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.

Previous Showing 11-20 of 46 results. Next

A372571 Lexicographically earliest infinite sequence such that a(i) = a(j) => A324198(i) = A324198(j) and A342671(i) = A342671(j), for all i, j >= 1.

Original entry on oeis.org

1, 2, 3, 1, 1, 2, 1, 2, 3, 4, 1, 1, 1, 2, 5, 1, 1, 2, 1, 6, 3, 2, 1, 7, 8, 2, 9, 1, 1, 2, 1, 10, 3, 2, 11, 1, 1, 2, 3, 12, 1, 13, 1, 2, 5, 2, 1, 1, 11, 14, 3, 1, 1, 7, 15, 13, 9, 2, 1, 16, 1, 2, 17, 1, 18, 2, 1, 10, 3, 19, 1, 7, 1, 2, 20, 1, 11, 2, 1, 4, 3, 2, 1, 11, 15, 2, 9, 10, 1, 2, 11, 2, 3, 2, 1, 10, 1, 21, 22, 23, 1, 2, 1, 2
Offset: 1

Views

Author

Antti Karttunen, May 24 2024

Keywords

Comments

Restricted growth sequence transform of the ordered pair [A324198(n), A342671(n)].

Crossrefs

Programs

  • PARI
    up_to = 65537;
    rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; };
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A324198(n) = { my(m=1, p=2, orgn=n); while(n, m *= (p^min(n%p, valuation(orgn, p))); n = n\p; p = nextprime(1+p)); (m); };
    Aux372571(n) = [A324198(n), gcd(sigma(n), A003961(n))];
    v372571 = rgs_transform(vector(up_to, n, Aux372571(n)));
    A372571(n) = v372571[n];

A246278 Prime shift array: Square array read by antidiagonals: A(1,col) = 2*col, and for row > 1, A(row,col) = A003961(A(row-1,col)).

Original entry on oeis.org

2, 4, 3, 6, 9, 5, 8, 15, 25, 7, 10, 27, 35, 49, 11, 12, 21, 125, 77, 121, 13, 14, 45, 55, 343, 143, 169, 17, 16, 33, 175, 91, 1331, 221, 289, 19, 18, 81, 65, 539, 187, 2197, 323, 361, 23, 20, 75, 625, 119, 1573, 247, 4913, 437, 529, 29, 22, 63, 245, 2401, 209, 2873, 391, 6859, 667, 841, 31
Offset: 2

Views

Author

Antti Karttunen, Aug 21 2014

Keywords

Comments

The array is read by antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc.
This array can be obtained by taking every second column from array A242378, starting from its column 2.
Permutation of natural numbers larger than 1.
The terms on row n are all divisible by n-th prime, A000040(n).
Each column is strictly growing, and the terms in the same column have the same prime signature.
A055396(n) gives the row number of row where n occurs,
and A246277(n) gives its column number, both starting from 1.
From Antti Karttunen, Jan 03 2015: (Start)
A252759(n) gives their sum minus one, i.e. the Manhattan distance of n from the top left corner.
If we assume here that a(1) = 1 (but which is not explicitly included because outside of the array), then A252752 gives the inverse permutation. See also A246276.
(End)

Examples

			The top left corner of the array:
   2,     4,     6,     8,    10,    12,    14,    16,    18, ...
   3,     9,    15,    27,    21,    45,    33,    81,    75, ...
   5,    25,    35,   125,    55,   175,    65,   625,   245, ...
   7,    49,    77,   343,    91,   539,   119,  2401,   847, ...
  11,   121,   143,  1331,   187,  1573,   209, 14641,  1859, ...
  13,   169,   221,  2197,   247,  2873,   299, 28561,  3757, ...
		

Crossrefs

First row: A005843 (the even numbers), from 2 onward.
Row 2: A249734, Row 3: A249827.
Column 1: A000040 (primes), Column 2: A001248 (squares of primes), Column 3: A006094 (products of two successive primes), Column 4: A030078 (cubes of primes).
Transpose: A246279.
Inverse permutation: A252752.
One more than A246275.
Arrays obtained by applying a particular function (given in parentheses) to the entries of this array. Cases where the columns grow monotonically are indicated with *: A249822 (A078898), A253551 (* A156552), A253561 (* A122111), A341605 (A017665), A341606 (A017666), A341607 (A006530 o A017666), A341608 (A341524), A341626 (A341526), A341627 (A341527), A341628 (A006530 o A341527), A342674 (A341530), A344027 (* A003415, arithmetic derivative), A355924 (A342671), A355925 (A009194), A355926 (A355442), A355927 (* sigma), A356155 (* A258851), A372562 (A252748), A372563 (A286385), A378979 (* deficiency, A033879), A379008 (* (probably), A294898), A379010 (* A000010, Euler phi), A379011 (* A083254).
Cf. A329050 (subtable).

Programs

  • Mathematica
    f[p_?PrimeQ] := f[p] = Prime[PrimePi@ p + 1]; f[1] = 1; f[n_] := f[n] = Times @@ (f[First@ #]^Last@ # &) /@ FactorInteger@ n; Block[{lim = 12}, Table[#[[n - k, k]], {n, 2, lim}, {k, n - 1, 1, -1}] &@ NestList[Map[f, #] &, Table[2 k, {k, lim}], lim]] // Flatten (* Michael De Vlieger, Jan 04 2016, after Jean-François Alcover at A003961 *)
  • Scheme
    (define (A246278 n) (if (<= n 1) n (A246278bi (A002260 (- n 1)) (A004736 (- n 1))))) ;; Square array starts with offset=2, and we have also tacitly defined a(1) = 1 here.
    (define (A246278bi row col) (if (= 1 row) (* 2 col) (A003961 (A246278bi (- row 1) col))))

Formula

A(1,col) = 2*col, and for row > 1, A(row,col) = A003961(A(row-1,col)).
As a composition of other similar sequences:
a(n) = A122111(A253561(n)).
a(n) = A249818(A083221(n)).
For all n >= 1, a(n+1) = A005940(1+A253551(n)).
A(n, k) = A341606(n, k) * A355925(n, k). - Antti Karttunen, Jul 22 2022

Extensions

Starting offset of the linear sequence changed from 1 to 2, without affecting the column and row indices by Antti Karttunen, Jan 03 2015

A341529 a(n) = sigma(n) * A003961(n), where A003961 shifts the prime factorization of n one step towards larger primes, and sigma is the sum of the divisors of n.

Original entry on oeis.org

1, 9, 20, 63, 42, 180, 88, 405, 325, 378, 156, 1260, 238, 792, 840, 2511, 342, 2925, 460, 2646, 1760, 1404, 696, 8100, 1519, 2142, 5000, 5544, 930, 7560, 1184, 15309, 3120, 3078, 3696, 20475, 1558, 4140, 4760, 17010, 1806, 15840, 2068, 9828, 13650, 6264, 2544, 50220, 6897, 13671, 6840, 14994, 3186, 45000, 6552, 35640
Offset: 1

Views

Author

Antti Karttunen, Feb 16 2021

Keywords

Comments

Question: Does the maximum value of ratio A341529(n)/A341528(n) stay below 2?
From Amiram Eldar and Antti Karttunen, Jan 28 2023: (Start)
Answer to the above question is yes: Sup_{n>=1} A341529(n)/A341528(n) = 2.
Proof:
f(n) = A341529(n)/A341528(n) is a multiplicative function with f(p^e) = (1 + 1/p + ... + 1/p^e)/(1 + 1/q + ... + 1/q^e), where q = nextprime(p).
First we prove a lemma which states that f(p^(1+e)) / f(p^e) > 1, for any prime p, and exponent e.
We note that (sigma(p^(1+e))/(p^(1+e))) / (sigma(p^e)/(p^e)) = (sigma(p^(1+e))/(p*sigma(p^e))) = sigma(p^(1+e)) / (sigma(p^(1+e)) - 1), so setting q = nextprime(p), we can write the ratio f(p^(1+e)) / f(p^e) as (sigma(p^(1+e))/(sigma(p^(1+e))-1)) / (sigma(q^(1+e))/(sigma(q^(1+e))-1)), and to prove this to be > 1, we just note that the denominator is less than the numerator, because sigma(p^e) is monotonically growing with respect to the increasing prime p.
Since q > p, we have f(p^e) > 1 for all p and all e>=1, and together with the above lemma this shows that f(n) <= f(n*m) for all m>=1.
Suppose n = Product_i p_i^e_i, and let pmax = max(p_i), emax = max(e_i), so n is a divisor of m = (pmax#)^emax, and f(n) < f(m), where p# = 2 * 3 * ... * p is the primorial of p, A034386(p).
Then f(m) = f(2^emax) * f(3^emax) * ... * f(pmax^emax) = (1 + 1/2 + ... + 1/2^emax)/(1 + 1/3 + ... + 1/3^emax)) * (1 + 1/3 + ... + 1/3^emax)/(1 + 1/5 + ... + 1/5^emax)) * ... * (1 + 1/p + ... + 1/p^emax)/(1 + 1/q + ... + 1/q^emax))[telescoping product] = (1 + 1/2 + ... + 1/2^emax)/(1 + 1/qmax + ... + 1/qmax^emax) <= (1 + 1/2 + ... + 1/2^emax) < 2, where qmax = nextprime(pmax).
So we have f(n) < 2 for all n.
To prove that 2 is the supremum, we have lim_{e,k -> oo) f(prime(k)#^e) = 2.
(End)

Crossrefs

Programs

  • Mathematica
    Array[DivisorSigma[1, #]*Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e > 0 :> {Prime[PrimePi@ p + 1], e}] - Boole[# == 1] &, 56] (* Michael De Vlieger, Feb 22 2021 *)
  • PARI
    A003961(n) = { my(f=factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
    A341529(n) = (sigma(n)*A003961(n));

Formula

Multiplicative with a(p^e) = q^e * (p^(e+1)-1)/(p-1), where q = nextPrime(p).
a(n) = A000203(n) * A003961(n).
For all n > 1, a(n) > A341528(n).
For all n >= 1, A072861(n) <= a(n) <= A003961(n)^2. [See A286385].
a(n) = A341528(n) + A341512(n) = A342671(n) * A342672(n) = A342661(A003961(n)). - Antti Karttunen, Mar 22 2021
Sum_{k=1..n} a(k) ~ c * n^3, where c = (1/3) * Product_{p prime} p^4*(p-1)/((p^3-nextprime(p))*(p^2-nextprime(p))) = 3.0664809..., where nextprime is A151800. - Amiram Eldar, Dec 08 2022

A349169 Numbers k such that k * gcd(sigma(k), A003961(k)) is equal to the odd part of {sigma(k) * gcd(k, A003961(k))}, where A003961 shifts the prime factorization one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 15, 105, 3003, 3465, 13923, 45045, 264537, 459459, 745875, 1541475, 5221125, 8729721, 10790325, 14171625, 29288025, 34563375, 57034575, 71430975, 99201375, 109643625, 144729585, 205016175, 255835125, 295708875, 356080725, 399242025, 419159475, 449323875, 928602675, 939495375, 1083656925, 1941623775, 1962350685, 2083228875
Offset: 1

Views

Author

Antti Karttunen, Nov 10 2021

Keywords

Comments

Numbers k such that A348990(k) [= k/gcd(k, A003961(k))] is equal to A348992(k), which is the odd part of A349162(k), thus all terms must be odd, as A348990 preserves the parity of its argument.
Equally, numbers k for which gcd(A064987(k), A191002(k)) is equal to A000265(gcd(A064987(k), A341529(k))).
Also odd numbers k for which A348993(k) = A319627(k).
Odd terms of A336702 are given by the intersection of this sequence and A349174.
Conjectures:
(1) After 1, all terms are multiples of 3. (Why?)
(2) After 1, all terms are in A104210, in other words, for all n > 1, gcd(a(n), A003961(a(n))) > 1. Note that if we encountered a term k with gcd(k, A003961(k)) = 1, then we would have discovered an odd multiperfect number.
(3) Apart from 1, 15, 105, 3003, 13923, 264537, all other terms are abundant. [These apparently are also the only terms that are not Zumkeller, A083207. Note added Dec 05 2024]
(4) After 1, all terms are in A248150. (Cf. also A386430).
(5) After 1, all terms are in A348748.
(6) Apart from 1, there are no common terms with A349753.
Note: If any of the last four conjectures could be proved, it would refute the existence of odd perfect numbers at once. Note that it seems that gcd(sigma(k), A003961(k)) < k, for all k except these four: 1, 2, 20, 160.
Questions:
(1) For any term x here, can 2*x be in A349745? (Partial answer: at least x should be in A191218 and should not be a multiple of 3). Would this then imply that x is an odd perfect number? (Which could explain the points (1) and (4) in above, assuming the nonexistence of opn's).

Crossrefs

Programs

  • Mathematica
    Select[Range[10^6], #1/GCD[#1, #3] == #2/(2^IntegerExponent[#2, 2]*GCD[#2, #3]) & @@ {#, DivisorSigma[1, #], Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]} &] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A000265(n) = (n >> valuation(n, 2));
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    isA349169(n) = { my(s=sigma(n),u=A003961(n)); (n*gcd(s,u) == A000265(s)*gcd(n,u)); }; \\ (Program simplified Nov 30 2021)

Formula

For all n >= 1, A007949(A000203(a(n))) = A007949(a(n)). [sigma preserves the 3-adic valuation of the terms of this sequence] - Antti Karttunen, Nov 29 2021

Extensions

Name changed and comment section rewritten by Antti Karttunen, Nov 29 2021

A349161 a(n) = A003961(n) / gcd(sigma(n), A003961(n)), where A003961 shifts the prime factorization of n one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 5, 9, 7, 5, 11, 9, 25, 7, 13, 45, 17, 11, 35, 81, 19, 25, 23, 3, 55, 13, 29, 9, 49, 17, 25, 99, 31, 35, 37, 27, 65, 19, 77, 225, 41, 23, 85, 21, 43, 55, 47, 39, 175, 29, 53, 405, 121, 49, 95, 153, 59, 25, 91, 99, 23, 31, 61, 15, 67, 37, 275, 729, 17, 65, 71, 19, 145, 77, 73, 45, 79, 41, 245, 207, 143, 85, 83
Offset: 1

Views

Author

Antti Karttunen, Nov 09 2021

Keywords

Comments

Numerator of ratio A003961(n) / A000203(n). Sequence A349162 gives the denominators.
Numerator of ratio A003961(n) / A161942(n). Sequence A348992 gives the denominators.
Both ratios are multiplicative because the constituent sequences are.
No 1's occur as terms after a(2), because for n > 2, sigma(n) < A003961(n). (See A286385).

Crossrefs

Programs

  • Mathematica
    Array[#2/GCD[##] & @@ {DivisorSigma[1, #], If[# == 1, 1, Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]]} &, 79] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A349161(n) = { my(u=A003961(n)); (u/gcd(u,sigma(n))); };
    
  • Python
    from math import prod, gcd
    from sympy import nextprime, factorint
    def A349161(n):
        f = factorint(n).items()
        a = prod(nextprime(p)**e for p, e in f)
        b = prod((p**(e+1)-1)//(p-1) for p, e in f)
        return a//gcd(a,b) # Chai Wah Wu, Mar 17 2023

Formula

a(n) = A003961(n) / A342671(n) = A003961(n) / gcd(A000203(n), A003961(n)).
a(n) = A003961(A349164(n)).

A349162 a(n) = sigma(n) / gcd(sigma(n), A003961(n)), where A003961 shifts the prime factorization of n one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 4, 7, 6, 4, 8, 5, 13, 6, 12, 28, 14, 8, 24, 31, 18, 13, 20, 2, 32, 12, 24, 4, 31, 14, 8, 56, 30, 24, 32, 7, 48, 18, 48, 91, 38, 20, 56, 10, 42, 32, 44, 28, 78, 24, 48, 124, 57, 31, 72, 98, 54, 8, 72, 40, 16, 30, 60, 8, 62, 32, 104, 127, 12, 48, 68, 14, 96, 48, 72, 13, 74, 38, 124, 140, 96, 56, 80, 62, 121, 42
Offset: 1

Views

Author

Antti Karttunen, Nov 09 2021

Keywords

Comments

Denominator of ratio A003961(n) / A000203(n).
Small values are rare, but are not limited to the beginning. For example in range 1 .. 2^25, a(n) = 4 at n = 3, 6, 24, 792, 2720, 122944, 31307472.
Question: Would it be possible to prove that a(n) > 1 for all n > 2?
Obviously, 1's may occur only on squares & twice squares (A028982). See also comments in A350072. - Antti Karttunen, Feb 16 2022

Crossrefs

Cf. A000203, A003961, A028982 (positions of odd terms), A319630, A336702, A342671, A348992 (the odd part), A348993, A349161 (numerators), A349163, A349164, A349627, A349628, A350072 [= a(n^2)].
Cf. also A349745, A351551, A351554.

Programs

  • Mathematica
    Array[#1/GCD[##] & @@ {DivisorSigma[1, #], If[# == 1, 1, Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]]} &, 82] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A349162(n) = { my(s=sigma(n)); (s/gcd(s,A003961(n))); };

Formula

a(n) = A000203(n) / A342671(n) = A000203(n) / gcd(A000203(n), A003961(n)).

A349745 Numbers k for which k * gcd(sigma(k), A003961(k)) is equal to sigma(k) * gcd(k, A003961(k)), where A003961 shifts the prime factorization one step towards larger primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 120, 216, 672, 2464, 22176, 228480, 523776, 640640, 837760, 5581440, 5765760, 7539840, 12999168, 19603584, 33860736, 38342304, 71344000, 95472000, 102136320, 197308800, 220093440, 345080736, 459818240, 807009280, 975576960, 1476304896, 1510831360, 1773584640
Offset: 1

Views

Author

Antti Karttunen, Nov 29 2021

Keywords

Comments

Numbers k for which k * A342671(k) = A000203(k) * A322361(k).
Numbers k such that gcd(A064987(k), A191002(k)) = gcd(A064987(k), A341529(k)).
Obviously, all odd terms in this sequence must be squares.
All the terms k of A005820 that satisfy A007949(k) < A007814(k) [i.e., whose 3-adic valuation is strictly less than their 2-adic valuation] are also terms of this sequence. Incidentally, the first six known terms of A005820 satisfy this condition, while on the other hand, any hypothetical odd 3-perfect number would be excluded from this sequence. Also, as a corollary, any hypothetical 3-perfect numbers of the form 4u+2 must not be multiples of 3 if they are to appear here. Similarly for any k which occurs in A349169, for 2*k to occur in this sequence, it shouldn't be a multiple of 3 and k should also be a term of A191218. See question 2 and its partial answer in A349169.
From Antti Karttunen, Feb 13-20 2022: (Start)
Question: Are all terms/2 (A351548) abundant, from n > 1 onward?
Note that of the 65 known 5-multiperfect numbers, all others except these three 1245087725796543283200, 1940351499647188992000, 4010059765937523916800 are also included in this sequence. The three exceptions are distinguished by the fact that their 3 and 5-adic valuations are equal. In 62 others the former is larger.
If k satisfying the condition were of the form 4u+2, then it should be one of the terms of A191218 doubled as only then both k and sigma(k) are of the form 4u+2, with equal 2-adic valuations for both. More precisely, one of the terms of A351538.
(End)

Crossrefs

Cf. also A349169, A349746, A351458, A351549 for other variants.
Subsequence of A351554 and also of its subsequence A351551.
Cf. A351459 (subsequence, intersection with A351458), A351548 (terms halved).

Programs

  • Mathematica
    f1[p_, e_] := (p^(e + 1) - 1)/(p - 1); f2[p_, e_] := NextPrime[p]^e; q[1] = True; q[n_] := n * GCD[(s = Times @@ f1 @@@ (f = FactorInteger[n])), (r = Times @@ f2 @@@ f)] == s*GCD[n, r]; Select[Range[10^6], q] (* Amiram Eldar, Nov 29 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    isA349745(n) = { my(s=sigma(n),u=A003961(n)); (n*gcd(s,u) == (s*gcd(n,u))); };

Formula

For all n >= 1, A007814(A000203(a(n))) = A007814(a(n)). [sigma preserves the 2-adic valuation of the terms of this sequence]

A349164 a(n) = A064989(A003961(n) / gcd(sigma(n), A003961(n))), where A003961 shifts the prime factorization of n one step towards larger primes, while A064989 shifts it back towards smaller primes, and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 3, 4, 5, 3, 7, 4, 9, 5, 11, 12, 13, 7, 15, 16, 17, 9, 19, 2, 21, 11, 23, 4, 25, 13, 9, 28, 29, 15, 31, 8, 33, 17, 35, 36, 37, 19, 39, 10, 41, 21, 43, 22, 45, 23, 47, 48, 49, 25, 51, 52, 53, 9, 55, 28, 19, 29, 59, 6, 61, 31, 63, 64, 13, 33, 67, 17, 69, 35, 71, 12, 73, 37, 75, 76, 77, 39, 79, 40, 81, 41, 83, 84
Offset: 1

Views

Author

Antti Karttunen, Nov 09 2021

Keywords

Crossrefs

Cf. A349144 and A349168 [positions where a(n) is / is not relatively prime with A349163(n) = n/a(n)].

Programs

  • Mathematica
    Array[Times @@ Map[If[#1 <= 2, 1, NextPrime[#1, -1]]^#2 & @@ # &, FactorInteger[#2/GCD[##]]] & @@ {DivisorSigma[1, #], Times @@ Map[NextPrime[#1]^#2 & @@ # &, FactorInteger[#]]} &, 84] (* Michael De Vlieger, Nov 11 2021 *)
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A064989(n) = { my(f=factor(n)); if((n>1 && f[1,1]==2), f[1,2] = 0); for(i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f); };
    A349164(n) = { my(u=A003961(n)); A064989(u/gcd(u,sigma(n))); };

Formula

a(n) = A064989(A349161(n)).
a(n) = n / A349163(n).

A349166 Numbers k such that sigma(k) and A003961(k) share a prime factor, where A003961(n) is fully multiplicative function with a(prime(k)) = prime(k+1).

Original entry on oeis.org

2, 6, 8, 10, 14, 18, 20, 22, 24, 26, 27, 30, 32, 34, 38, 40, 42, 44, 46, 50, 54, 56, 57, 58, 60, 62, 65, 66, 68, 70, 72, 74, 78, 80, 82, 86, 87, 88, 90, 92, 94, 96, 98, 99, 100, 102, 104, 106, 108, 110, 114, 116, 118, 120, 122, 126, 128, 130, 132, 134, 135, 136, 138, 140, 142, 146, 150, 152, 154, 158, 160, 162, 164
Offset: 1

Views

Author

Antti Karttunen, Nov 09 2021

Keywords

Comments

The only prime term is 2. A prime power prime(j)^k with k > 1 is a term if and only if k+1 is divisible by the multiplicative order of prime(j) mod prime(j+1). - Robert Israel, May 22 2025

Examples

			For n = 2, A000203(2) = A003961(2) = 3, therefore they share a prime factor 3, and 2 is included in this sequence.
For n = 10 = 2*5, sigma(10) = 18 = 2 * 3^2, while A003961(10) = 21 = 3*7, therefore 10 is included, as there is a shared prime factor (3).
		

Crossrefs

Positions of terms larger than ones in A342671, and also in A349163.
Positions of zeros in A349167.
Cf. A349165 (complement), A349168 (subsequence).

Programs

  • Maple
    filter:= proc(n) local F,a,b,t;
       F:= ifactors(n)[2];
       b:= convert(map(nextprime,F[..,1]),`*`);
       a:= mul((t[1]^(t[2]+1)-1)/(t[1]-1),t=F);
       igcd(a,b) <> 1
    end proc;
    select(filter, [$1..1000]); # Robert Israel, May 21 2025
  • PARI
    A003961(n) = { my(f = factor(n)); for (i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    isA349166(n) = (1!=gcd(sigma(n), A003961(n)));

A351546 a(n) is the largest unitary divisor of sigma(n) coprime with A003961(n), where A003961 is fully multiplicative with a(p) = nextprime(p), and sigma is the sum of divisors function.

Original entry on oeis.org

1, 1, 4, 7, 6, 4, 8, 5, 13, 2, 12, 28, 14, 8, 24, 31, 18, 13, 20, 2, 32, 4, 24, 4, 31, 14, 8, 56, 30, 8, 32, 7, 48, 2, 48, 91, 38, 20, 56, 10, 42, 32, 44, 28, 78, 8, 48, 124, 57, 31, 72, 98, 54, 8, 72, 40, 16, 10, 60, 8, 62, 32, 104, 127, 12, 16, 68, 14, 96, 16, 72, 13, 74, 38, 124, 140, 96, 56, 80, 62, 121, 14, 84
Offset: 1

Views

Author

Antti Karttunen, Feb 16 2022

Keywords

Examples

			For n = 672 = 2^5 * 3^1 * 7^1, and the largest unitary divisor of the sigma(672) [= 2^5 * 3^2 * 7^1] coprime with A003961(672) = 13365 = 3^5 * 5^1 * 11^1 is 2^5 * 7^1 = 224, therefore a(672) = 224.
		

Crossrefs

Programs

  • PARI
    A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
    A351546(n) = { my(f=factor(sigma(n)),u=A003961(n)); prod(k=1,#f~,f[k,1]^((0!=(u%f[k,1]))*f[k,2])); };

Formula

a(n) = Product_{p^e || A000203(n)} p^(e*[p does not divide A003961(n)]), where [ ] is the Iverson bracket, returning 0 if p is a divisor of A003961(n), and 1 otherwise. Here p^e is the largest power of each prime p dividing sigma(n).
a(n) = A000203(n) / A351544(n).
a(n) = A353666(n) * A353668(n) = A351547(n) / A354997(n). - Antti Karttunen, Jul 09 2022
Previous Showing 11-20 of 46 results. Next