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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A366825 Numbers of the form p^2 * m, squarefree m > 1, prime p < lpf(m), where lpf(m) = A020639(m).

Original entry on oeis.org

12, 20, 28, 44, 45, 52, 60, 63, 68, 76, 84, 92, 99, 116, 117, 124, 132, 140, 148, 153, 156, 164, 171, 172, 175, 188, 204, 207, 212, 220, 228, 236, 244, 260, 261, 268, 275, 276, 279, 284, 292, 308, 315, 316, 325, 332, 333, 340, 348, 356, 364, 369, 372, 380, 387
Offset: 1

Views

Author

Michael De Vlieger, Dec 15 2023

Keywords

Comments

Proper subset of A126706. Proper subset of A364996.
Prime signature of a(n) is 2 followed by at least one 1.
Numbers of the form A065642(A120944(k)) for some k.
The asymptotic density of this sequence is (6/Pi^2) * Sum_{p prime} (1/p^2) * (Product_{primes q <= p} (q/(q+1))) = 0.155068688392... . - Amiram Eldar, Dec 18 2023

Examples

			a(1) = 12 = 4*3 = p^2 * m, squarefree m > 1; sqrt(4) < lpf(3), i.e., 2 < 3.
a(5) = 45 = 9*5 = p^2 * m, squarefree m > 1; sqrt(9) < lpf(5), i.e., 3 < 5.
Prime powers p^k, k > 2, are not in the sequence since m = p^(k-2) is not squarefree and p = lpf(m).
		

Crossrefs

Programs

  • Mathematica
    Select[Select[Range[500], PrimeOmega[#] > PrimeNu[#] > 1 &], First[#1] == 2 && Union[#2] == {1} & @@ TakeDrop[FactorInteger[#][[All, -1]], 1] &]
  • PARI
    is(n) = {my(e = factor(n)[, 2]); #e > 1 && e[1] == 2 && vecmax(e[2..#e]) == 1;} \\ Amiram Eldar, Dec 18 2023

A285109 a(n) = n multiplied by its smallest prime factor; a(1) = 1.

Original entry on oeis.org

1, 4, 9, 8, 25, 12, 49, 16, 27, 20, 121, 24, 169, 28, 45, 32, 289, 36, 361, 40, 63, 44, 529, 48, 125, 52, 81, 56, 841, 60, 961, 64, 99, 68, 175, 72, 1369, 76, 117, 80, 1681, 84, 1849, 88, 135, 92, 2209, 96, 343, 100, 153, 104, 2809, 108, 275, 112, 171, 116, 3481, 120, 3721, 124, 189, 128, 325, 132, 4489, 136, 207
Offset: 1

Views

Author

Antti Karttunen, Apr 19 2017

Keywords

Crossrefs

Differs from A065642 for the first time at n=12. See A284342 for all the differing points.

Programs

  • Mathematica
    a[n_] := n * FactorInteger[n][[1, 1]]; Array[a, 100] (* Amiram Eldar, Jun 30 2022 *)
  • PARI
    a(n)=if(n==1, 1, n*factor(n)[1,1]); \\ Joerg Arndt, Oct 27 2021
  • Scheme
    (define (A285109 n) (* (A020639 n) n))
    

Formula

a(n) = A020639(n) * n.
Other identities. For all n >= 1:
a(A285100(n)) = A065642(A285100(n)). [Agrees with A065642 on all terms of A285100, but not on any other points.]

A285333 a(n) = A048675(A285332(n)).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 3, 6, 4, 9, 6, 5, 8, 8, 4, 15, 8, 12, 5, 14, 10, 27, 8, 10, 6, 25, 12, 7, 16, 16, 5, 210, 16, 45, 10, 35, 16, 18, 5, 105, 16, 28, 11, 462, 28, 81, 10, 21, 12, 20, 7, 154, 26, 125, 16, 30, 8, 49, 24, 11, 32, 32, 6, 10659, 212, 420, 17, 910, 46, 75, 10, 78, 36, 175, 20, 33, 20, 24, 6, 3094, 106, 315, 18, 385, 32, 56, 17, 780045
Offset: 0

Views

Author

Antti Karttunen, Apr 19 2017

Keywords

Comments

Following A285332, also this sequence can be represented in a form of a binary tree:
0
|
...................1...................
2 2
3......../ \........4 4......../ \........3
/ \ / \ / \ / \
/ \ / \ / \ / \
/ \ / \ / \ / \
6 4 9 6 5 8 8 4
15 8 12 5 14 10 27 8 10 6 25 12 7 16 16 5
etc.

Crossrefs

Cf. A001477, A048675, A109162, A285325, A285330, A285332 (even bisection).

Programs

Formula

a(n) = A048675(A285332(n)).
For all n >= 1, a(2n) = A285332(n).
a(2^n) = A109162(1+n). [The left edge of the tree.]
a(A000225(n)) = n. [The right edge of tree.]

A285325 Square array A(n,k) = A048675(A285321(n,k)), read by descending antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), ...

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 4, 4, 6, 4, 5, 8, 5, 8, 5, 6, 6, 12, 5, 10, 6, 7, 8, 7, 16, 6, 12, 7, 8, 8, 10, 9, 20, 6, 14, 8, 9, 16, 9, 10, 8, 24, 7, 16, 9, 10, 10, 24, 9, 12, 10, 28, 7, 18, 10, 11, 12, 11, 32, 11, 14, 9, 32, 7, 20, 11, 12, 12, 18, 17, 40, 10, 12, 11, 36, 8, 22, 12, 13, 16, 13, 14, 12, 48, 10, 14, 13, 40, 8, 24, 13
Offset: 1

Views

Author

Antti Karttunen, Apr 17 2017

Keywords

Examples

			The top left 15x6 corner of the array:
  1,  2, 3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 13, 14, 15
  2,  4, 4,  8,  6,  8,  8, 16, 10, 12, 12, 16, 14, 16, 16
  3,  6, 5, 12,  7, 10,  9, 24, 11, 18, 13, 20, 15, 18, 17
  4,  8, 5, 16,  9, 10,  9, 32, 17, 14, 13, 20, 17, 22, 17
  5, 10, 6, 20,  8, 12, 11, 40, 12, 20, 14, 24, 21, 18, 19
  6, 12, 6, 24, 10, 14, 10, 48, 18, 16, 19, 28, 16, 20, 18
		

Crossrefs

Row 1 & column 1: A000027.
Row 2: A285326, Row 3: A285327.

Programs

Formula

A(n,k) = A048675(A285321(n,k)).

A285326 a(0) = 0, for n > 0, a(n) = n + A006519(n).

Original entry on oeis.org

0, 2, 4, 4, 8, 6, 8, 8, 16, 10, 12, 12, 16, 14, 16, 16, 32, 18, 20, 20, 24, 22, 24, 24, 32, 26, 28, 28, 32, 30, 32, 32, 64, 34, 36, 36, 40, 38, 40, 40, 48, 42, 44, 44, 48, 46, 48, 48, 64, 50, 52, 52, 56, 54, 56, 56, 64, 58, 60, 60, 64, 62, 64, 64, 128, 66, 68, 68, 72, 70, 72, 72, 80, 74, 76, 76, 80, 78, 80, 80, 96, 82, 84, 84, 88, 86, 88, 88, 96, 90, 92, 92
Offset: 0

Views

Author

Antti Karttunen, Apr 19 2017

Keywords

Comments

From M. F. Hasler, Oct 19 2019: (Start)
This sequence is equal to itself multiplied by 2 and interleaved with the positive even numbers: We have a(2n-1) = 2n (n >= 1) from the very definition, since A006519(m) = 1 for odd m. And a(2n) = 2n + A006519(2n) = 2*a(n), using A006519(2n) = 2*A006519(n).
The sequence repeats the pattern [A, B, C, C] where in the n-th occurrence C = 4n, B = C - 2, A = C if n is even, A = C + 4 if n = 3 (mod 4), and A = 16*a((n-1)/4) otherwise. (End)

Crossrefs

Row 2 of A285325 (after the initial zero).
Cf. A109168 (same terms divided by 2), also A140472.

Programs

Formula

a(0) = 0; for n > 0, a(n) = n + A006519(n).
For n >= 1, a(n) = 2*A109168(n).
a(n) = 2*A140472(n) and a(2n) = 2*a(n) and a(2^n) = 2^(n+1) for all n >= 0, a(2n-1) = 2n for all n >= 1. - M. F. Hasler, Oct 19 2019

A286543 Restricted growth sequence of A286542.

Original entry on oeis.org

1, 2, 2, 3, 4, 3, 2, 5, 4, 6, 4, 5, 4, 3, 2, 7, 8, 6, 4, 6, 9, 6, 8, 7, 4, 6, 9, 5, 9, 3, 2, 10, 8, 11, 8, 6, 9, 6, 4, 12, 8, 13, 9, 12, 14, 11, 9, 10, 9, 6, 4, 12, 8, 13, 14, 7, 8, 13, 9, 5, 9, 3, 2, 15, 16, 11, 8, 11, 14, 11, 8, 12, 8, 13, 14, 6, 4, 6, 4, 17, 18, 11, 14, 13, 9, 13, 9, 6, 19, 20, 14, 11, 14, 13, 14, 15, 18, 13, 14, 6, 18, 6, 4, 6, 18, 11, 8
Offset: 0

Views

Author

Antti Karttunen, May 18 2017

Keywords

Crossrefs

Programs

  • PARI
    allocatemem(2^30);
    rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
    write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
    A000120(n) = hammingweight(n);
    A002110(n) = prod(i=1,n,prime(i));
    A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ This function from M. F. Hasler
    A007947(n) = factorback(factorint(n)[, 1]); \\ From Andrew Lelechenko, May 09 2014
    A065642(n) = { my(r=A007947(n)); if(1==n,n,n = n+r; while(A007947(n) <> r, n = n+r); n); };
    A285332(n) = { if(n<=1,n+1,if(!(n%2),A019565(A285332(n/2)),A065642(A285332((n-1)/2)))); };
    A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); };  \\ This function from Charles R Greathouse IV, Aug 17 2011
    A286542(n) = if(!n,1,if(!(n%2),A002110(A000120(A285332(n/2))),A046523(A285332(n))));
    write_to_bfile(0,rgs_transform(vector(8192,n,A286542(n-1))),"b286543.txt");

A081381 Numbers n such that n and tau(n) = A000005(n) have the same prime factors (ignoring multiplicity).

Original entry on oeis.org

1, 2, 8, 9, 12, 18, 72, 80, 96, 108, 128, 288, 448, 486, 625, 720, 768, 864, 972, 1152, 1200, 1250, 1620, 1944, 2000, 2025, 2560, 4032, 4050, 5000, 5625, 6144, 6561, 6912, 7500, 7776, 8748, 9408, 10800, 11250, 11264, 12960, 13122, 16200, 18000, 18432, 19440
Offset: 1

Views

Author

Labos Elemer, Mar 26 2003

Keywords

Examples

			n = 5000 = 2*2*2*5*5*5*5, tau(5000) = 20 = 2*2*5, common prime factors: {2,5}
		

Crossrefs

Programs

  • Mathematica
    ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ba[x_] := Table[Part[ffi[x], 2*w-1], {w, 1, lf[x]}] Do[s=ba[DivisorSigma[0, n]]; If[Equal[s, ba[n]], Print[n]], {n, 1, 10000}]
  • PARI
    is(n)=my(f=factor(n)); factor(numdiv(f))[,1]==f[,1] \\ Charles R Greathouse IV, Oct 19 2017

A081378 Numbers k for which the sums of prime factors (ignoring multiplicity) of sigma(k) and phi(k) are equal but the sets of prime factors of sigma and phi are different.

Original entry on oeis.org

412, 1142, 1236, 1328, 1339, 1703, 2855, 2875, 2884, 3406, 3426, 3668, 3708, 3984, 4017, 5109, 5356, 5710, 5750, 5924, 6003, 6281, 6399, 6413, 6640, 6812, 7994, 8054, 8318, 8515, 8565, 8611, 8625, 8652, 8843, 8858, 9373, 9707, 9991
Offset: 1

Views

Author

Labos Elemer, Mar 26 2003

Keywords

Examples

			k = 412 = 2*2*103: sigma(412) = 728 = 2*2*2*7*13, phi(412) = 204 = 2*2*3*17, the sums of prime factors are identical (2 + 7 + 13 = 22 = 2 + 3 + 17) but the prime sets are different: {2,7,13} vs. {2,7,17}.
		

Crossrefs

Programs

  • Mathematica
    ffi[x_] := Flatten[FactorInteger[x]]; lf[x_] := Length[FactorInteger[x]]; ba[x_] := Table[Part[ffi[x], 2*w-1], {w, 1, lf[x]}]; spf[x_] := Apply[Plus, ba[x]]; k=0; Do[s=ba[DivisorSigma[1, n]]; s1=ba[EulerPhi[n]]; s3=spf[DivisorSigma[1, n]]; s4=spf[EulerPhi[n]]; If[ !Equal[s, s1]&&Equal[s3, s4], k=k+1; Print[{n, s, s1, ba[n], s3}]], {n, 1, 10000}]
  • PARI
    is(n) = {my(f = factor(n), p1 = factor(sigma(f))[, 1], p2 = factor(eulerphi(f))[, 1]); p1 != p2 && vecsum(p1) == vecsum(p2) ;} \\ Amiram Eldar, Mar 25 2024

A081382 a(1) = 1, for n > 1 a(n) = Min{x > n, A008472(x) = A008472(n)}.

Original entry on oeis.org

1, 4, 9, 8, 6, 12, 10, 16, 27, 20, 121, 18, 22, 28, 45, 32, 210, 24, 34, 40, 30, 44, 273, 25, 36, 52, 81, 56, 399, 60, 58, 64, 70, 68, 42, 48, 435, 76, 55, 49, 651, 84, 82, 88, 75, 92, 777, 54, 50, 80, 91, 104, 903, 72, 66, 98, 85, 116, 1645, 63, 118, 124, 90, 128, 77, 117
Offset: 1

Views

Author

Labos Elemer, Mar 26 2003

Keywords

Comments

Differs from A065642, a(n) <= A065642(n).

Crossrefs

Programs

  • Haskell
    a081382 1 = 1
    a081382 n = head [x | let sopf = a008472 n, x <- [n+1..], a008472 x == sopf]
    -- Reinhard Zumkeller, Jun 12 2015
  • Mathematica
    ffi[x_] := Flatten[FactorInteger[x]] lf[x_] := Length[FactorInteger[x]] ba[x_] := Table[Part[ffi[x], 2*w-1], {w, 1, lf[x]}] spf[x_] := Apply[Plus, ba[x]] Table[Min[Flatten[Position[Table[spf[w], {w, n+1, n^2}]-spf[n], 0]]+n], {n, 1, 100}]

A285110 a(n) = A001222(A285323(n)).

Original entry on oeis.org

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

Views

Author

Antti Karttunen, Apr 19 2017

Keywords

Comments

The sequence is completely determined by the positions of two least significant 1-bits of n: After initial zero, if n is a power of two (only one 1-bit present) or if prime(1+A285099(n)) > prime(1+A007814(n))^2, a(n) = 2, otherwise a(n) = 1.

Crossrefs

Programs

  • Python
    from operator import mul
    from sympy import prime, primefactors
    from functools import reduce
    def a001222(n): return 0 if n<2 else a001222(n//min(primefactors(n))) + 1
    def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
    def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
    def a065642(n):
        if n==1: return 1
        r=a007947(n)
        n += r
        while a007947(n)!=r:
            n+=r
        return n
    def a285323(n): return a065642(a065642(a019565(n)))//a019565(n)
    def a(n): return a001222(a285323(n))
    print([a(n) for n in range(121)]) # Indranil Ghosh, Apr 20 2017
  • Scheme
    (define (A285110 n) (A001222 (A285323 n)))
    (define (A285110 n) (cond ((zero? n) n) ((or (= 1 (A000120 n)) (> (A000040 (+ 1 (A285099 n))) (A000290 (A000040 (+ 1 (A007814 n)))))) 2) (else 1)))
    

Formula

a(n) = A001222(A285323(n)).
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