A069638 -id:A069638 - cp's OEIS frontend

A108567 a(0) = 0, a(1) = a(2) = 1, a(3) = 2, a(4) = 4, a(5) = 8, a(6) = 16, for n>5: a(n+1) = SORT[ a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) + a(n-6)], where SORT places digits in ascending order and deletes 0's.

0, 1, 1, 2, 4, 8, 16, 23, 55, 19, 127, 225, 347, 128, 249, 115, 112, 133, 139, 1223, 299, 227, 2248, 1348, 1567, 157, 679, 2556, 2788, 11334, 2249, 1233, 2699, 23358, 12467, 12568, 5689, 2366, 368, 15559, 23577, 24579, 4678, 16678, 5788, 12279, 11338

Offset: 0

Jonathan Vos Post, Jun 11 2005

T. D. Noe found that the maximum is attained at a(4992871827) = 234444568999. The periodic part of this sequence begins at a(3544675600) and has length 5158842780.

			a(7) = SORT[a(0) + a(1) + a(2) + a(3) + a(4) + a(5) + a(6)] = SORT[0 + 1 + 1 + 2 + 4 + 8 + 16] = SORT[32] = 23.
a(8) = SORT[a(1) + a(2) + a(3) + a(4) + a(5) + a(6) + a(7)] = SORT[1 + 1 + 2 + 4 + 8 + 16 + 23] = SORT[55] = 55.
a(9) = SORT[a(2) + a(3) + a(4) + a(5) + a(6) + a(7) + a(8)] = SORT[1 + 2 + 4 + 8 + 16 + 23 + 55] = SORT[109] = 19.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Richard I. Hess, Problem 920: sorted Fibonacci sequence, Pi Mu Epsilon Journal, Vol. 10 (Fall 1998) No. 9, pp. 754-755.

Cf. A066178, A069638, A107281, A108564-A108566, A108568-A108573.

nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,FromDigits[Sort[ IntegerDigits[ a+b+c+d+e+f+g]/.(0->Nothing)]]}; NestList[nxt,{0,1,1,2,4,8,16},50][[All,1]] (* Harvey P. Dale, May 09 2020 *)

A346296 a(0) = 1; thereafter a(n) = 2*a(n-1) + 1, with digits rearranged into nondecreasing order.

Original entry on oeis.org

1, 3, 7, 15, 13, 27, 55, 111, 223, 447, 589, 1179, 2359, 1479, 2599, 1599, 1399, 2799, 5599, 11199, 22399, 44799, 58999, 117999, 235999, 147999, 259999, 159999, 139999, 279999, 559999, 1119999, 2239999, 4479999, 5899999, 11799999, 23599999, 14799999, 25999999

Offset: 0

Ctibor O. Zizka, Jul 13 2021

			a(3) = A004185(2*7+1) = A004185(15) = 15.
a(4) = A004185(2*15+1) = A004185(31) = 13.

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,100,-100).

Cf. A004185, A010888, A057615, A069638, A321542.

a[0] = 1; a[n_] := a[n] = FromDigits @ Sort @ IntegerDigits[2*a[n - 1] + 1]; Array[a, 45, 0] (* Amiram Eldar, Jul 13 2021 *)
NestList[FromDigits[Sort[IntegerDigits[2#+1]]]&,1,40] (* Harvey P. Dale, Oct 01 2023 *)

lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = fromdigits(vecsort(digits(2*va[n-1]+1)));); va;} \\ Michel Marcus, Aug 31 2021

from itertools import accumulate
def atis(anm1, _): return int("".join(sorted(str(2*anm1+1))))
print(list(accumulate([1]*39, atis))) # Michael S. Branicky, Aug 31 2021

a(n) = A004185(2*a(n-1)+1).
For k >= 1;
a(12*k-9) = 100^(k-1) *   16 - 1;
a(12*k-8) = 100^(k-1) *   14 - 1;
a(12*k-7) = 100^(k-1) *   28 - 1;
a(12*k-6) = 100^(k-1) *   56 - 1;
a(12*k-5) = 100^(k-1) *  112 - 1;
a(12*k-4) = 100^(k-1) *  224 - 1;
a(12*k-3) = 100^(k-1) *  448 - 1;
a(12*k-2) = 100^(k-1) *  590 - 1;
a(12*k-1) = 100^(k-1) * 1180 - 1;
a(12*k)   = 100^(k-1) * 2360 - 1;
a(12*k+1) = 100^(k-1) * 1480 - 1;
a(12*k+2) = 100^(k-1) * 2600 - 1.
G.f.: -(1800*x^15 -720*x^14 +1080*x^13 -1080*x^12 -590*x^11 -142*x^10 -224*x^9 -112*x^8 -56*x^7 -28*x^6 -14*x^5 +2*x^4 -8*x^3 -4*x^2 -2*x -1) / ((x-1)*(10*x^6-1)*(10*x^6+1)). - Alois P. Heinz, Aug 02 2021
a(n) = 100*a(n-12) + 99 for n >= 15. - Pontus von Brömssen, Sep 01 2021

A368559 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number not occurring earlier such that a(n) contains all the distinct digits of a(n-1) - a(n-2).

Original entry on oeis.org

1, 2, 10, 8, 12, 4, 18, 14, 24, 100, 67, 3, 46, 34, 21, 13, 28, 15, 31, 16, 51, 35, 61, 26, 53, 27, 62, 135, 37, 89, 25, 64, 39, 52, 103, 105, 20, 58, 38, 102, 146, 40, 106, 6, 101, 59, 42, 17, 125, 108, 71, 73, 22, 115, 93, 23, 70, 47, 32, 145, 113, 123, 104, 19, 85, 36, 49, 130, 81, 94

Offset: 1

Scott R. Shannon, Dec 30 2023

The sequence is conjectured to be a permutation of the positive integers. The fixed points begin 1, 2, 95, 122, 156, 318, 1644, 1964, 2189, 2740, 8264, 16904, ... although it is likely there are infinitely more.

			a(3) = 10 as a(2) - a(1) = 2 - 1 = 1, and 10 is the smallest unused number to contain 1.
a(11) = 67 as a(10) - a(9) = 100 - 24 = 76, and 67 is the smallest unused number to contain 7 and 6.

Scott R. Shannon, Table of n, a(n) for n = 1..10000
Scott R. Shannon, Image of the first 250000 terms. The green line is a(n) = n.

Cf. A362093, A362075, A184992, A069638.

A345902 a(0) = 1; for n >= 1, a(n) = A004185(a(n-1)*n).

Original entry on oeis.org

1, 1, 2, 6, 24, 12, 27, 189, 1125, 1125, 1125, 12357, 124488, 1134468, 12255588, 12333888, 12234789, 11234799, 22222368, 222224499, 444448899, 2333467899, 12333567789, 12234567789, 222336666999, 1445555667789, 12334444556778, 33333336, 33333489, 111666789

Offset: 0

Ctibor O. Zizka, Jun 29 2021

Rearranging of digits of a(n-1)*n in ascending order gives in step n the smallest number possible. For n >= 5; the number of digits of a(n) < the number of digits of n!.

			a(4) = A004185(4*6) = 24; a(5) = A004185(5*24) = 12; a(6) = A004185(6*12) = 27.

Cf. A000142, A004185, A057615, A069638.

f(n) = fromdigits(vecsort(digits(n))); \\ A004185
a(n) = if (n, f(a(n-1)*n), 1); \\ Michel Marcus, Jun 30 2021

def A004185(n):
    return 0 if n == 0 else int("".join(sorted(str(n))).strip('0'))
def aupton(nn):
    alst = [1]
    for n in range(1, nn+1): alst.append(A004185(alst[-1]*n))
    return alst
print(aupton(29)) # Michael S. Branicky, Jun 29 2021

cp's OEIS Frontend

A108567 a(0) = 0, a(1) = a(2) = 1, a(3) = 2, a(4) = 4, a(5) = 8, a(6) = 16, for n>5: a(n+1) = SORT[ a(n) + a(n-1) + a(n-2) + a(n-3) + a(n-4) + a(n-5) + a(n-6)], where SORT places digits in ascending order and deletes 0's.

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A346296 a(0) = 1; thereafter a(n) = 2*a(n-1) + 1, with digits rearranged into nondecreasing order.

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A368559 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number not occurring earlier such that a(n) contains all the distinct digits of a(n-1) - a(n-2).

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A345902 a(0) = 1; for n >= 1, a(n) = A004185(a(n-1)*n).

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Python