A261103 -id:A261103 - cp's OEIS frontend

A259934 Infinite sequence starting with a(0)=0 such that A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n - (number of divisors of n).

0, 2, 6, 12, 18, 22, 30, 34, 42, 46, 54, 58, 62, 70, 78, 90, 94, 102, 106, 114, 118, 121, 125, 129, 144, 152, 162, 166, 174, 182, 190, 194, 210, 214, 222, 230, 236, 242, 250, 254, 270, 274, 282, 294, 298, 302, 310, 314, 330, 342, 346, 354, 358, 366, 374, 390, 394, 402, 410, 418, 426, 434, 442, 446, 462, 466, 474, 486, 494, 510, 522, 530, 546, 558, 562, 566, 574, 582, 590

Offset: 0

Max Alekseyev, Jul 09 2015

Equivalently, satisfies the property: A000005(a(n)) = a(n)-a(n-1). The first differences a(n)-a(n-1) are given in A259935.
V. S. Guba (2015) proved that such an infinite sequence exists. Numerical evidence suggests that it may also be unique -- is it? All terms below 10^10 are defined uniquely.
If the current definition does not uniquely define the sequence, the "lexicographically earliest" condition may be added to make the sequence well-defined.
From Vladimir Shevelev, Jul 21 2015: (Start)
If a(k), a(k+1), a(k+2) is an arithmetic progression, then a(k+1) is in A175304.
Indeed, by the definition of this sequence, a(n)-a(n-1) = d(a(n)), for all n>=1, where d(n) = A000005(n). Hence, have a(k+1) - a(k) = a(k+2) - a(k+1) = d(a(k+1)) = d(a(k+2)). So a(k+1) + d(a(k+2)) = a(k+2) and a(k+1) + d(a(k+1)) = a(k+2).
Therefore, d(a(k+1) + d(a(k+1))) = d(a(k+2))= d(a(k+1)), i.e., a(k+1) is in A175304. Thus, if there are infinitely many pairs of the same consecutive terms of A259935, then A175304 is infinite (see there my conjecture). (End)
From Antti Karttunen, Nov 27 2015: (Start)
If multiple apparently infinite branches would occur at some point of computing, then even if the "lexicographically earliest" condition were then added to the definition, it would not help us much (when computing the sequence), as we would still not know which of the said branches were truly infinite. [See also Max Alekseyev's latter Jul 9 2015 posting on SeqFan-list, where he notes the same thing.] Note that many of the derived sequences tacitly assume that the uniqueness-conjecture is true. See also comments at A262693 and A262896.
One sufficient (but not a necessary) condition for the uniqueness of this sequence is that the sequence A262509 has infinite number of terms. Please see further comments there.
The graph of sequence exhibits two markedly different slopes, depending on whether it is on the "fast lane" of A049820 (even numbers) or the "slow lane" [odd numbers, for example when traversing the 1356 odd terms from 123871 to 113569 at range a(9859) .. a(8504)]. See A263086/A263085 for the "average cumulative speed difference" between the lanes. In general, slow and fast lane stay separate, except when they terminate into one of the squares (A262514) that work as "exchange ramps", forcing the parity (and thus the speed) to change. In average, the odd squares are slightly better than the even squares in attracting lanes going towards smaller numbers (compare A263253 to A263252). The cumulative effect of this bias is that the odd terms are much rarer in this sequence than the even terms (compare A263278 to A262516).
(End)

Robert Israel, Table of n, a(n) for n = 0..64800
M. Alekseyev et al., Apparently unique infinite sequences related to the sum of divisors, Discussion in SeqFan mailing list, 2015.
Michael De Vlieger, Poster illustrating A259934 and A263267
V. S. Guba et al., Sequence and divisors, 2015. (in Russian)

Cf. A000005, A049820, A060990, A259935 (first differences).
Topmost row of A263255. Cf. also irregular tables A263267 & A263265 and array A262898.
Cf. A262693 (characteristic function).
Cf. A155043, A262694, A262904 (left inverses).
Cf. A262514 (squares present), A263276 (their positions), A263277.
Cf. A262517 (odd terms).
Cf. A262509, A262510, A262897 (other subsequences).
Cf. also A175304, A260257, A262680.
Cf. A261089, A261103, A262503, A262506, A262516, A263279, A263280, A263085, A263086, A263253, A263257, A263278.
Cf. also A262679, A262896 (see the C++ program there).
No common terms with A045765 or A262903.
Positions of zeros in A262522, A262695, A262696, A262697, A263254.
Various metrics concerning finite side-trees: A262888, A262889, A262890.
Cf. also A262891, A262892 and A262895 (cf. its graph).
Cf. A260084, A260124 (variants).
Cf. also A179016 (a similar "beanstalk trunk sequence" but with more tractable and regular behavior).

N:= 10^4: # to get "guaranteed unique" terms <= N
S:= Vector(N,datatype=integer[1]):
for n from N+1 to 2*N do
  k:= n - numtheory:-tau(n);
  if k <= N then S[k]:= S[k]+1; B[k]:= n; fi;
od:
for n from N to 3 by -1 do
  if S[n] >= 1 then
    k:= n - numtheory:-tau(n);
    S[k]:= S[k]+1; B[k]:= n;
  fi
od:
A[0]:= 0: A[1]:= 2:
for n from 2 do
  b:= B[A[n-1]];
  if b > N or S[b] > 1 then break fi;
  A[n]:= b;
od:
seq(A[i],i=0..n-1); # Robert Israel, Jul 09 2015

NN = 10^4; (* to get "guaranteed unique" terms <= NN *)
Clear[A, B, S]; S[]=0; For[n = NN+1, n <= 2*NN, n++, k = n-DivisorSigma[0, n]; If[k <= NN, S[k] = S[k]+1; B[k]=n]]; For[n=NN, n >= 3, n--, If[S[n] >= 1 , k = n-DivisorSigma[0, n]; S[k] = S[k]+1; B[k]=n]]; A[0]=0; A[1]=2; For[n=2, True, n++, b = B[A[n-1]]; If[b>NN || S[b]>1, Break[]]; A[n]=b]; Table[A[i], {i, 0, n-1}] (* _Jean-François Alcover, Jul 22 2015, after Robert Israel *)

From Antti Karttunen, Nov 27 2015: (Start)
Other identities and observations. For all n >= 0:
a(n) = A262679(A262896(n)).
A155043(a(n)) = A262694(a(n)) = A262904(a(n)) = n.
A261089(n) <= a(n) <= A262503(n). [A261103 and A262506 give the distances of a(n) to these bounds.]
(End)

A261089 a(n) = least k such that A155043(k) = n; positions of records in A155043.

Original entry on oeis.org

0, 1, 3, 5, 7, 13, 17, 19, 23, 27, 29, 31, 35, 37, 41, 43, 51, 53, 57, 59, 61, 65, 67, 71, 73, 77, 79, 143, 149, 151, 155, 157, 161, 163, 173, 177, 179, 181, 185, 191, 193, 199, 203, 209, 211, 215, 219, 223, 231, 233, 237, 239, 241, 249, 251, 263, 267, 269, 271, 277, 285, 291, 293, 299, 303, 315, 317, 321, 327, 331, 335, 337, 341, 347, 349, 357, 359, 369, 515

Offset: 0

Antti Karttunen, Sep 23 2015

nonn

Note that there are even terms besides 0, and they all seem to be squares: a(915) = 7744 (= 88^2), a(41844) = 611524 (= 782^2), a(58264) = 872356 (= 934^2), a(66936) = 1020100 (= 1010^2), a(95309) = 1503076 (= 1226^2), a(105456) = 1653796 (= 1286^2), ...

Antti Karttunen, Table of n, a(n) for n = 0..110880

Cf. A000005, A155043, A049820, A259934, A261103.
Cf. A262503 (the last occurrence of n in A155043).
Cf. A262505 (difference between the last and the first occurrence).
Cf. A262507 (the number of occurrences of n in A155043).
Cf also A261085, A261088.

import Data.List (elemIndex); import Data.Maybe (fromJust)
a261089 = fromJust . (`elemIndex` a155043_list)
-- Reinhard Zumkeller, Nov 27 2015

lim = 80; a[0] = 0; a[n_] := a[n] = 1 + a[n - DivisorSigma[0, n]]; t = Table[a@ n, {n, 0, 12 lim}]; Table[First@ Flatten@ Position[t, n] - 1, {n, 0, lim}] (* Michael De Vlieger, Sep 29 2015 *)

allocatemem(123456789);
uplim = 2162160; \\ = A002182(41).
v155043 = vector(uplim);
v155043[1] = 1; v155043[2] = 1;
for(i=3, uplim, v155043[i] = 1 + v155043[i-numdiv(i)]);
A155043 = n -> if(!n,n,v155043[n]);
n=0; k=0; while(k <= 10000, if(A155043(n)==k, write("b261089.txt", k, " ", n); k++); n++;);

;; With Antti Karttunen's IntSeq-library, two variants.
(definec (A261089 n) (let loop ((k 0)) (if (= n (A155043 k)) k (loop (+ 1 k)))))
(define A261089 (RECORD-POS 0 0 A155043))

Other identities. For all n >= 0:

A155043(a(n)) = n.

A262505 Distance between the positions of the first and the last occurrence of n in A155043: a(n) = A262503(n) - A261089(n).

Original entry on oeis.org

0, 1, 3, 7, 11, 9, 13, 15, 19, 21, 31, 41, 49, 59, 67, 77, 81, 87, 55, 57, 65, 59, 63, 67, 77, 79, 89, 37, 27, 33, 37, 47, 55, 65, 67, 71, 85, 99, 65, 67, 77, 89, 93, 103, 95, 105, 109, 117, 121, 131, 135, 115, 117, 119, 133, 133, 153, 133, 143, 141, 147, 159, 175, 181, 201, 205, 223, 239, 245, 249, 259, 275, 269, 271, 273, 271, 289, 303, 77

Offset: 0

Antti Karttunen, Sep 25 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..110880

Cf. A155043, A259934, A261089, A261103, A262503, A262506.
Cf. A262508 (positions of zeros).
Cf. A262507 (the number of occurrences of n in A155043).

(define (A262505 n) (- (A262503 n) (A261089 n)))

a(n) = A262503(n) - A261089(n).

a(n) = A261103(n) + A262506(n).

A263279 a(n) = A263259(A259934(n)); one-based position of A259934(n) at row n of table A263265.

Original entry on oeis.org

1, 2, 3, 5, 4, 5, 6, 4, 4, 3, 6, 3, 2, 4, 5, 6, 5, 5, 5, 7, 4, 4, 5, 6, 9, 6, 7, 5, 6, 4, 4, 3, 5, 5, 4, 5, 3, 3, 4, 4, 5, 5, 5, 5, 5, 4, 4, 3, 5, 7, 4, 4, 5, 7, 8, 5, 7, 5, 7, 8, 8, 6, 8, 6, 7, 7, 7, 9, 10, 10, 11, 9, 10, 9, 11, 10, 9, 7, 8, 9, 7, 6, 9, 8, 10, 6, 5, 6, 5, 9, 8, 8, 8, 7, 6, 4, 3, 3, 3, 5, 4, 7, 6, 6, 7, 9, 5, 5, 10, 10, 11, 6, 7, 9, 9, 7, 9, 13, 7, 6, 8

Offset: 0

Antti Karttunen, Nov 24 2015

nonn

a(n) gives the number of integers k <= A259934(n) for which A155043(k) = n = A155043(A259934(n)).

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A155043, A259934, A263259, A263265, A263280.

Cf. also A261103.

(define (A263267 n) (A263259 (A259934 n)))

a(n) = A263259(A259934(n)).

A262506 a(n) = A262503(n) - A259934(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 14, 22, 26, 30, 30, 38, 38, 6, 2, 8, 3, 5, 9, 6, 4, 6, 14, 2, 2, 2, 10, 6, 14, 18, 18, 28, 38, 0, 4, 0, 14, 14, 18, 8, 18, 18, 26, 22, 22, 26, 0, 0, 2, 10, 6, 26, 0, 4, 0, 6, 16, 26, 34, 42, 54, 66, 74, 78, 70, 72, 82, 64, 60, 60, 62, 74, 90, 2, 0, 2, 0, 2, 10, 0, 4, 14, 18, 22, 18, 22, 40, 58, 30, 2, 8, 22, 38, 42, 0, 0, 0, 4, 0, 0, 0, 4, 0, 2

Offset: 0

Antti Karttunen, Sep 25 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 0..64800
A. Karttunen, Ratio A261103(n)/a(n) drawn with the help of OEIS Plot2-script

Cf. A259934, A261089, A261103, A262503, A262505.

(define (A262506 n) (- (A262503 n) (A259934 n)))

a(n) = A262503(n) - A259934(n).

cp's OEIS Frontend

A259934 Infinite sequence starting with a(0)=0 such that A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n - (number of divisors of n).

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A261089 a(n) = least k such that A155043(k) = n; positions of records in A155043.

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A262505 Distance between the positions of the first and the last occurrence of n in A155043: a(n) = A262503(n) - A261089(n).

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A263279 a(n) = A263259(A259934(n)); one-based position of A259934(n) at row n of table A263265.

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A262506 a(n) = A262503(n) - A259934(n).

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