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 41-49 of 49 results.

A253790 a(1) = 1; thereafter, odd numbers such that A055396(n) = A001511(n-1).

Original entry on oeis.org

1, 3, 5, 15, 27, 39, 51, 63, 75, 85, 87, 99, 111, 123, 125, 135, 147, 159, 171, 183, 195, 205, 207, 209, 217, 219, 231, 243, 245, 255, 267, 279, 291, 303, 315, 325, 327, 329, 339, 351, 363, 365, 375, 387, 399, 411, 423, 435, 445, 447, 459, 471, 481, 483, 485, 495, 507, 519, 531, 543, 553, 555
Offset: 1

Views

Author

Antti Karttunen, Jan 13 2015

Keywords

Comments

After 1, all such odd numbers whose smallest prime factor A020639(n) = A000040(k+1), where k = A007814(n-1), the 2-adic valuation of the preceding even number.
Any odd number present in A253789 must be one of these terms.

Links

Crossrefs

Cf. A000040, A001511, A007814, A020639, A029747, A055396, A253789.

A332216 Fixed points of A332221: Numbers k such that A156552(sigma(k)) is equal to k.

Original entry on oeis.org

2, 3, 5, 7, 11, 19, 23, 31, 47, 55, 79, 87, 127, 191, 383, 1279, 5119, 6143, 8191, 20479, 81919, 131071, 524287, 786431, 1310719, 2147483647
Offset: 1

Views

Author

Antti Karttunen, Feb 10 2020

Keywords

Comments

Equally, numbers k such that sigma(k) is equal to A005940(1+k).
The primes in this sequence are obtained by subtracting 1 from those terms of A029747 that are one more than a prime.
Questions: Are there other composite terms than 55 and 87? Are there other even terms than 2? (All such even terms should also occur in A332218).

Links

Crossrefs

Cf. A000203, A005940, A029747, A156552, A332217, A332218, A332221.
Subsequences: A000668, A007505, A050522.

Programs

  • Mathematica
    Select[Range[10^5], DivisorSigma[1, #] == Block[{p = Partition[Split[Join[IntegerDigits[#, 2], {2}]], 2], q}, Times @@ Flatten[Table[q = Take[p, -i]; Prime[Count[Flatten[q], 0] + 1]^q[[1, 1]], {i, Length[p]}]]] &] (* Michael De Vlieger, Feb 12 2020, after Robert G. Wilson v at A005940 *)

A356450 Positions of numbers m = A005940(n+1) such that m < n.

Original entry on oeis.org

8, 16, 17, 32, 33, 34, 35, 64, 65, 66, 67, 68, 69, 71, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 139, 143, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 269, 271, 272, 273, 275, 279, 287, 288, 384, 512, 513, 514, 515, 516, 517, 518, 519, 520
Offset: 1

Views

Author

Michael De Vlieger, Aug 07 2022

Keywords

Comments

This sequence contains 2^k for k >= 3. Powers of 2 expressed in binary consist of a 1 followed by k zeros. Therefore, A005940(2^k) = prime(k+1)^1. For k >= 3, 2^k > prime(k+1).

Examples

			34 is in the sequence since A005940(34) = A005940("100010"_2) = prime(1+1)^1 * prime(4+1)^1 = 33, and 33 < 34.

Links

  • Michael De Vlieger, Table of n, a(n) for n = 1..13175 (terms m < 2^21)
  • Michael De Vlieger, Fan style binary tree diagram of b(n) for n = 1..2^14-1, where b(n) = A005940(n), highlighting terms such that b(n+1) < n in red, b(n+1) = n in yellow, and b(n+1) > n in blue. Positions of the terms shown in red are in this sequence, while b(A029747(n)+1) = A029747(n) appears in yellow.

Crossrefs

Cf. A005940, A029747, A356455.

Programs

  • Mathematica
    nn = 2^10; a[0] = 1; Reap[Do[k = Prime[1 + DigitCount[n, 2, 0]]*a[n - 2^Floor@ Log2@ n]; Set[a[n], k]; If[k < n, Sow[n]], {n, nn}]][[-1, -1]] (* Michael De Vlieger, Aug 07 2022 *)

A364960 Numbers k such that A005940(k) is either k itself or its descendant in Doudna-tree, A005940.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 16, 20, 24, 25, 32, 40, 45, 48, 49, 64, 80, 96, 128, 160, 192, 256, 320, 384, 512, 640, 768, 1024, 1280, 1536, 2048, 2560, 3072, 4096, 5120, 6144, 8192, 10240, 12288, 16384, 20480, 24576, 32768, 40131, 40960, 49152, 50575, 65536, 79625, 81920, 98304, 131072, 163840, 196608, 262144, 327680
Offset: 1

Views

Author

Antti Karttunen, Aug 14 2023

Keywords

Comments

Numbers k such that A252464(k) is equal to A364569(k).

Crossrefs

Positions of 0's in A364570.
Cf. A252464, A364569.
Subsequences: A029747, A364961 (odd terms).

Programs

  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };
A064989(n) = {my(f); 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)};
A252463(n) = if(!(n%2),n/2,A064989(n));
isA364960(n) = { my(k=A005940(n)); while(k>n, k = A252463(k)); (k==n); };
  • PARI
    Abincompreflen(n, m) = { my(x=binary(n), y=binary(m), u=min(#x, #y)); for(i=1, u, if(x[i]!=y[i], return(i-1))); (u); };
A156552(n) = {my(f = factor(n), p, 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
A364569(n) = Abincompreflen(A156552(n), (n-1));
A061395(n) = if(n>1, primepi(vecmax(factor(n)[, 1])), 0);
A252464(n) = if(1==n,0,(bigomega(n) + A061395(n) - 1));
isA364960(n) = (A252464(n)==A364569(n));

A364961 Odd numbers k such that A005940(k) is either k itself or its descendant in Doudna-tree, A005940.

Original entry on oeis.org

1, 3, 5, 25, 45, 49, 40131, 50575, 79625, 1486485, 1872507, 3403125
Offset: 1

Views

Author

Antti Karttunen, Aug 14 2023

Keywords

Comments

Odd numbers k such that A252464(k) is equal to A364569(k).
Apparently, A364960 without the even terms of A029747.
Note that 1, 25, 45, 49 are so far the only known integers x for which A005940(x) = A003961(x).

Examples

			Term (and its factorization)            A005940(term) (and its factorization)
      1                              -> 1
      3                              -> 3
      5                              -> 5
     25 = 5^2                        -> 49        = 7^2
     45 = 3^2 * 5                    -> 175       = 5^2 * 7
     49 = 7^2                        -> 121       = 11^2
  40131 = 3^2 * 7^3 * 13             -> 100847877 = 3 * 13^2 * 19^3 * 29
  50575 = 5^2 * 7 * 17^2             -> 22467159  = 3^3 * 11^2 * 13 * 23^2
  79625 = 5^3 * 7^2 * 13             -> 787365187 = 7 * 19^3 * 23^2 * 31
1486485 = 3^3 * 5 * 7 * 11^2 * 13    -> 25468143451205
                                        = 5 * 7 * 13 * 17^3 * 19 * 23 * 29^2 * 31
1872507 = 3 * 7 * 13 * 19^3          -> 240245795625
                                        = 3 * 5^4 * 11 * 17 * 23 * 31^3,
3403125 = 3^2 * 5^5 * 11^2           -> 2394659631669305
                                        = 5 * 7^3 * 11 * 13^2 * 17^5 * 23^2.
See also examples in A364959.

Crossrefs

Odd terms in A364960.
Cf. A003961, A005940, A029747, A156552, A252464, A364569, A364570.
Cf. also A364956 (odd terms there), A364959, A364962.

Programs

  • PARI
    Abincompreflen(n, m) = { my(x=binary(n), y=binary(m), u=min(#x, #y)); for(i=1, u, if(x[i]!=y[i], return(i-1))); (u); };
A156552(n) = {my(f = factor(n), p, 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
A364569(n) = Abincompreflen(A156552(n), (n-1));
A061395(n) = if(n>1, primepi(vecmax(factor(n)[, 1])), 0);
A252464(n) = if(1==n,0,(bigomega(n) + A061395(n) - 1));
isA364961(n) = ((n%2)&&(A252464(n)==A364569(n)));
  • PARI
    A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); (t); };
A064989(n) = {my(f); 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)};
A252463(n) = if(!(n%2),n/2,A064989(n));
isA364961(n) = if(!(n%2),0,my(k=A005940(n)); while(k>n, k = A252463(k)); (k==n));

A320674 Positive integers m with binary expansion (b_1, ..., b_k) (where k = A070939(m)) such that b_i = [m == 0 (mod prime(i))] for i = 1..k (where prime(i) denotes the i-th prime number and [] is an Iverson bracket).

Original entry on oeis.org

2, 4, 6, 8, 10, 12, 16, 20, 24, 32, 40, 48, 64, 80, 96, 128, 160, 192, 256, 320, 384, 512, 640, 768, 1024, 1280, 1536, 2048, 2560, 3072, 4096, 5120, 6144, 8192, 10240, 12288, 16384, 20480, 24576, 32768, 40960, 49152, 65536, 81920, 98304, 131072, 163840, 196608
Offset: 1

Views

Author

Rémy Sigrist, Oct 19 2018

Keywords

Comments

In other words, the 1's in the binary representation of a term of this sequence encode the first prime divisors of this term.
All terms are even.
All even terms in A029747 belong to this sequence.
The term a(71) = 33554434 is the first one that does not belong to A029747.
See A320673 for similar sequences.

Examples

			The initial terms, alongside their binary representation and the prime divisors encoded therein, are:
  n   a(n)      bin(a(n))                   First prime divisors
  --  --------  --------------------------  --------------------
   1         2  10                          2
   2         4  100                         2
   3         6  110                         2, 3
   4         8  1000                        2
   5        10  1010                        2, 5
   6        12  1100                        2, 3
   7        16  10000                       2
   8        20  10100                       2, 5
   9        24  11000                       2, 3
  ...
  71  33554434  10000000000000000000000010  2, 97
  ...
33554434 is in the sequence because its binary expansion 10000000000000000000000010 of length 26 has a 1 in the 1st place and in the 25th place from the left and 0 elsewhere. As it is divisible by the 1st and 25th prime and by no other prime with index <= 26, 33554434 in the sequence. - _David A. Corneth_, Oct 20 2018

Crossrefs

Cf. A029747, A070939, A320673.

Programs

  • Mathematica
    selQ[n_] := With[{bb = IntegerDigits[n, 2]}, (Prime /@ Flatten[Position[bb, 1]]) == FactorInteger[n][[All, 1]]];
Select[Range[2, 200000], selQ] (* Jean-François Alcover, Nov 01 2018 *)
  • PARI
    is(n) = my (b=binary(n)); b==vector(#b, k, n%prime(k)==0)

A324107 Fixed points of A324106, where A324106 is a multiplicative function with A324106(p^e) = A005940(p^e).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 32, 40, 48, 60, 63, 64, 80, 96, 120, 126, 128, 160, 192, 240, 252, 256, 315, 320, 384, 480, 504, 512, 630, 640, 768, 960, 1008, 1024, 1260, 1280, 1536, 1920, 2016, 2048, 2520, 2560, 3072, 3840, 4032, 4096, 5040, 5120, 6144, 7680, 8064, 8192, 10080, 10240, 12288, 15360, 16128, 16384
Offset: 1

Views

Author

Antti Karttunen, Feb 15 2019

Keywords

Comments

Numbers n such that A324106(n) = n.

Examples

			For n = 63 = 3^2 * 7^1, we find that A005940(9) = 7 and A005940(7) = 9, thus A324106(63) = 7*9 = 63, and 63 is a member of this sequence.

Crossrefs

Cf. A005940, A070776, A324106, A324109.
Cf. A029747 (a subsequence).

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
A324106(n) = { my(f=factor(n)); prod(i=1, #f~, A005940(f[i,1]^f[i,2])); };
isA324107(n) = (n==A324106(n));
for(n=1,16384,if(isA324107(n), print1(n,", ")))

A356455 Numbers m = A005940(n) such that A005940(n) < (n-1), listed in order of appearance in A005940.

Original entry on oeis.org

7, 11, 14, 13, 22, 33, 28, 17, 26, 39, 44, 65, 66, 56, 19, 34, 51, 52, 85, 78, 117, 88, 119, 130, 132, 112, 23, 38, 57, 68, 95, 102, 153, 104, 133, 170, 255, 156, 234, 176, 209, 238, 260, 264, 224, 247, 361, 29, 46, 69, 76, 115, 114, 171, 136, 161, 190, 285, 204
Offset: 1

Views

Author

Michael De Vlieger, Aug 07 2022

Keywords

Comments

Primes p > 5 are in the sequence. We may rewrite the k-th prime p instead as prime(k)^1, and thus convert it to a single run of 1 ones followed by (k-1) zeros, which becomes a binary number consisting of 1 followed by k-1 zeros. It is clear that for p > 5, prime(k) < 2^(k-1).

Examples

			7 is in the sequence since 7 = prime(3+1)^1, which we write as 1 following 3 zeros when approached from the least significant digit, i.e., "1000"_2 = 8, thus A005940(8) = 7; and 7 < 8.
5 is not in the sequence since 5 = prime(2+1)^1 -> "100"_2 = 4, and 5 > 4.
14 is in the sequence since 14 = prime(0+1)^1 * prime(3+1)^1, which we can express as a binary number with singleton 1s following 0 and 3 zeros, i.e., "10001"_2 = 17, hence A005940(17) = 14 and we see 14 < 17.
33 is in the sequence since 33 = prime(1+1)^1 * prime(4+1)^1 = A005940("100010"_2) = A005940(34) = 33, and we see 33 < 34.

Links

  • Michael De Vlieger, Table of n, a(n) for n = 1..13175 (terms in A005940 for n < 2^21)
  • Michael De Vlieger, Fan style binary tree diagram of b(n) for n = 1..2^14-1, where b(n) = A005940(n), highlighting terms such that b(n+1) < n in red, b(n+1) = n in ochre, and b(n+1) > n in blue. Terms shown in red appear in this sequence, while b(A029747(n)+1) = A029747(n) appears in ochre.

Crossrefs

Cf. A005940, A029747, A356450.

Programs

  • Mathematica
    nn = 2^10; a[0] = 1; Reap[Do[k = Prime[1 + DigitCount[n, 2, 0]]*a[n - 2^Floor@ Log2@ n]; Set[a[n], k]; If[k < n, Sow[k]], {n, nn}]][[-1, -1]]

A364962 Odd numbers k such that A005941(k) is either k itself or its descendant in Doudna-tree, A005940.

Original entry on oeis.org

1, 3, 5, 11, 19, 23, 31, 37, 41, 43, 47, 53, 59, 61, 71, 73, 79, 83, 85, 89, 97, 101, 103, 107, 109, 113, 127
Offset: 1

Views

Author

Antti Karttunen, Aug 14 2023

Keywords

Comments

Questions: Is 85 the only composite in this sequence? (See also A364565). Are there any more terms after 127, or is the sequence finite?

Examples

			85 = 5*17 is a term, because A005941(85) = 133 = 7*19 = A003961(85), thus 133 is a left hand side child of 85 in the tree depicted in A005940, and therefore 85 is included in this sequence. (See also the last example in A364959).

Crossrefs

Cf. A005940, A005941, A029747, A156552, A252464.
Cf. also A364959, A364961.

Programs

  • PARI
    A005941(n) = { my(f=factor(n), p, 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]); (1+res) }; \\ (After David A. Corneth's program for A156552)
A064989(n) = {my(f); 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)};
A252463(n) = if(!(n%2),n/2,A064989(n));
isA364962(n) = if(!(n%2),0,my(k=A005941(n)); while(k>n, k = A252463(k)); (k==n));
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