A262909 -id:A262909 - cp's OEIS frontend

A263081 a(n) = largest k for which A155043(k) < A262508(n); a(n) = A262509(n) + A262909(n).

124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 124340, 24684000, 24684000, 24684000, 24684000, 24684000, 24684000, 24684000

Offset: 1

Antti Karttunen, Oct 09 2015

nonn

a(n) = largest k for which A155043(k) < A155043(A262509(n)).
If a(n) > A262509(n) then it must be a leaf (see comments in A262909 for why). Particularly, we have A045765(40722) = 124340, A045765(8191770) = 24684000.
Terms of sequence (together with the corresponding values in A262508) give particularly clean values for the boundaries that are used for example in the C++-program which computes A262896.

Antti Karttunen, Table of n, a(n) for n = 1..68

Cf. A045765, A155043, A262508, A262509, A262896, A262909, A263077, A263083.

(define (A263081 n) (+ (A262509 n) (A262909 n)))

a(n) = A263077(A262509(n)).

a(n) = A262509(n) + A262909(n).

A045765 k - d(k) never takes these values, where d(k) = A000005(k).

Original entry on oeis.org

7, 8, 13, 19, 20, 24, 25, 28, 33, 36, 37, 40, 43, 49, 50, 52, 55, 56, 63, 64, 66, 67, 68, 74, 75, 79, 85, 88, 98, 100, 103, 108, 109, 112, 113, 116, 117, 123, 124, 126, 131, 132, 133, 134, 136, 140, 143, 145, 150, 153, 156, 159, 160, 163, 164, 167, 168

Offset: 1

David W. Wilson

nonn

Complement of A236562. - Jaroslav Krizek, Feb 09 2014
Positions of zeros in A060990, leaf-nodes in the tree generated by edge-relation A049820(child) = parent. - Antti Karttunen, Oct 06 2015
Since A000005(x) <= 1 + x/2, k is in the sequence if there are no x <= 2*(k+1) with k = x - d(x). - Robert Israel, Oct 12 2015
This can be improved as: k is in the sequence if there are no x <= k + A002183(2+A261100(k)) with k = x - d(x). Cf. also A070319, A262686. - Antti Karttunen, Oct 12 2015
Luca (2005) proved that this seqeunce is infinite. - Amiram Eldar, Jul 26 2025

Antti Karttunen, Table of n, a(n) for n = 1..10000
Florian Luca, On numbers not of the form n-omega(n), Acta Mathematica Hungarica, Vol. 106, No. 1 (2005), pp. 117-135.

Top row of A262898.
Cf. A000005, A002183, A049820, A060990, A070319, A236562, A236565, A261100, A262511, A262686, A262901, A262902, A262903, A262909, A263081.
Cf. A263091 (primes in this sequence), A263095 (squares).
Cf. A259934 (gives the infinite trunk of the same tree, conjectured to be unique).

N:= 1000: # to get all terms <= N
sort(convert({$1..N} minus {seq(x - numtheory:-tau(x), x=1..2*(1+N))},list)); # Robert Israel, Oct 12 2015

lim = 10000; Take[Complement[Range@ lim, Sort@ DeleteDuplicates@ Table[n - DivisorSigma[0, n], {n, lim}]], 57] (* Michael De Vlieger, Oct 13 2015 *)

allocatemem((2^31)+(2^30));
uplim = 36756720 + 640; \\ = A002182(53) + A002183(53).
v060990 = vector(uplim);
for(n=3, uplim, v060990[n-numdiv(n)]++);
A060990 = n -> if(!n,2,v060990[n]);
uplim2 = 36756720;
n=0; k=1; while(n <= uplim2, if(0==A060990(n), write("b045765_big.txt", k, " ", n); k++); n++;);
\\ Antti Karttunen, Oct 09 2015

(define A045765 (ZERO-POS 1 1 A060990))
;; Using also IntSeq-library of Antti Karttunen, Oct 06 2015

A263078 a(n) = greatest k for which A155043(n+k) < A155043(n); a(n) = A263077(n)-n.

Original entry on oeis.org

-1, -2, -1, -2, 1, -4, 5, -2, -3, -4, 1, -6, 5, -2, 3, 2, 5, -6, 11, -2, 9, -4, 11, -2, -3, -4, 15, -6, 19, -8, 29, -2, 27, -4, 37, 12, 47, -4, 45, -6, 55, -8, 65, -2, 51, -4, 61, -6, -1, -2, 69, -4, 79, -6, 77, -8, 83, 2, 81, -12, 79, 10, 77, 76, 75, 6, 73, 16, 71, 14, 69, -12, 67, 22, 65, 20, 73, 18, 77, 16, 27, 26, 37, -12, 35, 34, 45, 20, 51, 18, 49, 40, 47, 26, 45, -12, 43, 42, 41, 40, 39, 30

Offset: 1

Antti Karttunen, Oct 09 2015

sign

			For n=1 we have A049820(1) = 0, thus A155043(1) = 1, and 0 is the only (and thus the largest) number from which zero can be reached with less steps (namely in zero steps, A155043(0) = 0), thus a(1) = 0 - 1 = -1.
For n=7, we have A155043(7) = 4 [as A049820(7) = 5, A049820(5) = 3, A049820(3) = 1, A049820(1) = 0], but there exists x=12 for which we have A049820(12) = 6, A049820(6) = 2, A049820(2) = 0, and this is the largest x such that A155043(x) < A155043(7), thus a(7) = 12 - 7 = 5.

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

Cf. A155043, A261089, A262503, A262909, A263077.

Cf. A263079 (indices of the negative terms), A263080 (of the positive terms).

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, {n, 102}] (* Michael De Vlieger, Oct 13 2015 *)

A263078 = n -> A263077(n) - n;
for(n=1,124340,write("b263078.txt",n," ",A263078(n)));
\\ Other code as in A263077

a(n) = A263077(n)-n.

A262908 a(n) = largest k such that A049820(k + A262509(n)) <= A262509(n).

Original entry on oeis.org

53, 49, 69, 55, 53, 31, 47, 39, 25, 35, 31, 39, 37, 51, 33, 43, 33, 69, 65, 57, 43, 41, 57, 49, 33, 33, 43, 41, 37, 33, 37, 39, 35, 27, 41, 27, 43, 75, 177, 171, 173, 155, 45, 133, 107, 121, 111, 139, 78, 119, 123, 47, 65, 79, 77, 97, 81, 151, 149, 145, 111, 197, 375, 71, 59, 81, 259, 257

Offset: 1

Antti Karttunen, Oct 08 2015

nonn

For all nonzero terms a(n), A263083(n) = a(n) + A262509(n) and A155043(A263083(n)) < A155043(A262509(n)) because at each A262509(n) the "distance to zero", A155043 obtains a unique value A262508(n), thus no A049820-iteration trajectory starting from any k larger than A262509(n) and using a greater or equal number of steps to reach zero may bypass A262509(n) [i.e., without going through A262509(n)], because then A262508(n) would not be unique anymore. See also comments in A262909.

Cf. A049820, A155043, A262508, A262509, A262909, A263083.

(define (A262908 n) (let ((t (A262509 n))) (let loop ((k (A002183 (+ 2 (A261100 t))))) (cond ((<= (A049820 (+ t k)) t) k) (else (loop (- k 1)))))))

Other identities. For all n >= 1:

a(n) <= A262909(n).

cp's OEIS Frontend

A263081 a(n) = largest k for which A155043(k) < A262508(n); a(n) = A262509(n) + A262909(n).

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A045765 k - d(k) never takes these values, where d(k) = A000005(k).

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A263078 a(n) = greatest k for which A155043(n+k) < A155043(n); a(n) = A263077(n)-n.

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A262908 a(n) = largest k such that A049820(k + A262509(n)) <= A262509(n).

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