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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A332461 a(n) = Product_{d|n, d>1} A000040(A297113(d)), where A000040(n) gives the n-th prime, and A297113(n) = the excess of n plus the index of the largest dividing prime (A046660 + A061395).

Original entry on oeis.org

1, 2, 3, 6, 5, 18, 7, 30, 15, 50, 11, 270, 13, 98, 75, 210, 17, 450, 19, 1050, 147, 242, 23, 9450, 35, 338, 105, 3234, 29, 11250, 31, 2310, 363, 578, 245, 47250, 37, 722, 507, 57750, 41, 43218, 43, 9438, 2625, 1058, 47, 727650, 77, 2450, 867, 17238, 53, 22050, 605, 210210, 1083, 1682, 59, 8268750, 61, 1922, 8085, 30030
Offset: 1

Views

Author

Antti Karttunen, Feb 22 2020

Keywords

Crossrefs

Programs

  • PARI
    A297113(n) = if(1==n, 0, (primepi(vecmax(factor(n)[, 1])) + (bigomega(n)-omega(n))));
    A332461(n) = if(1==n,1, my(m=1); fordiv(n,d,if(d>1, m *= prime(A297113(d)))); (m));

Formula

a(n) = Product_{d|n, d>1} A000040(A297113(d)).
a(p) = p for all primes p.
For all n >= 0, a(2^n) = A002110(n).
For all n >= 1:
A046523(a(n)) = A324202(n).
A048675(a(n)) = A156552(n).
A097248(a(n)) = A332462(n).

A332462 a(n) = A019565(A156552(n)).

Original entry on oeis.org

1, 2, 3, 6, 5, 10, 7, 30, 15, 14, 11, 42, 13, 22, 21, 210, 17, 70, 19, 66, 33, 26, 23, 330, 35, 34, 105, 78, 29, 110, 31, 2310, 39, 38, 55, 462, 37, 46, 51, 390, 41, 130, 43, 102, 165, 58, 47, 2730, 77, 154, 57, 114, 53, 770, 65, 510, 69, 62, 59, 546, 61, 74, 195, 30030, 85, 170, 67, 138, 87, 182, 71, 4290, 73, 82, 231, 174, 91
Offset: 1

Views

Author

Antti Karttunen, Feb 22 2020

Keywords

Crossrefs

Permutation of squarefree numbers, A005117.

Programs

Formula

a(n) = A019565(A156552(n)) = A097248(A332461(n)).
a(p) = p for all primes p.
For all n >= 1, A048675(a(n)) = A156552(n).
For all n >= 0, a(A005940(1+n)) = A019565(n).
For all n >= 0, a(2^n) = A002110(n).

A331751 Numbers k such that A048675(sigma(k)) is equal to A048675(2*k).

Original entry on oeis.org

2, 6, 27, 28, 84, 270, 496, 1053, 1120, 1488, 1625, 1638, 3360, 3780, 4875, 8128, 10530, 24384, 66960, 147420, 167400, 406224, 611226, 775000, 872960, 943250, 1097280, 1245699, 1255338, 1303533, 1464320, 1686400, 1740024, 1922375, 1952500, 2011625, 2193408, 2325000, 2611440, 2618880, 2829750, 2941029, 4392960
Offset: 1

Views

Author

Antti Karttunen, Feb 05 2020

Keywords

Comments

Numbers k such that A097248(sigma(k)) is equal to A097248(2*k).
Numbers k such that A331750(k) is equal to 1+A048675(k), which in turn is equal to A048675(A225546(2*k)) = A048675(2*A225546(k)).
Among the first 60 terms, 15 are odd: 27, 1053, 1625, 4875, 1245699, 1303533, 1922375, 2011625, 2941029, 5767125, 6034875, 12733875, 17137575, 26316675, 29362905, and only 1053 = 3^4 * 13 is in A228058.
Note that the condition A090880(sigma(k)) == A090880(2*k) appears to be much more constrained.

Examples

			For n = 1053 = 3^4 * 13^1, A331750(1053) = A331750(81) + A331750(13) = 32+9 = 41, while A048675(2*1053) = A048675(2)+A048675(81)+A048675(13) = 1+8+32 = 41 also, thus 1053 is included in this sequence.
For n = 3360 = 2^5 * 3^1 * 5^1 * 7^1, A331750(3360) = A331750(32)+A331750(3)+A331750(5)+A331750(7) = 12+2+3+3 = 20, while A048675(2*3360) = A048675(2)+A048675(32)+A048675(3)+A048675(5)+A048675(7) = 1+5+2+4+8 = 20 also, thus 3360 is included in this sequence.
		

Crossrefs

Programs

A097249 a(n) is the number of times we must iterate A097246, starting at n, before the result is squarefree.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 2, 0, 0, 0, 2, 0, 1, 0, 1, 0, 0, 0, 2, 1, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 3, 0, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 3, 2, 0, 0, 2, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 2, 0, 1, 1, 1, 0, 0, 0, 1, 0
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 03 2004

Keywords

Comments

a(n) = Min{k: r(n,k)=r(n,k+1)}, where r(n,k)=A097246(r(n,k-1)), r(n,0)=n;
a(A005117(n))=0; a(A097250(n))=n and a(m)A097250(n).

Crossrefs

Programs

  • Mathematica
    f[n_] := Product[{p, e} = pe; NextPrime[p]^Quotient[e, 2] p^Mod[e, 2], {pe, FactorInteger[n]}];
    a[n_] := (NestWhileList[f, n, !SquareFreeQ[#]&] // Length) - 1;
    Array[a, 105] (* Jean-François Alcover, Nov 18 2021 *)
  • PARI
    A097246(n) = { my(f=factor(n)); prod(i=1, #f~, (nextprime(f[i,1]+1)^(f[i,2]\2))*((f[i,1])^(f[i,2]%2))); };
    A097249(n) = if(issquarefree(n),0,1+A097249(A097246(n))); \\ Antti Karttunen, Jul 29 2018

Formula

If A008966(n) = 1 [when n is in A005117], a(n) = 0, otherwise a(n) = 1 + a(A097246(n)). - Antti Karttunen, Jul 29 2018

Extensions

Edited by Sam Alexander, Jan 05 2005

A334110 The squares of squarefree numbers (A062503), ordered lexicographically according to their prime factors. a(n) = Product_{k in I} prime(k+1)^2, where I are the indices of nonzero binary digits in n = Sum_{k in I} 2^k.

Original entry on oeis.org

1, 4, 9, 36, 25, 100, 225, 900, 49, 196, 441, 1764, 1225, 4900, 11025, 44100, 121, 484, 1089, 4356, 3025, 12100, 27225, 108900, 5929, 23716, 53361, 213444, 148225, 592900, 1334025, 5336100, 169, 676, 1521, 6084, 4225, 16900, 38025, 152100, 8281, 33124, 74529, 298116, 207025, 828100, 1863225, 7452900, 20449, 81796, 184041
Offset: 0

Views

Author

Antti Karttunen and Peter Munn, May 01 2020

Keywords

Comments

For the lexicographic ordering, the prime factors must be written in nonincreasing order. If we write the factors in nondecreasing order, we get a lexicographically ordered set with an order type that is greater than a natural number index - the resulting sequence does not include all qualifying numbers. (Note also that the symbols used for the lexicographic order are the prime numbers, not their digits.)
a(n) is the n-th power of 4 in the monoid defined in A331590.
Conjecture: a(n) is the position of the first occurrence of n in A334109.

Examples

			The initial terms are shown below, equated with the product of their prime factors to exhibit the lexicographic ordering. The list starts with 1, since 1 is factored as the empty product and the empty list is first in lexicographic order.
    1 = .
    4 = 2*2.
    9 = 3*3.
   36 = 3*3*2*2.
   25 = 5*5.
  100 = 5*5*2*2.
  225 = 5*5*3*3.
  900 = 5*5*3*3*2*2.
   49 = 7*7.
  196 = 7*7*2*2.
  441 = 7*7*3*3.
		

Crossrefs

Cf. A000079, A019565 (square roots), A048675, A097248, A225546, A267116, A332382, A334109 (a left inverse).
Column 2 of A329332. Permutation of A062503.
After 1, the right children of the leftmost edge of A334860, or respectively, the left children of the rightmost edge of A334866.
Subsequences: A001248, A061742, A166329.
Subsequence of A052330.
A003961, A003987, A059897, A331590 are used to express relationship between terms of this sequence.

Programs

  • Mathematica
    Array[If[# == 0, 1, Times @@ Flatten@ Map[Function[{p, e}, Map[Prime[Log2@ # + 1]^(2^(PrimePi@ p - 1)) &, DeleteCases[NumberExpand[e, 2], 0]]] @@ # &, FactorInteger[3^#]]] &, 51, 0] (* Michael De Vlieger, May 26 2020 *)
  • PARI
    A334110(n) = { my(p=2,m=1); while(n, if(n%2, m *= p^2); n >>= 1; p = nextprime(1+p)); (m); };

Formula

a(n) = A019565(n)^2.
For n >= 1, a(A000079(n-1)) = A001248(n).
For all n >= 0, A334109(a(n)) = n.
a(n+k) = A331590(a(n), a(k)).
a(n XOR k) = A059897(a(n), a(k)), where XOR denotes bitwise exclusive-or, A003987.
a(n) = A225546(3^n).
a(2n) = A003961(a(n)).
a(2n+1) = 4 * a(2n).
a(2^k-1) = A061742(k).
A267116(a(n)) = 2.
A048675(a(n)) = 2n.
A097248(a(n)) = A332382(n) = A019565(2n).

A322869 a(n) = A000120(A048675(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 2, 1, 2, 2, 3, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 1, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 3, 1, 2, 2, 2, 1, 3, 1, 3, 3
Offset: 1

Views

Author

Antti Karttunen, Dec 31 2018

Keywords

Crossrefs

Cf. A002110 (position of the first occurrence of n).

Programs

  • Mathematica
    Array[If[# == 1, 0, DigitCount[Total@ Map[#2*2^(PrimePi@ #1 - 1) & @@ # &, FactorInteger[#]], 2, 1]] &, 105]  (* Michael De Vlieger, Dec 31 2018 *)
  • PARI
    A048675(n) = { my(f = factor(n)); sum(k=1, #f~, f[k, 2]*2^primepi(f[k, 1]))/2; }; \\ From A048675
    A322869(n) = hammingweight(A048675(n));

Formula

a(n) = A000120(A048675(n)) = A000120(A322821(n)).
a(n) = A001221(A097248(n)) = A001222(A097248(n)).
If n is squarefree, then a(n) = A001221(n) = A322862(n).
a(A002110(n)) = n.

A332824 a(n) = Product_{d|n} A019565(phi(d)), where phi is Euler totient function A000010.

Original entry on oeis.org

2, 4, 6, 12, 10, 36, 30, 60, 90, 100, 42, 540, 70, 900, 210, 420, 22, 8100, 66, 2100, 3150, 1764, 330, 18900, 550, 4900, 2970, 94500, 770, 44100, 2310, 4620, 6930, 484, 11550, 4252500, 130, 4356, 16170, 115500, 182, 9922500, 546, 291060, 242550, 108900, 2730, 1455300, 8190, 302500, 858, 1131900, 1430, 8820900, 19110
Offset: 1

Views

Author

Antti Karttunen, Feb 25 2020

Keywords

Crossrefs

Cf. A048675 (a left inverse).

Programs

  • PARI
    A019565(n) = {my(j,v); factorback(Mat(vector(if(n, #n=vecextract(binary(n), "-1..1")), j, [prime(j), n[j]])~))}; \\ From A019565
    A332824(n) = { my(m=1); fordiv(n,d,m *= A019565(eulerphi(d))); (m); };

Formula

a(n) = Product_{d|n} A332825(d).
a(n) = A318834(n) * A332825(n).
A048675(a(n)) = n.
A097248(a(n)) = A019565(n).

A376406 a(0) = 1, and for n > 0, a(n) = A019565(Sum_{i=0..n-1} a(i)), where A019565 is the base-2 exp-function.

Original entry on oeis.org

1, 2, 6, 14, 330, 10166, 12075690, 1174153011328084322, 73582975079922326904310062621361286633125176265747127754
Offset: 0

Views

Author

Antti Karttunen, Nov 04 2024

Keywords

Comments

a(9) has 272 digits and a(10) has 1523 digits.
The lexicographically earliest infinite sequence x for which A048675(x(n)) gives the partial sums of x (shifted right once). This follows because the "least k" condition in the alternative formula also ensures that each k is squarefree, as we have A097248(n) = A019565(A048675(n)) <= n for all n, with equivalence only when n is squarefree.
Compare also to A376408.

Examples

			Starting with a(0) = 1, we take partial sums of previous terms, and apply A019565 to get the next term, and in the rightmost column, we "unbox" that term by applying A048675 to get A376407(n), which thus gives the partial sums of a(0)..a(n-1):
a(0)                               = 1          -> 0
a(1) = A019565(1)                  = 2,         -> 1     = 1
a(2) = A019565(1+2)                = 6,         -> 3     = 1+2
a(3) = A019565(1+2+6)              = 14,        -> 9     = 1+2+6
a(4) = A019565(1+2+6+14)           = 330,       -> 23    = 1+2+6+14
a(5) = A019565(1+2+6+14+330)       = 10166,     -> 353   = 1+2+6+14+330
a(6) = A019565(1+2+6+14+330+10166) = 12075690,  -> 10519 = 1+2+6+14+330+10166
etc.
		

Crossrefs

Cf. A376407 (= A048675(a(n)), also gives the partial sums from its second term onward).
Subsequence of A005117.
Cf. also analogous sequences A002110 (for A276085), A093502 (for A056239), A376399 (for A276075).

Programs

  • PARI
    up_to = 12;
    A019565(n) = { my(m=1, p=1); while(n>0, p = nextprime(1+p); if(n%2, m *= p); n >>= 1); (m); };
    A376406list(up_to) = { my(v=vector(up_to), s=1); v[1]=1; for(n=2,up_to,v[n] = A019565(s); s += v[n]); (v); };
    v376406 = A376406list(1+up_to);
    A376406(n) = v376406[1+n];

Formula

a(n) = A019565(A376407(n)) = A019565(Sum_{i=0..n-1} a(i)).
a(0) = 1, and for n > 0, a(n) is the least k such that A048675(k) = a(n-1) + A048675(a(n-1)), where A048675 is the base-2 log-function.
For n > 0, a(n) <= a(n-1) * A019565(a(n-1)).

A277886 If n is squarefree, a(n) = n, else a(n) = A000040(1+A277885(n)) * (n/(A249739(n)^2)).

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 6, 5, 10, 11, 9, 13, 14, 15, 12, 17, 10, 19, 15, 21, 22, 23, 18, 7, 26, 15, 21, 29, 30, 31, 24, 33, 34, 35, 27, 37, 38, 39, 30, 41, 42, 43, 33, 25, 46, 47, 36, 11, 14, 51, 39, 53, 30, 55, 42, 57, 58, 59, 45, 61, 62, 35, 48, 65, 66, 67, 51, 69, 70, 71, 54, 73, 74, 21, 57, 77, 78, 79, 60, 45
Offset: 1

Views

Author

Antti Karttunen, Nov 08 2016

Keywords

Comments

If n has non-unitary prime divisors, then divide it by the square of the smallest of them and multiply by a single instance of the next larger prime.
This differs from related A097246 for the first time at n=16. For both sequences A097248 gives the eventual stable points reached when starting iterating from n.

Examples

			For n = 12 = 2*2*3, the smallest non-unitary prime divisor (and in this case the only one) is 2, thus we divide with 2^2 and multiply with the next larger prime 3, to get ((2^2 * 3)/(2^2))*3 = 3*3, thus a(12) = 9.
For n = 16 = 2^4, we divide two instances of 2 out and multiply by a single instance of 3 to get 2*2*3 = 12.
		

Crossrefs

Programs

  • Mathematica
    Table[If[SquareFreeQ@ n, n, Prime[1 + PrimePi@ Min[Select[FactorInteger[n][[All, 1]], ! CoprimeQ[#, n/#] &] /. {} -> 0]] (n/If[SquareFreeQ@ n, 1, p = 2; While[! Divisible[n, p^2], p = NextPrime@ p]; p]^2)], {n, 81}] (* Michael De Vlieger, Nov 15 2016 *)
  • Scheme
    (define (A277886 n) (if (zero? (A277885 n)) n (* (A000040 (+ 1 (A277885 n))) (/ n (expt (A249739 n) 2)))))

Formula

If A277885(n) = 0 [when n is squarefree], then a(n) = n, otherwise a(n) = A000040(1+A277885(n)) * (n/(A249739(n)^2)).
Other identities. For all n >= 1:
A048675(a(n)) = A048675(n).

A097247 A097246(A097246(n)).

Original entry on oeis.org

1, 2, 3, 3, 5, 6, 7, 6, 5, 10, 11, 5, 13, 14, 15, 5, 17, 10, 19, 15, 21, 22, 23, 10, 7, 26, 15, 21, 29, 30, 31, 10, 33, 34, 35, 15, 37, 38, 39, 30, 41, 42, 43, 33, 7, 46, 47, 15, 11, 14, 51, 39, 53, 30, 55, 42, 57, 58, 59, 25, 61, 62, 35, 15, 65, 66, 67, 51, 69, 70, 71, 30, 73, 74
Offset: 1

Views

Author

Reinhard Zumkeller, Aug 03 2004

Keywords

Comments

This sequence is not multiplicative, see example.

Examples

			m=20 and n=63 are coprime:
a(20) = A097246(A097246(2*2*5)) = A097246(3*5)=15,
a(63) = A097246(A097246(3*3*7)) = A097246(5*7)=35 and
a(20*63) = A097246(A097246(2*2*3*3*5*7)) = A097246(3*5*5*7) =
3*7*7 = 147 <> a(20)*a(63) = 15*35 = 525: the sequence is not multiplicative.
		

Crossrefs

Cf. A097248.
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