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 1-5 of 5 results.

A337339 Denominator of (1+sigma(s)) / ((s+1)/2), where s is the square of n prime-shifted once (s = A003961(n)^2 = A003961(n^2)).

Original entry on oeis.org

1, 5, 13, 41, 25, 113, 61, 365, 313, 221, 85, 1013, 145, 109, 613, 3281, 181, 2813, 265, 1985, 1513, 761, 421, 9113, 1201, 1301, 7813, 377, 481, 5513, 685, 29525, 2113, 1625, 2965, 25313, 841, 2381, 3613, 17861, 925, 13613, 1105, 6845, 15313, 3785, 1405, 82013, 7321, 10805, 4513, 11705, 1741, 70313, 4141, 8821, 6613, 865
Offset: 1

Views

Author

Antti Karttunen, Aug 24 2020

Keywords

Comments

All terms are members of A007310, because all terms of A337336 and A337337 are.
No 1's after the initial one at a(1) => No quasiperfect numbers. See comments in A336700 & A337342.
If any quasiperfect numbers qp exist, they must occur also in A325311.
Question: Is there any reliable lower bound for this sequence? See A337340, A337341.
Duplicate values are rare, but at least two cases exist: a(21) = a(74) = 1513 and a(253) = a(554) = 71065. - Antti Karttunen, Jan 03 2024

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A337339(n) = { my(s=(A003961(n)^2),u=(s+1)/2); (u/gcd(1+sigma(s), u)); };
    \\ Or alternatively as:
    A337339(n) = { my(s=A003961(n^2)); denominator((1+sigma(s))/((s+1)/2)); };

Formula

a(n) = A337336(n) / A337337(n) = A048673(n^2) / gcd(A048673(n^2), A336844(n^2)).
a(n) = A337336(n) / gcd(A337336(n), 1+A003973(n^2)).

A332223 a(1) = 1, and for n > 1, a(n) = A005940(1+sigma(A156552(n))).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 16, 7, 25, 18, 32, 25, 64, 21, 21, 49, 128, 27, 256, 35, 40, 121, 512, 49, 125, 385, 49, 121, 1024, 13, 2048, 13, 225, 1573, 105, 77, 4096, 57, 187, 343, 8192, 63, 16384, 65, 55, 4693, 32768, 121, 625, 32, 15625, 85, 65536, 81, 180, 91, 253, 9945, 131072, 175, 262144, 508079, 625, 847, 729, 169, 524288, 2057, 2601, 105, 1048576
Offset: 1

Views

Author

Antti Karttunen, Feb 12 2020

Keywords

Comments

From Antti Karttunen, Jul 31 - Aug 06 2020: (Start)
As a curiosity, like with sigma, also here a(14) = a(15). [Cf. also A003973 and A341512]
Question: is it possible that a(k) = 2*k for any k? If not, then the deficiency (A033879) cannot be -1, and there are no quasiperfect numbers. If there were such cases, then A156552(k) = q would be an instance of quasiperfect number, which should also be an odd square, thus k would need to be of the form 4u+2.
In range n <= 10000, a(n) is a nontrivial multiple of n only at n = [25, 35, 343, 539, 847, 3315] with a(n) = [125, 105, 2401, 2695, 2541, 9945]. The quotients are thus also odd: 5, 3, 7, 5, 3, 3.
This rather meager empirical evidence motivates a conjecture that no quotient a(n)/n may be an even integer, and particularly, never a power of 2 larger than one, which (when translated back to the ordinary, unconjugated sigma) claims that it is not possible that sigma(n) = 2^k * n + 2^k - 1, for any n > 1, k > 0. See also A336700 and A336701, where this leads to a rather surprising empirical observation.
(End)

Crossrefs

Cf. A003961, A332449, A332450, A332451, A332460 (for other functions similarly conjugated).

Programs

  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; \\ From A005940
    A156552(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ From A156552
    A332223(n) = if(1==n,n,A005940(1+sigma(A156552(n))));
    
  • PARI
    A332223(n) = if(1==n,n,A005940(1+sumdiv(A156552(n),d,d))); \\ Antti Karttunen, Aug 04 2020

Formula

For n > 1, a(n) = A005940(1+A000203(A156552(n))) = A005940(1+A323243(n)).
a(A324201(n)) = A003961(A324201(n)). [It's an open problem whether A324201 gives all such solutions]
For n > 1, a(n) = A005940(1 + (Sum_{d|A156552(n)} d)). - Antti Karttunen, Aug 04 2020

A325311 Odd abundant numbers k for which sigma(k) == 3 (mod 4).

Original entry on oeis.org

11025, 99225, 245025, 540225, 893025, 1334025, 2205225, 3980025, 4862025, 5832225, 6890625, 8037225, 8555625, 9828225, 10595025, 10989225, 12006225, 14402025, 19847025, 20385225, 24354225, 26163225, 26471025, 29648025, 31979025, 35820225, 38378025, 43758225, 46580625, 49491225, 50339025, 52490025, 55577025, 57836025, 60140025
Offset: 1

Views

Author

Antti Karttunen, Apr 20 2019

Keywords

Comments

These are all squares. Square roots are in A325312.

Crossrefs

Cf. A000203, A324647, A325312 (square roots).
Intersection of A005231 and A324899.
Subsequence of A156942.

Programs

  • Mathematica
    Select[Range[1, 7755, 2]^2, Mod[(s = DivisorSigma[1, #]), 4] == 3 && s > 2*# &] (* Amiram Eldar, Apr 05 2024 *)
  • PARI
    isA325311(n) = (n%2 && (3==sigma(n)%4) && sigma(n)>(2*n));

Formula

a(n) = A325312(n)^2. - Amiram Eldar, Apr 05 2024

A324909 Odd numbers k for which sigma(k^2) == 3 (mod 4).

Original entry on oeis.org

5, 13, 15, 17, 29, 35, 37, 39, 41, 45, 51, 53, 55, 61, 73, 87, 89, 91, 95, 97, 101, 105, 109, 111, 113, 115, 117, 119, 123, 125, 135, 137, 143, 149, 153, 155, 157, 159, 165, 173, 181, 183, 187, 193, 197, 203, 215, 219, 229, 233, 235, 241, 245, 247, 257, 259, 261, 267, 269, 273, 277, 281, 285, 287, 291, 293, 295, 299, 303, 313
Offset: 1

Views

Author

Antti Karttunen, Apr 20 2019

Keywords

Comments

Odd numbers with an odd number of prime factors congruent to 1 modulo 4, or odd members of A209922. - Charlie Neder, Apr 24 2019

Crossrefs

Cf. A325312 (a subsequence).
Cf. A209922 (a supersequence).

Programs

  • Mathematica
    Select[Range[1, 315, 2], Mod[DivisorSigma[1, #^2], 4] == 3 &] (* Amiram Eldar, Apr 05 2024 *)
  • PARI
    isA324909(n) = ((n%2)&&3==(sigma(n*n)%4));

Formula

a(n) = sqrt(A324899(n)).

A337342 Numbers k such that A048673(k) divides 1+A003973(k).

Original entry on oeis.org

1, 10, 584, 3824, 23008, 5033216
Offset: 1

Views

Author

Antti Karttunen, Aug 24 2020

Keywords

Comments

Numbers k such that A048673(k) = A337335(k). Equivalently, numbers k such that (A003961(k)+1)/2 divides 1+A003973(k).
No squares larger than one in this sequence => No quasiperfect numbers. See also A337339. For any x corresponding to a quasiperfect number qp = A003961(x), the quotient (1+A003973(x)) / A048673(x) should be 4. Thus that A003961(x) should also be a member of A325311.
At least for the terms x = a(2) .. a(6) here, the quotient (1+A003973(x)) / A048673(x) = 3. The terms for which the quotient is 3 are precisely those which by prime shifting become the terms of A007593 (that are all odd), thus the terms y = A064989(A007593(n)), for n >= 1, form a subsequence of this sequence.
a(7) > 2^28.
Terms 65810851904356352, 30943274395471606363637940224, 40102483616531202199118491418624 are also in the sequence, but their positions are unknown. (Adapted from Jud McCranie's Dec 16 1999 comment in A007593).

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    isA337342(n) = { my(s=A003961(n)); !((1+sigma(s))%((1+s)/2)); };
Showing 1-5 of 5 results.