more than that of the preceding sequence, whence also a(2) = 587 = a(1)+1.
Starting with 8, we get 8 -> 80 (since the largest cube <= 8 is 8, at distance 0) -> 6416 (since the cube 64 is at distance 16, two less than in 1's orbit) -> 5832584 (since 5832 = 18^3 is at distance 584, again 2 less than in 1's orbit) -> 5832000584 (since 180^3 is again at distance 584) -> ... We see that in this sequence each term is 2 (resp. 8) less than the corresponding term of 1's (resp. 7's) orbit (with the initial term deleted). Hence also a(8) = 584 = a(7)-8 = a(1)-2. From here on subsequent terms will again increase by 1 up to n = 17.
Starting with 18, we get 18 -> 810 (since the largest cube <= 18 is 8, at distance 10) -> 72981 (since the cube 729 is at distance 81) -> 689214060 (since 68921 = 41^3 is at distance 4060) -> 688465387748673 (since 688465387 = 883^3 is at distance 748673), from where on the cube roots get multiplied by 10 and the distance from the cubes remains the same, a(18) = 748673.
For n = 64 -> 640 (= 8^3 + 128) -> 512128 = 80^3 + 128, we have a(n) = 128.

Eric Angelini, Pseudo-loops with cubes, and post to math-fun discussion list; April 1, 2024.

Cf. A000578 (cubes), A048766 (cube root), A048762 (largest cube <= n), A055400 (cube excess), A055642 (length of n in base 10), A122840 (10-valuation of n).

Cf. A369860 (a-values)

A369861(n)={until(, my(c=sqrtnint(n,3), v=valuation(c,10), L=logint(max(n-c^3,1),10)+1); L==v*3 && return(n-c^3); n += c^3*(10^L-1))}

import sympy
def A369861(n: int):
    while True:
        C = sympy.integer_nthroot(n, 3)[0]**3; L = A055642(n-C)
        if sympy.multiplicity(10, C) == L: return n-C
        n += C * (10**L - 1)

A373205 Numbers m such that m^m == m (mod 10^(len(m) + 1)), where len(m) is the number of digits of m (A055642).

Original entry on oeis.org

1, 51, 57, 101, 151, 176, 201, 301, 351, 401, 501, 551, 576, 601, 625, 701, 751, 801, 901, 951, 976, 1001, 1376, 2001, 2057, 2751, 3001, 4001, 4193, 4751, 5001, 5376, 6001, 6249, 6751, 7001, 8001, 8751, 9001, 9375, 9376, 10001, 10751, 11001, 12001, 13001

Offset: 1

Marco Ripà, May 27 2024

By definition, the present sequence is a subsequence of A082576.
For each integer r >= 2 this sequence contains 10^r + 1.
All terms > 1 end in 01, 25, 49, 51, 57, 75, 76, or 93.

			51 is a term since 51 is a 2-digit number and 51^51 == 5051 (mod 10^4) and thus 51^51 == 51 (mod 10^(2 + 1)).

Index entries for sequences related to automorphic numbers

Cf. A055642, A082576, A373206.

for (len_m = 1, 5, for (m = 10^(len_m - 1), 10^len_m - 1, if (m == Mod(m, 10^(len_m + 1))^m, print1(m, ", "))))

from itertools import count
def A373205_gen(): # generator of terms
    for i in count(0,100):
        for j in (1, 25, 49, 51, 57, 75, 76, 93):
            m = i+j
            if pow(m,m,10*10**(len(str(m)))) == m:
                yield m
A373205_list = list(islice(A373205_gen(),20)) # Chai Wah Wu, Jun 02 2024

A087095 a(n) = A077503(n) - n*10^d, where d = n-A055642(n), A055642(n) = number of digits in n.

Original entry on oeis.org

0, 5, 24, 96, 176, 625, 1316, 13025, 0, 14129, 24161, 580921, 1310976, 4588964, 7007601, 0, 233634401, 244436761, 2180575376, 3762025, 23492302009, 71829444801, 293908889924, 357018382249, 2562501107225, 2194311769649

Offset: 1

Ray Chandler, Aug 11 2003

			a(7) = 7001316 - 7000000 = 1316.

Cf. A055642, A077503.

ceil[n_]:=Module[{d=n-IntegerLength[n],c},c=n*10^d;Ceiling[Sqrt[c]]^2-c]; Array[ceil,30] (* Harvey P. Dale, Sep 03 2018 *)

a(n) = ceiling(sqrt(n*10^d))^2 - n*10^d, where d = n-digit(n), digit(n) = number of digits in n.

Edited by Charles R Greathouse IV, Aug 03 2010

A100470 n appears A055642(n) times (appearances equal number of decimal digits).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 10, 11, 11, 12, 12, 13, 13, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35, 35, 36, 36, 37, 37, 38, 38, 39, 39, 40, 40, 41

Offset: 0

Rick L. Shepherd, Nov 21 2004

Lexicographically smallest nondecreasing left inverse to A117804: a(A117804(n)) = n. Gives the number to which belongs the digit A007376(n). - M. F. Hasler, Oct 23 2019

Cf. A055642, A030530 (n's appearances equal its number of bits).

A270142 a(n) = product of first k composites, with the i-th composite raised to the d-th power, where k = A055642(n) and d is the i-th digit of n.

Original entry on oeis.org

4, 16, 64, 256, 1024, 4096, 16384, 65536, 262144, 4, 24, 144, 864, 5184, 31104, 186624, 1119744, 6718464, 40310784, 16, 96, 576, 3456, 20736, 124416, 746496, 4478976, 26873856, 161243136, 64, 384, 2304, 13824, 82944, 497664, 2985984, 17915904, 107495424

Offset: 1

Felix Fröhlich, Mar 12 2016

All terms are multiples of 4, since A002808(1) = 4 and the most significant digit of n is always nonzero.
Does a term exist such that a(n) = n? Such a number would be the analog of a Meertens number when raising composites to the powers of the digits of n instead of raising primes to the powers of the digits.
From Chai Wah Wu, Dec 15 2022: (Start)
If a(n) is defined using digits of n in base b, then there are bases b and numbers n such that a(n) = n. For instance:
base b             n
------------------------------------------------
2                  4, 24, 36, 24192000, 85155840
3                  2592
4                  4, 103680
6                  20736
8                  16, 256, 13824
12                 1327104
16                 21233664
23                 24
24                 746496
(End)

			a(12) = 144, since A002808(1) = 4, A002808(2) = 6 and 4^1 * 6^2 = 144.

Felix Fröhlich, Table of n, a(n) for n = 1..10000

Cf. A189398, A246532.

composite(n) = my(i=0, c=2); while(1, if(!ispseudoprime(c), i++); if(i==n, return(c)); c++)
compopowerprod(n) = my(d=digits(n)); for(k=1, #d, p=prod(i=1, #d, composite(i)^d[i])); p
a(n) = compopowerprod(n)

from math import prod
from sympy import composite
def A270142(n): return prod(composite(i)**int(d) for i, d in enumerate(str(n),1)) # Chai Wah Wu, Dec 09 2022

A270542 Numbers of the form (pq)^2, where p is the number of digits of n (A055642) and q is the sum of the digits of n (A007953).

Original entry on oeis.org

0, 1, 2704, 5184, 7744

Offset: 1

José de Jesús Camacho Medina, Mar 18 2016

			2704 is a term because 2704 = [4*(2+7+0+4)]^2;
5184 is a term because 5184 = [4*(5+1+8+4)]^2.

Cf. A007953, A055642, A257784.

Position[ Table[ IntegerLength[k] Sum[( Floor[k/10^n] - 10 Floor[k/10^(n + 1)]), {n, 0, IntegerLength@ k^2}] - k, {k, 1, 10^6}], 0] // Flatten = {1, 1232, 4100, 268542}
Select[Range[10^3]^2,With[{id=IntegerDigits[#]},#==(Length[id]*Plus@@id)^2]&] (* Ray Chandler, Apr 01 2016 *)

isok(n) = d = digits(n); (#d*vecsum(d))^2 == n; \\ Michel Marcus, Mar 26 2016

cp's OEIS Frontend

A224927 Denominator of lexicographically least fraction f satisfying floor(f*10^A055642(n)) = n.

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A308314 Decimal expansion of Sum_{k>=1} (1/A055642(k)^A055642(k)) where A055642(k) is the number of digits of the integer k.

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Maple

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A330814 a(1) = 1; a(n+1) = Sum_{k=1..n} {q(a(k)): q(a(k)) = q(a(n))}, where q(n) = A007953(n) + A055642(n).

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Magma

Mathematica

A345672 a(n) is the least m such that the decimal expansion of n appears in the concatenation of m, ..., m+k for some k >= 0 and Sum_{i = m..m+k-1} A055642(i) < A055642(n) <= Sum_{i = m..m+k} A055642(i).

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A369861 The orbit of n under iterations of x -> concatenate(A048762(x), A055400(x)) enters a pseudo-loop x(k) = a^3 * 10^((k-k0)*A055642(b)) + b for k > k0. This sequence lists the b-value.

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PARI

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A373205 Numbers m such that m^m == m (mod 10^(len(m) + 1)), where len(m) is the number of digits of m (A055642).

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PARI

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A087095 a(n) = A077503(n) - n*10^d, where d = n-A055642(n), A055642(n) = number of digits in n.

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A100470 n appears A055642(n) times (appearances equal number of decimal digits).

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