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 1-10 of 19 results. Next

A324201 a(n) = A062457(A000043(n)) = prime(A000043(n))^A000043(n), where A000043 gives the exponent of the n-th Mersenne prime.

Original entry on oeis.org

9, 125, 161051, 410338673, 925103102315013629321, 1271991467017507741703714391419, 49593099428404263766544428188098203, 165163983801975082169196428118414326197216835208154294976154161023
Offset: 1

Views

Author

Antti Karttunen, Feb 18 2019

Keywords

Comments

If there are no odd perfect numbers, then the terms give all solutions n > 1 to A323244(n) = 0.
Conversely, if these are all numbers k > 1 that satisfy A323244(k) = 0 (which can be proved if one can show, for example, that no number in A007916 can satisfy the equation), then no odd perfect numbers exist. See also A336700. - Antti Karttunen, Jan 12 2024

Crossrefs

Subsequence of A001597.
Cf. also A336700, A368989.

Programs

  • Mathematica
    Prime[#]^#&/@MersennePrimeExponent[Range[8]] (* Harvey P. Dale, Mar 15 2024 *)

Formula

a(n) = A062457(A000043(n)).
A323244(a(n)) = 0.
a(n) = A005940(1+A000396(n)). [Provided no odd perfect numbers exist]

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

A336701 Numbers k for which A000265(1+A000265(sigma(k))) is equal to A000265(1+k).

Original entry on oeis.org

1, 3, 7, 15, 31, 127, 1023, 8191, 34335, 57855, 131071, 524287, 2147483647
Offset: 1

Views

Author

Antti Karttunen, Aug 02 2020

Keywords

Comments

Numbers k such that A336698(k) [= A000265(1+A161942(k))] is equal to A000265(1+k).
Numbers k such that A337194(k) = 2^e * A000265(1+k), for some e >= 1, where that e = A337195(k).
Any odd perfect number would trivially satisfy this condition.
Also, all hypothetical quasiperfect numbers, numbers k that satisfy sigma(k) = 2k+1, would be members.
Question: Is A066175 a subsequence of this sequence?
From Antti Karttunen, Aug 23 2020: (Start)
Numbers k such that (1+k) = 2^e * A336698(k), for some e >= 0.
Thus numbers k such that for some e >= 0, (1+k) = 2^(e-A337195(k)) * A337194(k), or equally, that A337194(k) = 2^(A337195(k)-e) * (1+k).
Conjecture: There are no even terms. This is equivalent to claim that there are no k such that A336698(k) = 1+k: If we assume that k is even, then in above equations we set e=0, and the requirement will then become that A337194(k) = 2^A337195(k)*(1+k), thus 1+k = A336698(k) = A000265(1+A000265(sigma(k))).
(End)

Crossrefs

Subsequence of A336700.
Cf. A000668 (a subsequence).
See also comments in A326042, A332223.

Programs

  • Mathematica
    Block[{f}, f[n_] := n/2^IntegerExponent[n, 2]; Select[Range[2^20], f[1 + f[DivisorSigma[1, #]]] == f[1 + #] &] ] (* Michael De Vlieger, Aug 22 2020 *)
  • PARI
    A000265(n)  = (n>>valuation(n,2));
    isA336701(n) = (A000265(1+A000265(sigma(n))) == A000265(1+n));

A348742 Odd numbers k for which A161942(k) >= k, where A161942 is the odd part of sigma.

Original entry on oeis.org

1, 9, 25, 49, 81, 121, 169, 225, 289, 361, 441, 529, 625, 729, 841, 961, 1089, 1225, 1369, 1521, 1681, 1849, 2025, 2205, 2209, 2401, 2601, 2809, 3025, 3249, 3481, 3721, 3969, 4225, 4489, 4761, 5041, 5329, 5625, 5929, 6241, 6561, 6889, 7225, 7569, 7921, 8281, 8649, 9025, 9409, 9801, 10201, 10609, 11025, 11449, 11881, 12321
Offset: 1

Views

Author

Antti Karttunen, Nov 02 2021

Keywords

Comments

All odd squares (A016754) are present, but not all terms are squares. A348743 gives the nonsquare terms.
Odd terms of A336702 form a subsequence. Also all odd terms of A005820 would be present here, as well as any hypothetical quasi-perfect numbers (see comments and references in A332223, A336700), both in A016754. - Antti Karttunen, Nov 28 2024

Crossrefs

Union of A016754 and A348743.
Cf. A161942, A162284 (subsequence), A336702, A348741 (complement among the odd numbers).

Programs

  • Maple
    q:= n-> (t-> is(t/2^padic[ordp](t,2)>=n))(numtheory[sigma](n)):
    select(q, [2*i-1$i=1..10000])[];  # Alois P. Heinz, Nov 28 2024
  • Mathematica
    odd[n_] := n/2^IntegerExponent[n, 2]; Select[Range[1, 10^4, 2], odd[DivisorSigma[1, #]] >= # &] (* Amiram Eldar, Nov 02 2021, edited (because of the changed definition) by Antti Karttunen, Nov 28 2024 *)
  • PARI
    A000265(n) = (n >> valuation(n, 2));
    isA348742(n) = ((n%2)&&A000265(sigma(n))>=n); \\ revised by Antti Karttunen, Nov 28 2024

Extensions

a(1) = 1 inserted as the initial term, because of the changed definition (from > to >=) - Antti Karttunen, Nov 28 2024

A336699 a(n) = A000265(1+A000265(sigma(A000265(n)))), where A000265(k) gives the odd part of k, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 5, 7, 3, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 5, 1, 7, 5, 3, 1, 1, 11, 1, 3, 1, 5, 1, 1, 1, 29, 1, 5, 1, 7, 3, 5, 1, 3, 1, 1, 1, 1, 1, 7, 1, 11, 1, 9, 5, 1, 1, 5, 7, 19, 5, 1, 3, 1, 1, 3, 1, 61, 11, 11, 1, 7, 3, 1, 1, 23, 5, 1, 1, 1, 1, 1, 1, 25, 29, 5, 1, 13, 5, 7, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 02 2020

Keywords

Comments

See the "lacunae" in the scatter plot. - Antti Karttunen, Mar 27 2022

Crossrefs

Programs

Formula

a(n) = A000265(1+A000265(A000593(n))) = A000265(1+A161942(A000265(n))).
a(n) = A336698(A000265(n)).
From Antti Karttunen, Mar 27 2022: (Start)
a(n) = A351565(A000593(n)).
[The following formulas were discovered by Sequence Machine]:
a(n) = A351565(A002131(n)) = A000265(1+A000265(A002131(n))).
a(n) = A336698(1+A322250(n)).
a(n) = A171435(A000593(n)+A082903(n)).
(End)

A336697 a(n) = gcd((s+1)/2, 1+sigma(s)), where s is the n-th odd square, (2n-1)^2.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Antti Karttunen, Aug 18 2020

Keywords

Comments

See comments in A336700, A337337, and A337339.

Crossrefs

Programs

  • Mathematica
    Array[GCD[(# + 1)/2, 1 + DivisorSigma[1, #]] &[(2 # - 1)^2] &, 120] (* Michael De Vlieger, Aug 24 2020 *)
  • PARI
    A336697(n) = { my(s=((n+n-1)^2)); gcd((s+1)/2,1+sigma(s)); };

Formula

For all n >= 1, a(A048673(n)) = A337337(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)); };

A353365 Numbers k such that the odd part of sigma(sigma(k)) is equal to the odd part of sigma(k).

Original entry on oeis.org

1, 5, 12, 427, 9120, 9180, 9504, 9720, 9960, 10296, 10620, 10740, 10824, 11070, 11310, 11480, 11484, 11556, 11628, 11748, 11934, 11960, 12024, 12036, 12072, 12084, 12376, 12460, 12510, 12570, 12640, 12924, 12980, 13000, 13216, 13340, 13554, 13804, 13806, 13962, 13984, 14022, 14056, 14094, 14178, 14212, 14336, 14380
Offset: 1

Views

Author

Antti Karttunen, Apr 17 2022

Keywords

Comments

Numbers k such that sigma(sigma(k)) = 2^e * sigma(k), for some e >= 0.
Numbers k such that sigma(k) is in A336702.
Numbers k for which A000265(A051027(k)) = A161942(k).
If there existed any hypothetical 3-perfect number (A005820) of the form x = 4u+2 and not divisible by 3, then x would be also included in this sequence, as then sigma(sigma(x)) = 12*x = 4*sigma(x). Such x would be also a term of A349745 and of A351458, and x/2 would be a rare odd term of A000396, and also in A336702. See also the diagram in A347392.

Crossrefs

Programs

A379505 a(n) is the number of ways of partitioning the divisors of n into two disjoint sets with equal sum, when an extra 1-divisor is added to the divisor set, and the two 1-divisors are considered indistinguishable.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 5, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 07 2025

Keywords

Comments

Conjecture 1: For all n >= 1, a(A156942(n)) > 0. Also, if a(A156942(n)) > 1 is true for all n, it would imply that there are no quasiperfect numbers, numbers x with sigma(x) = 2x+1, as such numbers must all reside in A156942 and have a(x) = 1. (See references in A336700).
Conjecture 2: a(n) = 1 if and only if n = 2^k, with k >= 0. This claim is equal to the statement that there are neither quasiperfect numbers nor almost perfect (least deficient) numbers, numbers x with sigma(x) = 2x-1, others than those given by A000079.

Examples

			a(18) = 2 as its divisor set with an extra 1 is [1, 1, 2, 3, 6, 9, 18], and this can be partitioned to two sets with equal sums either as 1+1+3+6+9 = 2+18 or as 2+3+6+9 = 1+1+18.
a(36) = 5 as its divisor set with an extra 1 is [1, 1, 2, 3, 4, 6, 9, 12, 18, 36], and this can be partitioned in any of the following five ways, when two 1's are considered indistinguishable:
  1+1+2+6+36  = 3+4+9+12+18,
  1+2+3+4+36  = 1+6+9+12+18,
  1+3+6+36    = 1+2+4+9+12+18,
  1+9+36      = 1+2+3+4+6+12+18,
  4+6+36      = 1+1+2+3+9+12+18,
where each sum on the left and right hand side gives (sigma(36)+1)/2 = 46.
There are 42 partitions of (sigma(72)+1)/2 = 98 into the divisors of 72, [1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72], with an extra 1 allowed:
  [2, 24, 72],
  [1, 1, 24, 72],
  [8, 18, 72],
  [2, 6, 18, 72],
  [1, 1, 6, 18, 72],
  [1, 3, 4, 18, 72],
  [1, 1, 2, 4, 18, 72],
  [1, 4, 9, 12, 72],
  [2, 3, 9, 12, 72],
  [1, 1, 3, 9, 12, 72],
  [6, 8, 12, 72],
  [2, 4, 8, 12, 72],
  [1, 1, 4, 8, 12, 72],
  [1, 2, 3, 8, 12, 72],
  [1, 3, 4, 6, 12, 72],
  [1, 1, 2, 4, 6, 12, 72],
  [3, 6, 8, 9, 72],
  [1, 2, 6, 8, 9, 72],
  [2, 3, 4, 8, 9, 72],
  [1, 1, 3, 4, 8, 9, 72],
  [1, 1, 2, 3, 4, 6, 9, 72],
  [8, 12, 18, 24, 36],
  [2, 6, 12, 18, 24, 36],
  [1, 1, 6, 12, 18, 24, 36],
  [1, 3, 4, 12, 18, 24, 36],
  [1, 1, 2, 4, 12, 18, 24, 36],
  [3, 8, 9, 18, 24, 36],
  [1, 2, 8, 9, 18, 24, 36],
  [1, 4, 6, 9, 18, 24, 36],
  [2, 3, 6, 9, 18, 24, 36],
  [1, 1, 3, 6, 9, 18, 24, 36],
  [1, 1, 2, 3, 4, 9, 18, 24, 36],
  [2, 4, 6, 8, 18, 24, 36],
  [1, 1, 4, 6, 8, 18, 24, 36],
  [1, 2, 3, 6, 8, 18, 24, 36],
  [3, 6, 8, 9, 12, 24, 36],
  [1, 2, 6, 8, 9, 12, 24, 36],
  [2, 3, 4, 8, 9, 12, 24, 36],
  [1, 1, 3, 4, 8, 9, 12, 24, 36],
  [1, 1, 2, 3, 4, 6, 9, 12, 24, 36],
  [2, 3, 4, 6, 8, 9, 12, 18, 36],
  [1, 1, 3, 4, 6, 8, 9, 12, 18, 36],
therefore a(72) = 42/2 = 21.
		

Crossrefs

Cf. A083206, A103977, A156942, A336700, A379502, A379503 (positions of nonzero terms), A379504 (version where two 1's are considered distinct).
Cf. A000079 (conjectured to give the positions of 1's).

Programs

  • PARI
    partitions_into_distinct_parts_with_extra1allowed(n, parts, from=1) = if(n<=1, 1, if(from>#parts, 0, my(s=0); for(i=from, #parts, if(parts[i]<=n, s += partitions_into_distinct_parts_with_extra1allowed(n-parts[i], parts, i+1))); (s)));
    A379505(n) = if(1==n, n, if(!issquare(n) && !issquare(2*n), 0, my(divs=divisors(n), s=sigma(n)); partitions_into_distinct_parts_with_extra1allowed((s+1)/2, vecsort(divs,,4))/2));

Formula

a(n) <= A379504(n).
A103977(n) = 1 <=> a(n) > 0.
Showing 1-10 of 19 results. Next