cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A262680 Number of squares encountered before zero is reached when iterating A049820 starting from n: a(0) = 0 and for n >= 1, a(n) = A010052(n) + a(A049820(n)).

Original entry on oeis.org

0, 1, 0, 1, 2, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1
Offset: 0

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Author

Antti Karttunen, Oct 03 2015

Keywords

Comments

Number of perfect squares (A000290) encountered before zero is reached when starting from k = n and repeatedly applying the map that replaces k by k - d(k), where d(k) is the number of divisors of k (A000005). This count includes n itself if it is a square, but excludes the final zero.
Also number of times the parity (of numbers encountered) changes until zero is reached when iterating A049820. This count includes also the last parity change 1 - d(1) -> 0 if coming to zero through 1.
There is a lower bound for this sequence that grows without limit if and only if either (1) A259934 is indeed the unique sequence (satisfying its given condition) and it contains an infinite number of squares (see A262514), or (2) more generally, if each one of all (hypothetically multiple) infinite branches of the tree (defined by parent-child relation A049820(child) = parent) contains an infinite number of squares. See also comments in A262509.

Examples

			For n=1, we subtract 1 - A000005(1) = 0, thus we reach zero in one step, and the starting value 1 is a square, thus a(1) = 1. Also, the parity changes once, from odd to even as we go from 1 to 0.
For n=24, when we start repeatedly subtracting the number of divisors (A000005), we obtain the following numbers: 24 - A000005(24) = 24 - 8 = 16, 16 - A000005(16) = 16 - 5 = 11, 11 - 2 = 9, 9 - 3 = 6, 6 - 4 = 2, 2 - 2 = 0. Of these numbers, 16 and 9 are squares larger than zero, thus a(24)=2. Also, we see that the parity changes twice: from even to odd at 16 and then back from odd to even at 9.
		

Crossrefs

Bisections: A262681, A262682.
Cf. A262687 (positions of records).

A263095 Squares in A045765; numbers n^2 such that there is no such k for which k - d(k) = n^2, where d(k) is the number of divisors of k (A000005).

Original entry on oeis.org

25, 36, 49, 64, 100, 196, 256, 324, 400, 484, 676, 729, 784, 1156, 1225, 1369, 2116, 2209, 2500, 2916, 3136, 3364, 3481, 3600, 3844, 4489, 5329, 6084, 6724, 7225, 7921, 8100, 8836, 9025, 9604, 10000, 10609, 10816, 11236, 12100, 12996, 13456, 13924, 14884, 15376, 15625, 15876, 16129, 16384, 16900, 18225, 19600, 19881, 20164, 21904, 22500, 24025, 24964, 25921, 26896
Offset: 1

Views

Author

Antti Karttunen, Oct 10 2015

Keywords

Comments

Some of the terms are shared with A262687, but none with A262514.

Crossrefs

Cf. A263093 (gives the square roots).
Intersection of A000290 and A045765.
Cf. also A263091.

Programs

  • Mathematica
    lim = 40000; Take[Select[Complement[Range@ lim, Sort@ DeleteDuplicates@ Table[n - DivisorSigma[0, n], {n, lim}]], IntegerQ@ Sqrt@ # &], 60] (* Michael De Vlieger, Oct 13 2015 *)
  • PARI
    \\ See code in A263093.
    
  • Scheme
    ;; With Antti Karttunen's IntSeq-library.
    (define A263095 (MATCHING-POS 1 1 (lambda (n) (and (= 1 (A010052 n)) (zero? (A060990 n))))))

A262688 a(n) = square root of the index where A262680 reaches n for the first time.

Original entry on oeis.org

0, 1, 2, 13, 18, 39, 354, 355, 652, 1309, 1310, 2227, 4976
Offset: 0

Views

Author

Antti Karttunen, Oct 03 2015

Keywords

Comments

Natural numbers whose squares set new records for A262680.
Even and odd terms alternate.

Crossrefs

Programs

Formula

a(n) = A000196(A262687(n)).
Showing 1-3 of 3 results.