A268710 -id:A268710 - cp's OEIS frontend

A268395 Partial sums of A268389.

0, 0, 0, 1, 1, 3, 4, 4, 4, 5, 7, 7, 8, 8, 8, 11, 11, 15, 16, 16, 18, 18, 18, 19, 20, 20, 20, 22, 22, 23, 26, 26, 26, 27, 31, 31, 32, 32, 32, 34, 36, 36, 36, 37, 37, 40, 41, 41, 42, 42, 42, 47, 47, 48, 50, 50, 50, 52, 53, 53, 56, 56, 56, 57, 57, 59, 60, 60, 64, 64, 64, 65, 66, 66, 66, 69, 69, 70, 72, 72, 74, 74, 74, 75, 75, 81

Offset: 0

Antti Karttunen, Feb 10 2016

nonn

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A048631, A268389, A268672.
Cf. A268678 (with duplicates removed), A268677 (numbers that do not occur here).
Cf. A268708, A268709, A268710.
Cf. also A054861.

f[n_] := Which[n == 1, 0, OddQ@ #, 0, EvenQ@ #, 1 + f[#/2]] &@ Fold[BitXor, n, Quotient[n, 2^Range[BitLength@ n - 1]]]; Accumulate@ Array[f, {85}] (* Michael De Vlieger, Feb 12 2016, after Jan Mangaldan at A006068 *)

a(0) = 0, for n >= 1, a(n) = A268389(n) + a(n-1).
Other identities. For all n >= 0:
a(n) = A268389(A048631(n)).
a(n) = n - A268672(n).

A268708 Number of iterations of A268395 needed to reach zero: a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).

Original entry on oeis.org

0, 1, 1, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 18

Offset: 0

Antti Karttunen, Feb 11 2016

nonn

Set k = n, take the k-th Xfactorial A048631(k) and find what is the maximum exponent h so that polynomial (X+1)^h still divides it (over GF(2), this can be computed with A268389, maybe also more directly). Set this h as a new value of k, and repeat. a(n) tells how many steps are needed before zero is reached.

Note that so far it is just an empirical observation that A268672(n) > 0 for all n > 0.

Antti Karttunen, Table of n, a(n) for n = 0..65537

Cf. A048631, A268389, A268395, A268672.

Cf. also A268709, A268710.

a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).

A268709 Number of iterations of A268395 needed to reach zero from 2^n: a(n) = A268708(2^n).

Original entry on oeis.org

1, 1, 2, 3, 5, 9, 15, 25, 40, 75, 134, 246, 428, 802, 1453, 2643, 4587, 8851, 16849, 32368, 60503, 117343

Offset: 0

Antti Karttunen, Feb 11 2016

Cf. A000079, A268395, A268708, A268710.

(define (A268709 n) (A268708 (A000079 n)))

a(n) = A268708(2^n).

cp's OEIS Frontend

A268395 Partial sums of A268389.

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A268708 Number of iterations of A268395 needed to reach zero: a(0) = 0, for n >= 1, a(n) = 1 + a(A268395(n)).

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A268709 Number of iterations of A268395 needed to reach zero from 2^n: a(n) = A268708(2^n).

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