A155043 -id:A155043 - cp's OEIS frontend

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

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.

A262503 a(n) = largest k such that A155043(k) = n.

Original entry on oeis.org

0, 2, 6, 12, 18, 22, 30, 34, 42, 48, 60, 72, 84, 96, 108, 120, 132, 140, 112, 116, 126, 124, 130, 138, 150, 156, 168, 180, 176, 184, 192, 204, 216, 228, 240, 248, 264, 280, 250, 258, 270, 288, 296, 312, 306, 320, 328, 340, 352, 364, 372, 354, 358, 368, 384, 396, 420, 402, 414, 418, 432, 450, 468, 480, 504, 520, 540, 560, 572, 580, 594, 612, 610, 618, 622, 628, 648, 672, 592

Offset: 0

Antti Karttunen, Sep 24 2015

nonn

The first odd terms occur as a(121) = 1089, a(123) = 1093, a(349) = 3253, a(717) = 7581, a(807) = 8685, a(1225) = 13689, etc.

Antti Karttunen, Table of n, a(n) for n = 0..110880
A. Karttunen, Ratio A262502(n+2)/a(n) drawn with the help of OEIS Plot2-script
A. Karttunen, Ratio a(n)/A261089(n) drawn with the help of OEIS Plot2-script

Cf. A261089 (gives the first occurrence of n in A155043).
Cf. A262507 (gives the number of times n occurs in A155043).
Cf. A259934, A262502.

lim = 80; a[0] = 0; a[n_] := a[n] = 1 + a[n - DivisorSigma[0, n]]; t = Table[a@ n, {n, 0, 12 lim}]; Last@ Flatten@ Position[t, #] - 1 & /@ Range[0, lim] (* Uses the product of a limit and an arbitrary coefficient (12) based on observation of output for low values (n < 500). This might need to be adjusted for large n to give correct values of a(n). - Michael De Vlieger, Sep 29 2015 *) (* Note: one really should use a general safe limit, like A262502(n+2) I use in my Scheme-program. - Antti Karttunen, 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]);
uplim2 = 110880; \\ = A002182(30).
v262503 = vector(uplim2);
for(i=1, uplim, if(v155043[i] <= uplim2, v262503[v155043[i]] = i));
A262503 = n -> if(!n,n,v262503[n]);
for(n=0, uplim2, write("b262503.txt", n, " ", A262503(n)));

(define (A262503 n) (let loop ((k (A262502 (+ 2 n)))) (if (= (A155043 k) n) k (loop (- k 1)))))

Other identities and observations. For all n >= 0:

A262502(n+2) > a(n). [Not rigorously proved, but empirical evidence and common sense agrees.]

A262509 Numbers n such that there is no other number k for which A155043(k) = A155043(n).

Original entry on oeis.org

0, 119143, 119147, 119163, 119225, 119227, 119921, 119923, 120081, 120095, 120097, 120101, 120281, 120293, 120349, 120399, 120707, 120747, 120891, 120895, 120903, 120917, 120919, 121443, 121551, 121823, 122079, 122261, 122263, 122273, 122277, 122813, 122961, 123205, 123213, 123223, 123237, 123257, 123765, 24660543, 24660549, 24662311, 24662329, 24663759, 24664997, 24665023, 24665351

Offset: 0

Antti Karttunen, Sep 25 2015

nonn

Starting offset is zero, because a(0) = 0 is a special case in this sequence.
Numbers where A155043 takes a unique value. (Those values are given by A262508.)
Numbers n such that there does not exist any other number h from which one could reach zero in exactly the same number of steps as from n by repeatedly applying the map where k is replaced by k - A000005(k) = A049820(k). Thus in the tree where zero is the root and parent-child relation is given by A049820(child) = parent, all the numbers > n+t (where t is a small value depending on n) have n as their common ancestor. As it is guaranteed that there is at least one infinite path in such a tree, any n in this sequence can be neither a leaf nor any other vertex in a finite side-tree, as then at least one node in the infinite part would have the same distance to the root, thus it must be that n itself is in the infinite part and thus has an infinite number of descendant vertices. Also, for the same reason, the tree cannot branch to two infinite parts from any ancestor of n (which are nodes nearer to the root of the tree, zero).
From the above it follows that if this sequence is infinite, then A259934 is guaranteed to be the only infinite sequence which starts with a(0)=0 and satisfies the condition A049820(a(k)) = a(k-1) for all k>=1, where A049820(n) = n-d(n) and d(n) is the number of divisors of n (A000005).  This is a sufficient condition for the uniqueness of A259934, although not necessary. See e.g. A179016 which is the unique infinite solution to a similar problem, even though A086876 contains no ones after its two initial terms.
Is it possible for any even terms to occur after zero? If not, then apart from zero, this would be a subsequence of A262517.

Antti Karttunen, Table of n, a(n) for n = 0..68

Cf. A000005, A049820, A070319, A155043, A262508.
Subsequence of A259934, A261089, A262503 and A262513.
Cf. A262510 (gives the parent nodes for the terms a(1) onward), A262514, A262516, A262517.
Cf. also A086876, A179016.

\\ Compute A262508 and A262509 at the same time:
allocatemem((2^31)+(2^30));
\\ The limits used are quite ad hoc. Beware of the horizon-effect if you change these.
\\ As a post-check, test that A262509(n) = A259934(A262508(n)) for all the terms produced by this program.
uplim1 = 43243200 + 672; \\ = A002182(54) + A002183(54).
uplim2 = 36756720; \\ = A002182(53).
uplim3 = 10810800; \\
v155043 = vector(uplim1);
v262503 = vector(uplim3);
v262507 = vector(uplim3);
v155043[1] = 1; v155043[2] = 1;
for(i=3, uplim1, v155043[i] = 1 + v155043[i-numdiv(i)]);
A155043 = n -> if(!n,n,v155043[n]);
for(i=1, uplim1, v262503[v155043[i]] = i; v262507[v155043[i]]++; if(!(i%1048576),print1(i,", ")));
A262503 = n -> if(!n,n,v262503[n]);
A262507 = n -> if(!n,1,v262507[n]);
k=0; for(n=0, uplim3, if((1==A262507(n)) && (A262503(n) <= uplim2), write("b262508.txt", k, " ", n); write("b262509.txt", k, " ", A262503(n)); k++));

(define (A262509 n) (A261089 (A262508 n)))

a(n) = A261089(A262508(n)) = A262503(A262508(n)) = A259934(A262508(n)).

A262507 a(n) = number of times n occurs in A155043.

Original entry on oeis.org

1, 2, 3, 5, 4, 5, 6, 4, 4, 4, 8, 4, 4, 5, 8, 7, 7, 7, 7, 8, 5, 6, 6, 8, 10, 7, 8, 7, 7, 5, 5, 6, 6, 8, 6, 7, 7, 7, 4, 5, 5, 6, 6, 8, 7, 5, 5, 6, 7, 11, 5, 4, 5, 8, 12, 7, 9, 5, 8, 8, 9, 10, 14, 11, 12, 11, 9, 11, 13, 12, 12, 11, 11, 11, 12, 12, 10, 9, 9, 9, 8, 6, 10, 9, 10, 8, 7, 7, 8, 11, 10, 10, 12, 9, 7, 6, 5, 5, 5, 5, 4, 7, 8, 6, 7, 9, 7, 5, 11, 13, 13, 8, 10, 12, 13, 10, 12, 16, 9, 8, 12

Offset: 0

Antti Karttunen, Sep 25 2015

nonn

Records are: 1, 2, 3, 5, 6, 8, 10, 11, 12, 14, 16, 17, 19, 21, 22, 24, 25, 26, 27, 31, 35, 39, 44, ... and they occur at positions: 0, 1, 2, 3, 6, 10, 24, 49, 54, 62, 117, 236, 445, 484, 892, 893, 1022, 1784, 1911, 1912, 1913, 20600, 50822, ...

a(n) gives the length of each row of irregular table A263265.

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

Cf. A000005, A060990, A155043, A262502, A262505, A263265, A263270, A263279, A263280.
Cf. A261089, A262503.
Cf. A262508 (positions of ones).
Cf. A263260 (partial sums).

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)]);
uplim2 = 110880; \\ = A002182(30).
v262507 = vector(uplim2);
for(i=1, uplim, if(v155043[i] <= uplim2, v262507[v155043[i]]++));
A262507 = n -> if(!n,1,v262507[n]);
for(n=0, uplim2, write("b262507.txt", n, " ", A262507(n)));

(define (A262507 n) (add (lambda (k) (if (= (A155043 k) n) 1 0)) n (A262502 (+ 2 n))))
;; Auxiliary function add implements sum_{i=lowlim..uplim} intfun(i)
(define (add intfun lowlim uplim) (let sumloop ((i lowlim) (res 0)) (cond ((> i uplim) res) (else (sumloop (1+ i) (+ res (intfun i)))))))

a(n) = Sum_{k=n..A262502(2+n)} [A155043(k) == n]. (Here [...] denotes the Iverson bracket, resulting 1 when A155043(k) is n and 0 otherwise.)
Other identities. For all n >= 0:
a(n) = A263279(n) + A263280(n).

A262508 Numbers that occur only once in A155043; positions of zeros in A262505, ones in A262507.

Original entry on oeis.org

0, 9236, 9237, 9238, 9247, 9248, 9330, 9331, 9353, 9356, 9357, 9358, 9385, 9388, 9399, 9407, 9446, 9453, 9476, 9477, 9478, 9480, 9481, 9547, 9561, 9590, 9626, 9652, 9653, 9655, 9656, 9722, 9743, 9775, 9776, 9778, 9781, 9786, 9844, 1308289, 1308290, 1308465, 1308468, 1308592, 1308713, 1308717, 1308750, 1308809, 1308815, 1309104, 1309162, 1309214, 1309299, 1309397, 1309464, 1309465, 1309536, 1309537, 1309640, 1309641, 1309642, 1309648, 1309675, 1309714, 1309751, 1309879, 1309883, 1310010, 1310011

Offset: 0

Antti Karttunen, Sep 25 2015

nonn

Numbers n for which there exists exactly one natural number x from which one can reach zero in n steps by setting first k = x and then repeatedly applying the map where k is replaced with k - A000005(k). See A262509 for the corresponding x's and implications concerning A259934.

Starting offset is zero, because a(0) = 0 is a special case in this sequence.

Cf. A000005, A049820, A155043, A259934, A262505, A262507.

Cf. A262509, A262510.

\\ See the Pari-program given in A262509, which also computes the terms of this sequence at the same time.

A263265 Irregular triangle T(n,k), n >= 0, k = 1 .. A262507(n), read by rows, where each row n lists in ascending order all integers x for which A155043(x) = n.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 5, 8, 9, 10, 12, 7, 11, 14, 18, 13, 15, 16, 20, 22, 17, 24, 25, 26, 28, 30, 19, 21, 32, 34, 23, 38, 40, 42, 27, 44, 46, 48, 29, 36, 49, 50, 52, 54, 56, 60, 31, 33, 58, 72, 35, 62, 66, 84, 37, 39, 68, 70, 96, 41, 45, 74, 76, 78, 80, 104, 108, 43, 47, 81, 82, 88, 90, 120, 51, 83, 85, 86, 94, 128, 132, 53, 55, 87, 92, 102, 136, 140

Offset: 0

Antti Karttunen, Nov 24 2015

			Rows 0 - 8 of the triangle:
0;
1, 2;
3, 4, 6;
5, 8, 9, 10, 12;
7, 11, 14, 18;
13, 15, 16, 20, 22;
17, 24, 25, 26, 28, 30;
19, 21, 32, 34;
23, 38, 40, 42;
Row n contains A262507(n) terms, the first of which is A261089(n) and the last of which is A262503(n). For all terms on row n, A155043(n) = n.

Inverse: A263266.
Cf. A261089 (left edge), A262503 (right edge), A262507 (number of terms on each row).
Cf. A263279 (gives the positions of terms of A259934 on each row), A263280 (and their distance from the right edge).
Cf. A155043, A263259, A263270.
Cf. also permutations A263267 & A263268 and A263255 & A263256.
Differs from A263267 for the first time at n=31, where a(31) = 38, while A263267(31) = 40.

Other identities. For all n >= 0:

A155043(a(n)) = A263270(n).

A263260 a(n) = number of nonnegative integers k for which A155043(k) <= n; partial sums of A262507.

Original entry on oeis.org

1, 3, 6, 11, 15, 20, 26, 30, 34, 38, 46, 50, 54, 59, 67, 74, 81, 88, 95, 103, 108, 114, 120, 128, 138, 145, 153, 160, 167, 172, 177, 183, 189, 197, 203, 210, 217, 224, 228, 233, 238, 244, 250, 258, 265, 270, 275, 281, 288, 299, 304, 308, 313, 321, 333, 340, 349, 354, 362, 370, 379, 389, 403

Offset: 0

Antti Karttunen, Nov 24 2015

nonn

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

Cf. A000005, A049820, A155043, A262507, A263265, A263270.

a(0) = 1; for n >= 1, a(n) = A262507(n) + a(n-1).

A262518 Even bisection of A155043.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 4, 5, 4, 5, 5, 6, 6, 6, 6, 7, 7, 10, 8, 8, 8, 9, 9, 9, 10, 10, 10, 10, 11, 10, 12, 19, 12, 13, 13, 11, 14, 14, 14, 14, 15, 12, 16, 15, 15, 17, 16, 13, 18, 19, 17, 14, 18, 14, 18, 18, 19, 19, 20, 15, 21, 21, 20, 16, 22, 16, 23, 17, 23, 17, 24, 24, 25, 25, 24, 25, 26, 25, 27, 26, 26, 28, 27, 26, 27, 28, 28, 28, 29, 27, 29, 29

Offset: 0

Antti Karttunen, Oct 02 2015

nonn

Number of steps needed to reach zero when starting from k = 2*n and repeatedly applying the map that replaces k by k - d(k), where d(k) is the number of divisors of k (A000005).

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

Cf. A000005, A049820, A155043, A262519, A262520, A262682.

Scheme
```
(define (A262518 n) (A155043 (+ n n)))
```

a(n) = A155043(2*n).

A262519 Odd bisection of A155043.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Oct 02 2015

nonn

Number of steps needed to reach zero when starting from k = 2*n + 1 and repeatedly applying the map that replaces k by k - d(k), where d(k) is the number of divisors of k (A000005).

Antti Karttunen, Table of n, a(n) for n = 0..10000
A. Karttunen, Ratio a(n)/A262518(n) drawn with the help of OEIS Plot2-script

Cf. A000005, A049820, A155043, A262518, A262520, A262521, A262681.

Table[Length[NestWhileList[#-DivisorSigma[0,#]&,n,#!=0&]]-1,{n,1,200,2}] (* Harvey P. Dale, Aug 31 2017 *)

(define (A262519 n) (A155043 (+ n n 1)))

a(n) = A155043(2*n + 1).

A263077 a(n) = greatest k where A155043(k) < A155043(n).

Original entry on oeis.org

0, 0, 2, 2, 6, 2, 12, 6, 6, 6, 12, 6, 18, 12, 18, 18, 22, 12, 30, 18, 30, 18, 34, 22, 22, 22, 42, 22, 48, 22, 60, 30, 60, 30, 72, 48, 84, 34, 84, 34, 96, 34, 108, 42, 96, 42, 108, 42, 48, 48, 120, 48, 132, 48, 132, 48, 140, 60, 140, 48, 140, 72, 140, 140, 140, 72, 140, 84, 140, 84, 140, 60, 140, 96, 140, 96, 150, 96, 156, 96, 108, 108, 120, 72, 120, 120, 132, 108, 140, 108, 140, 132, 140, 120, 140, 84

Offset: 1

Antti Karttunen, Oct 09 2015

nonn

Antti Karttunen, Table of n, a(n) for n = 1..124340

Cf. A155043, A261089, A262503, A263078, A263079, A263080, A263082.

a[0] = 0; a[n_] := a[n] = 1 + a[n - DivisorSigma[0, n]]; Table[k = 3 n;
While[a@ k >= a@ n, k--]; k, {n, 96}] (* Michael De Vlieger, Oct 13 2015 *)

allocatemem((2^31)+(2^30));
uplim1 = 36756720 + 640; \\ = A002182(53) + A002183(53).
uplim2 = 36756720; \\ = A002182(53).
uplim3 = 32432400; \\ = A002182(52). Really just some Ad Hoc value smaller than above.
v155043 = vector(uplim1);
vother = vector(uplim3); \\ Contains A262503 and A263082 in succession.
v155043[1] = 1; v155043[2] = 1;
for(i=3, uplim1, v155043[i] = 1 + v155043[i-numdiv(i)]; if(!(i%1048576),print1(i,", ")));
A155043 = n -> if(!n,n,v155043[n]);
maxlen = 0; for(i=1, uplim2, len = v155043[i]; vother[len] = i; maxlen = max(maxlen,len); if(!(i%1048576),print1(i,", "))); \\ First it will be A262503.
print("uplim2=", uplim2, " uplim3=", uplim3, " maxlen=", maxlen);
\\ Then we convert it to A263082:
m = 0; for(i=1, maxlen, m = max(m, vother[i]); vother[i] = m; if(!(i%1048576),print1(i,", ")));
A263082 = n -> if(!n,n,vother[n]);
A263077 = n -> A263082(A155043(n)-1);
\\ Finally we can compute A263077:
for(i=1, uplim3, write("b263077.txt", i, " ", A263077(i)); );

a(n) = A263082(A155043(n)-1).

cp's OEIS Frontend

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

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A262503 a(n) = largest k such that A155043(k) = n.

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A262509 Numbers n such that there is no other number k for which A155043(k) = A155043(n).

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A262507 a(n) = number of times n occurs in A155043.

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A262508 Numbers that occur only once in A155043; positions of zeros in A262505, ones in A262507.

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A263265 Irregular triangle T(n,k), n >= 0, k = 1 .. A262507(n), read by rows, where each row n lists in ascending order all integers x for which A155043(x) = n.

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A263260 a(n) = number of nonnegative integers k for which A155043(k) <= n; partial sums of A262507.

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A262518 Even bisection of A155043.

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