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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.
+20
2
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
COMMENTS
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.
EXAMPLE
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.
PROG
(PARI) 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))} (Python)
import sympy
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)
Number of positive cubes needed to sum to n using the greedy algorithm.
+10
14
0, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 1, 2, 3, 4, 5, 6, 7, 8, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7, 8, 9, 10, 4, 5, 6, 2, 3, 4, 5, 6, 7, 8, 9, 3, 4, 5, 6, 7
COMMENTS
Define f(n) = n - k^3 where (k+1)^3 > n >= k^3; a(n) = number of steps such that f(f(...f(n)))= 0.
Also sum of digits when writing n in base where place values are positive cubes, cf. A000433. [ Reinhard Zumkeller, May 08 2011]
FORMULA
a(0) = 0; for n >= 1, a(n) = a(n-floor(n^(1/3))^3)+1 = a( A055400(n))+1 = a(n- A048762(n))+1.
EXAMPLE
a(32)=6 because 32=27+1+1+1+1+1 (not 32=8+8+8+8).
a(33)=7 because 33=27+1+1+1+1+1+1 (not 33=8+8+8+8+1).
MAPLE
f:= proc(n, k) local m, j;
if n = 0 then return 0 fi;
for j from k by -1 while j^3 > n do od:
m:= floor(n/j^3);
m + procname(n-m*j^3, j-1);
end proc:
seq(f(n, floor(n^(1/3))), n=0..100); # Robert Israel, Aug 17 2015
MATHEMATICA
a[0] = 0; a[n_] := {n} //. {b___, c_ /; !IntegerQ[c^(1/3)], d___} :> {b, f = Floor[c^(1/3)]^3, c - f, d} // Length; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Aug 17 2015 *)
PROG
(PARI)
F=vector(30, n, n^3); /* modify to get other sequences of "greedy representations" */ last_leq(v, F)=
{ /* Return last element <=v in sorted array F[] */
local(j=1);
while ( F[j]<=v, j+=1 );
return( F[j-1] );
}
greedy(n, F)=
{
local(v=n, ct=0);
while ( v, v-=last_leq(v, F); ct+=1; );
return(ct);
}
vector(min(100, F[#F-1]), n, greedy(n, F)) /* show terms */
/* Joerg Arndt, Apr 08 2011 */
(Haskell)
a055401 n = s n $ reverse $ takeWhile (<= n) $ tail a000578_list where
s _ [] = 0
s m (x:xs) | x > m = s m xs
| otherwise = m' + s r xs where (m', r) = divMod m x
(Scheme, with memoization-macro definec)
CROSSREFS
Cf. A002376 (least number of positive cubes needed to represent n; differs from this sequence for the first time at n=32, where a(32)=6, while A002376(32)=4).
a(0)=1, a(1)=1, a(n) = largest prime <= a(n-1) + a(n-2).
+10
13
1, 1, 2, 3, 5, 7, 11, 17, 23, 37, 59, 89, 139, 227, 359, 577, 929, 1499, 2423, 3919, 6337, 10253, 16573, 26821, 43391, 70207, 113591, 183797, 297377, 481171, 778541, 1259701, 2038217, 3297913, 5336129, 8633983, 13970093, 22604069
COMMENTS
Or might be called Ishikawa primes, as he proved that prime(n+2) < prime(n) + prime(n+1) for n > 1. This improves on Bertrand's Postulate (Chebyshev's theorem), which says prime(n+2) < prime(n+1) + prime(n+1). - Jonathan Sondow, Sep 21 2013
FORMULA
a(n) is asymptotic to C*phi^n where phi = (1+sqrt(5))/2 and C = 0.41845009129953131631777132510164822489... - Benoit Cloitre, Apr 21 2003 a(n) >= prime(n-1) for n > 1, by Ishikawa's theorem. - Jonathan Sondow, Sep 21 2013
EXAMPLE
a(8) = 23 because 23 is largest prime <= a(7) + a(6) = 17 + 11 = 28.
MATHEMATICA
PrevPrim[n_] := Block[ {k = n}, While[ !PrimeQ[k], k-- ]; Return[k]]; a[1] = a[2] = 1; a[n_] := a[n] = PrevPrim[ a[n - 1] + a[n - 2]]; Table[ a[n], {n, 1, 42} ]
(* Or, if version >= 6 : *)a[0] = a[1] = 1; a[n_] := a[n] = NextPrime[ a[n-1] + a[n-2] + 1, -1]; Table[a[n], {n, 0, 100}](* Jean-François Alcover, Jan 12 2012 *) nxt[{a_, b_}]:={b, NextPrime[a+b+1, -1]}; Transpose[NestList[nxt, {1, 1}, 40]] [[1]] (* Harvey P. Dale, Jul 15 2013 *)
PROG
(Haskell)
a055500 n = a055500_list !! n
a055500_list = 1 : 1 : map a007917
(zipWith (+) a055500_list $ tail a055500_list)
(Python)
from sympy import prevprime; L = [1, 1]
for _ in range(36): L.append(prevprime(L[-2] + L[-1] + 1))
Cube excess of the n-th prime.
+10
4
1, 2, 4, 6, 3, 5, 9, 11, 15, 2, 4, 10, 14, 16, 20, 26, 32, 34, 3, 7, 9, 15, 19, 25, 33, 37, 39, 43, 45, 49, 2, 6, 12, 14, 24, 26, 32, 38, 42, 48, 54, 56, 66, 68, 72, 74, 86, 7, 11, 13, 17, 23, 25, 35, 41, 47, 53, 55, 61, 65, 67, 77, 91, 95, 97, 101, 115, 121, 4, 6, 10, 16, 24, 30
EXAMPLE
a(48) = 7 because the 48th prime is 223 and 223 - 6^3 = 7, while 223 - 7^3 = -120.
MATHEMATICA
#-Floor[Surd[#, 3]]^3&/@Prime[Range[80]] (* Harvey P. Dale, Feb 15 2018 *)
Distance to nearest cube different from n.
+10
3
1, 1, 1, 2, 3, 3, 2, 1, 7, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 8, 7, 6, 5, 4, 3, 2, 1, 19, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 37, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16
COMMENTS
Equals A074989(n) if this is not zero, else 1+ A055400(n-1), the distance to the nearest cube < n.
EXAMPLE
a(8)=7, because the two cubes below and above 8 are 1^3=1 and 3^3=27, and the distance to 1 is smaller, namely 8-1=7.
MAPLE
distNearstDiffCub := proc(n) local iscbr ; iroot(n, 3, 'iscbr') ; if iscbr then 1+ A055400(n-1); else A074989(n) ; end if; end proc;
MATHEMATICA
dnc[n_]:=Module[{c=Surd[n, 3]}, If[IntegerQ[c], n-(c-1)^3, Min[n-Floor[ c]^3, Ceiling[c]^3-n]]]; Array[dnc, 90, 0] (* Harvey P. Dale, Mar 30 2019 *)
The orbit of n under iterations of x -> c(x)*(10^L(x-c(x))-1) + x, where c(x) = floor(x^(1/3))^3, L(x) = floor(log_10(max(x,1))+1), enters a pseudo-loop x(k) = a^3 * 10^((k-k0)*L(b)) + b beyond some k0. This sequence lists the a-values.
+10
2
18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 18, 883, 883, 883, 883, 883, 883, 883, 883, 883, 581, 581, 581, 581, 581, 581, 581, 581, 581, 581, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 8, 8, 8, 8
COMMENTS
The iterated function can also be defined as x -> concatenate(c(x), A055400(x)), where c = A048762 gives the largest perfect cube <= x and x-c(x) = A055400(x) is the "cube excess" of x. L = A055642 gives the number of decimal digits. The corresponding b-values are listed in A369861.
EXAMPLE
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, as does the cube root a(1) = 18, up to factors 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). We see that in this sequence each term is just one more than that of the preceding sequence, so the cube root remains the same, a(2) = a(1) = 18.
For n = 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 root a(18) = 883 gets an additional factor 10 at each step, but the cube excess A055400 remains the same, A369861(18) = 748673.
PROG
(PARI) A369860(n)={until(, my(c=sqrtnint(n, 3), v=valuation(c, 10), L=logint(max(n-c^3, 1), 10)+1); L==v*3 && return(c/10^v); n += c^3*(10^L-1))} (Python)
while True:
C = sympy.integer_nthroot(n, 3)[0]; L = A055642(n-C**3) if sympy.multiplicity(10, C)*3 == L: return C//10**(L//3)
n += C**3 * (10**L - 1)
Hexagonal excess: smallest amount by which n exceeds a hexagonal number (2k^2-k, A000384).
+10
1
0, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 6, 7, 8, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20
EXAMPLE
a(19)=4 since 15(=2*3^2-3) is the largest hexagonal less than or equal to 19 and 19-15=4.
MATHEMATICA
Flatten[Range[#]&/@Differences[Array[#(2#-1)&, 10, 0]]-1] (* Harvey P. Dale, Jun 05 2013 *)
1, 8, 8, 1000, 64, 8, 729, 27, 232, 1728, 64, 216, 1728, 512, 8000, 4913, 729, 27, 125, 512, 64, 5832, 13331, 216, 13580, 125, 4913, 1000, 1856, 3375, 13824, 7073, 343, 2547, 8, 1331, 12167, 512, 1728, 8000, 13824, 13768, 24389, 9736, 16496, 216, 12167, 13824, 19683, 1
COMMENTS
Noncube terms of this sequence are 232, 13331, 13580, 1856, 7073, 2547, ...
How is the distribution of noncube terms in this sequence? See also A273592 that is related with this question.
PROG
(PARI) T = thueinit(x^3+1, 1);
isA001235(n) = my(v=thue(T, n)); sum(i=1, #v, v[i][1]>=0 && v[i][2]>=v[i][1])>1;
lista(nn) = for(n=1, nn, if(isA001235(n), print1(n-sqrtnint(n, 3)^3, ", ")));
Difference between smallest cube > n and n.
+10
1
1, 7, 6, 5, 4, 3, 2, 1, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45
COMMENTS
a(n) is the smallest positive number k such that n + k is a cube.
FORMULA
a(n) = floor(n^(1/3) + 1)^3 - n.
MATHEMATICA
Table[Floor[n^(1/3) + 1]^3 - n, {n, 0, 80}]
Numbers n for which the cube excess of the n-th prime is prime.
+10
0
2, 5, 6, 8, 10, 19, 20, 23, 26, 28, 31, 48, 49, 50, 51, 52, 55, 56, 57, 59, 61, 65, 66, 99, 100, 105, 110, 112, 114, 117, 121, 125, 127, 170, 171, 173, 178, 184, 185, 186, 190, 192, 194, 196, 200, 201, 206, 208, 214, 270, 271, 272, 274, 277, 278, 279, 280, 282
EXAMPLE
99 is an element of this sequence because the 99th prime is 523, 523 - 8^3 = 523-512 = 11 and 11 is prime. 100 is in this sequence because the 100th prime is 541 and 541-8^3 = 29, which is prime.
MATHEMATICA
f[n_] := Block[{k = 1, p = Prime[n]}, While[k^3 < p, k++ ]; p - (k - 1)^3]; Select[ Range[ 284], PrimeQ[ f[ # ]] &] (* Robert G. Wilson v, Mar 19 2005 *)
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