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.

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A353308 Numbers k for which A046523(A332223(k)) is equal to A046523(k).

Original entry on oeis.org

1, 2, 9, 14, 15, 38, 39, 57, 68, 70, 92, 106, 110, 111, 125, 129, 130, 156, 170, 183, 190, 213, 230, 242, 245, 267, 275, 338, 350, 380, 393, 416, 441, 455, 494, 518, 522, 532, 572, 579, 585, 590, 595, 627, 638, 646, 650, 682, 686, 722, 740, 754, 782, 790, 850, 855, 879, 902, 946, 950, 957, 969, 994, 1090, 1118, 1227
Offset: 1

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Author

Antti Karttunen, Apr 17 2022

Keywords

Comments

Numbers k such that A348717(A332223(k)) = A348717(k) form a subsequence of this sequence. As its subsequence, we further have sequences A005940(1+A336702(n)) and A005940(1+A027687(n)), computed for n >= 1, and sorted into ascending order.

Crossrefs

Cf. A324201 (subsequence), A353363.
Cf. also A019278, A323653.

A355279 Numbers k such that S(S(S(k))) = k, with S(n) = sigma(n)/4: 1/4-sociable numbers of order 1 or 3.

Original entry on oeis.org

30240, 32760, 2178540, 23569920, 45532800, 46475520, 48933360, 50995620, 60933600, 69995520, 72807696, 142990848
Offset: 1

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Author

M. F. Hasler, Sep 25 2022

Keywords

Comments

a(6..11) = { 46475520, 48933360, 50995620, ..., 72807696 } are the first terms not in A027687 (4-perfect numbers), which is obviously a subsequence.
a(11..12) = { 72807696, 142990848 } are the first terms that are not multiples of 60.

Crossrefs

A027687 is a subsequence.
Cf. A113286 (1/4 sociable numbers of order <= 2), A113546 (1/3-sociable numbers).

Programs

  • PARI
    is_A355279(n, f(x)=if(x && x%4==0, sigma(x\4)))= f(f(sigma(n)))==n*4

A227882 Known number of n_multiperfect numbers that can produce an hemiperfect of abundancy (2*n-1)/2.

Original entry on oeis.org

1, 3, 19, 0, 87, 117, 0, 30, 0, 0
Offset: 2

Views

Author

Michel Marcus, Oct 25 2013

Keywords

Comments

The hemiperfect that are obtained are coprime to p = 2*n-1.
When p=2*n-1 is prime, if m is a n-multiperfect is such that valuation(m, p) = 1, then let's define k = m/p, sigma(k) = sigma(m/p) = sigma(m)/sigma(p) = (n*m)/(p+1) = (n*m)/(2*n) = m/2. So sigma(k)/k = m/(2*k) = (k*p)/(2*k) = p/2 = (2*n-1)/2.

Examples

			a(2) = 1, since the only perfect number multiple of 3 is 6, and 6/3=2 has abundancy 3/2.
a(3) = 3, since the 3 known hemiperfect of abundancy 5/2 are coprime to 5.
a(5) = a(8) = a(11) = 0, since for those n, 2*n-1 is not prime.
a(10) is also 0, since all known 10-multiperfect are at least divisible by 19^2.
		

Crossrefs

Cf. A000396 (2), A005820 (3), A027687 (4), A046060 (5), A046061 (6), A007691 (integer abundancy).
Cf. A141643 (5/2), A055153 (7/2), A141645 (9/2), A159271 (11/2), A160678 (13/2), A159907 (half-integer abundancy).
Cf. A006254.
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